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Diffraction of Light - Young’s Single Slit Experiment

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Single Slit Diffraction

Single Slit Diffraction of Light

Diffraction is the line of study that helped in the development of precise spectrometers, hence helping Astronomy take great leaps. This was possible in the early nineteenth century when two great scientists were struggling in their lives separately in Italy and Germany. These two weak youngsters were still struggling to get educated. Fraunhofer was the son of a maker of decorative glasses and mirrors in Bavaria, while Fresnel was the son of an architect in Normandy.

What is Diffraction of Light?

Diffraction is the bending of light around the sharp corner of an obstacle. When light is incident on a slit, with a size comparable to the wavelength of light , an alternating dark and bright pattern can be observed. This phenomenon is called the single slit diffraction. According to Huygens’ principle, when light is incident on the slit, secondary wavelets generate from each point. These wavelets start out in phase and propagate in all directions. Each wavelet travels a different distance to reach any point on the screen. Due to the path difference, they arrive with different phases and interfere constructively or destructively.

Diffraction Due to Single Slit

When light is incident on the sharp edge of an obstacle, a faint illumination can be found within the geometrical shadow of the obstacle. This suggests that light bends around a sharp corner. The effect becomes significant when light passes through an aperture having a dimension comparable to the wavelength of light.

If light is incident on a slit having a width comparable to the wavelength of light, an alternating dark and bright pattern can be seen if a screen is placed in front of the slit. This phenomenon is known as single slit diffraction.

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Young’s Single Slit Experiment

Thomas Young’s double-slit experiment, performed in 1801, demonstrates the wave nature of light. In this experiment, monochromatic light is shone on two narrow slits. The waves, after passing through each slit, superimpose to give an alternate bright and dark distribution on a distant screen. All the bright fringes have the same intensity and width.

In a single slit experiment, monochromatic light is passed through one slit of finite width and a similar pattern is observed on the screen. Unlike the double-slit diffraction pattern, the width and intensity in the single-slit diffraction pattern reduce as we move away from the central maximum.

Explanation of The Phenomenon and Diffraction Formula

According to Huygens’ principle, when light is incident on the slit, secondary wavelets generate from each point. These wavelets start out in phase and propagate in all directions. Each wavelet travels a different distance to reach any point on the screen. Due to the path difference, they arrive with different phases and interfere constructively or destructively.

If a monochromatic light of wavelength λ falls on a slit of width a, the intensity on a screen at a distance L from the slit can be expressed as a function of θ.

Here, θ is the angle made with the original direction of light. It is given by,

I\[\left ( \theta \right )\] = I\[_{0}\] \[\frac{sin^{2}\alpha }{\alpha ^{2}}\]

Here, \[\alpha\] = \[\frac{\Pi }{\lambda }\] sin\[\theta\] and I₀ is the intensity of the central bright fringe, located at θ =0.

Diffraction Maxima and Minima

Bright fringes appear at angles,

\[\theta\] \[\rightarrow\] 0,\[\theta\]\[\rightarrow\] sin-1 \[\left ( \pm \frac{3\lambda }{2} \right )\], \[\theta\] \[\rightarrow\] sin-1 \[\left ( \pm \frac{5\lambda }{2} \right )\]

\[\theta\] \[\rightarrow\] 0 is the central maximum

Dark fringes correspond to the condition,

a sin\[\theta\] = m\[\lambda\] with m = ±1, ±2, ±3…

In a double slit arrangement, diffraction through single slits appears as an envelope over the interference pattern between the two slits.

Fringe width

The angular distance between the two first order minima (on either side of the center) is called the angular width of central maximum, given by

2\[\theta\] = \[\frac{2\lambda }{a}\]

The linear width is as follows,

\[\Delta\] = L.2\[\theta\] = \[\frac{2L\lambda }{a}\]

The width of the central maximum in the diffraction formula is inversely proportional to the slit width. If the slit width decreases, the central maximum widens, and if the slit width increases, it narrows down. It can be inferred from this behaviour that light bends more as the dimension of the aperture becomes smaller.

Conditions for Diffraction

The incident light should be monochromatic.

The slit width should be comparable to the wavelength of incident light.

Types of Diffraction

Fresnel Diffraction: The light source and the screen both are at finite distances from the slit. The incident waves are not parallel.

Fraunhofer Diffraction: The light source and the screen both are infinitely away from the slit such that the incident light rays are parallel.

Solved Examples

1. Fraunhofer diffraction at a single slit is performed using a 700 nm light. If the first dark fringe appears at an angle 30 ° , find the slit width.

Solution: Using the diffraction formula for a single slit of width a, the n th dark fringe occurs for,

a sin\[\theta\] = nλ

At angle θ=30°, the first dark fringe is located. Using n=1 and

λ = 700 nm=700 X 10⁻⁹m,

a sin 30°=1 X 700 X 10⁻⁹m

a=14 X 10⁻⁷m

The slit width is 1400 nm.

2. Find the angular width of central maximum for Fraunhofer diffraction due to a single slit of width 0.1 m, if the frequency of incident light is 5 X 10¹⁴ Hz.

Solution: wavelength of the incident light is,

\[\lambda\] = \[\frac{c}{\nu }\]

Here, c=3 X 10⁸m/s is the speed of light in vacuum and =5 X 10¹⁴Hz is the frequency.

The angular width of the central maximum is,

2\[\theta\] = \[\frac{2c}{\nu a}\]

Using c = 3×10⁸m/s, = 5×10¹⁴Hz a = 0.1m,

2\[\theta\] = 1.2×10⁻⁴rad

The angular width is 1.210⁻⁴rad

About Fraunhofer and Fresnel

Fraunhofer and Fresnel, both kids faced immense hardships in life, partly because of their weak constitutions. At the end of the first decade of the nineteenth century, the building where Fraunhofer lived collapsed. Fraunhofer was the only one who survived. Maximilian IV Joseph, Prince-Elector of Bavaria, who was busy in the rescue efforts, noticed Fraunhofer and decided to take him under his wing. With the entry of Maximilian into his life, Fraunhofer’s life took a turn for the better.

In the neighbouring country of Italy simultaneously, Fresnel was still struggling to read at the age of eight. Surrounded by his three siblings, Fresnel had toys aplenty. At the age of nine, he managed to transform a harmless toy into such a lethal weapon that elders in his society were forced to ban the use of the now-dangerous toy. This caused no end of worry and consternation to his parents and grown-ups, but his genius was finally recognized.

He was taught by fine tutors and by the age of 16, he was admitted to the École Polytechnique. Here, an eminent mathematician, Adrien-Marie Legendre noticed Fresnel’s answer paper of striking ingenuity. Fresnel was encouraged in his studies of Maths and Science. He then graduated from École des Ponts et Chaussées as an Engineer.

Fraunhofer, The Scientist

Somewhere down a few longitudes, Maximilian took charge of the education of Fraunhofer and put him under the care of Joseph von Utzschneider, an attorney with a unique business mind. In those days industrious Utzschneider was busy partnering with Georg Reichenbach and J. Leibherr to open the Mathematical Mechanical Institute Reichenbach Utzschneider Liebherr. This institute was established to produce surveying instruments with high-quality lenses – incidentally, just the thing that Fraunhofer had a knack for.

After a failed stint at establishing his own business, Fraunhofer was offered a position in the Institute by Utzschneider. Here, under the guidance of expert glassmaker Pierre Guinand, Fraunhofer shined. Within two years he was offered a junior partnership in the Institute. Within a decade, Fraunhofer had turned the Institute around and was producing quality refractor lenses with a diameter of seven inches. This was an achievement in those times and was ably equipped with a decent-sized telescope.

Fresnel, The Scientist

On the other side, Fresnel was following the footsteps of his father, the architect. Having a few false starts mostly because he supported the monarchy that kept changing at that point in history, Fresnel settled as an engineer for the roads of Paris.

By now the careers of both scientists were converging as both were taking a keen interest in optics. By the middle of the second decade, both Fresnel and Fraunhofer were knee-deep in their research, resulting in scintillating findings that are studied in Optics for class 12 and higher classes.

Fraunhofer had a knack for handling glass and optics. A major challenge at his time was accurately polishing spherical surfaces of large glass objects. With his ingenuity, Fraunhofer improved upon and also invented machines that did more accurate grinding and polishing of glass surfaces. This greatly improved the kinds of lenses being manufactured and thus improved the quality of optical instruments being used.

In 1811, he constructed the furnace that would make the perfect large-sized lenses required in larger telescopes. With this new furnace, Fraunhofer created his own crown glass that had no irregular refractive power as larger crown glasses of that time tended to have.

In 1814, Fraunhofer invented the modern spectroscope through a series of experiments. By observing the orange fixed light in the spectrum of fire, he researched the dark lines of the solar spectrum, as well as in several bright stars, thus initiating research in stellar spectroscopy.

In 1821, he invented the diffraction grating and became the first scientist to get line spectra and then measure the wavelengths of spectral lines. His passion always remained practical optics so he kept developing and bettering optical instruments. He also created the Dorpat Refractor ad and Bessel Heliometer.

The most practical and popular invention of Fresnel was Fresnel Lens. He created this for Napoleon. It is used in theatrical and cinematic lighting fixtures and lighthouses to shine a narrow beam in all directions simultaneously. These are made of a lot of complicated wedge-shaped parts. Fresnel lenses magnify a light beam about 150 times and shine it, thus making it very practical in lighthouses in his time. Today these lenses are popular in film-making and theatre.

Did you know?

In the diffraction pattern of white light , the central maximum is white but the other maxima becomes colored with red being the farthest away.

Diffraction patterns can be obtained for any wave. Subatomic particles like electrons also show similar patterns like light. This observation led to the concept of a particle’s wave nature and it is considered as one of the keystones for the advent of quantum mechanics.

The interatomic distances of certain crystals are comparable with the wavelength of X-rays. Using X-ray diffraction patterns, the crystal structures of different materials are studied in condensed matter physics.

FAQs on Diffraction of Light - Young’s Single Slit Experiment

1. What is Optics?

Optics is a branch of Physics in which the properties of light and how it interacts with other objects, are studied. In everyday life, the uses of Optics can be seen in mirrors, light tunnels, lenses, spectacles, microscopes, and fibre optics.

2. How does Optics connect to Astronomy?

Since travelling enormous distances in Astronomy is not practical, scientists study the light from different astronomical bodies in space to gauge their distance from earth and to each other. Optics helps scientists unravel the secrets of outer space since not much other than light reaches the earth from outer space. The distance in space is measured in Light Years. One light-year is the distance travelled by light in a span of one year.

3. How do we measure the intensity of light?

We can measure how bright a light is. The unit of measuring the luminous intensity of light is Candela. A small candle produces about 1 Cd (Candela) of light. The amount of light falling on a surface is measured in lux. 1 Lu is the amount of light from a 1 Cd strong source 1 meter away. To be able to read one needs 500 lux of light.

4. How do we make the light turn the opposite way from where it is coming?

We can make light change direction by using a mirror. This simple method is used in making light tunnels in newly constructed houses to save energy. A light tunnel captures sunlight and reflects it along a tunnel fitted with mirrors in precise angles so that the other end of the tunnel reflects light inside the house.

5. What is the difference between Fresnel and Fraunhofer class of diffraction?

The light source and the screen both are at finite distances from the slit for Fresnel diffraction whereas the distances are infinite for Fraunhofer diffraction. The incident light rays are parallel (plane wavefront) for the latter. For Fresnel diffraction, the incident light can have a spherical or cylindrical wavefront.

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Single-Slit Diffraction

Definition: what is the single-slit diffraction.

Single-slit diffraction is the most straightforward experimental setup where diffraction effects can be observed. When light passes through a slit whose width is on the order of the wavelength of light, a distinct diffraction pattern is observed on a screen that is kept at a certain distance from the slit. The intensity is a function of the angle through which the rays bend.

Huygens’ Principle

According to Huygens’ principle, every unobstructed point on a wavefront will act as a source of secondary spherical waves. The new wavefront is the surface tangent to all the secondary spherical waves. Therefore, each part of the slit can be thought of as an emitter of waves.

Single Slit Diffraction Pattern

All the waves passing through the slit interfere to produce a diffraction pattern consisting of bright and dark fringes. The bright fringes are due to constructive interference, and the dark areas are due to destructive interference. A number represents the order of the bright and dark fringes. The intensities of the fringes consist of a central maximum surrounded by maxima and minima on its either side. The central maximum is brighter than the other maxima. The maxima rapidly decrease as one moves further from the center.

Single Slit Diffraction Equation

In order to study the diffraction pattern on a screen, the single-slit experiment is employed. Consider a monochromatic source of light that passes through a slit AB of width a as shown in the figure. At point P on the screen, the secondary waves interfere destructively and produce a dark fringe. Let D be the distance between the slit and the screen, and y be the distance between point P and point O, the center of the screen. AC is perpendicular to BP. Let θ be the angle of diffraction, and θ’ is the angle BAC.

We assume that the screen is at a considerable distance from the slit, i.e., D >> a . Hence,

θ = θ’

sin θ ≈ tan θ ≈ θ = y/D

The path difference between the two rays AP and BP is given by,

Δ = BP – AP = BC

In the right-angled triangle BCA,

sin θ’ = sin θ = BC/BA

BC = BA sin θ = a sin θ

Δ = a sin θ

Diffraction Minima

The condition for minima or dark fringe is,

Path difference = integral multiple of wavelength

or, Δ = nλ (n=±1, ±2, ±3, … , etc.)

or, a sin θ = nλ

or, ay/D = nλ

or, y n = nλD/a

This equation gives the distance of the n-th dark fringe from the center.

The fringe width is given by,

β = y n+1 – y n = (n+1)λD/a – nλD/a

or, β = λD/a

Diffraction Maxima

The condition for maxima or bright fringe is,

Path difference = non-integral multiple of wavelength

or, Δ = (n+1/2)λ (n=±1, ±2, ±3, … , etc.)

or, a sin θ = (n+1/2)λ

or, ay/D = (n+1/2)λ

or, y n = (n+1/2)λD/a

The intensity of single-slit diffraction is given by,

I = I 0 [sin (π a sin θ/λ)/( π a sin θ/λ)] 2

Difference Between Single- and Double-Slits Diffraction



Consist of one slit	Consist of two slits
Waves that originate within the same slit interfere	Slits are so small that each one is considered as a single light source, and the interference of waves originating within the same slit can be neglected
The fringes are broad and not as sharp as double-slit	The fringes are narrow and sharp
All bright fringes are visible	Some bright fringes are missing since they suppressed by the minima of a single-slit interference pattern
Intensity: [sin (π a sin θ/λ)/( π a sin θ/λ)]	Intensity: cos [π d sin θ/λ] [sin (π a sin θ/λ)/( π a sin θ/λ)]

Single-Slit Diffraction – Opentextbc.ca
Single-Slit Diffraction – Labman.phys.utk.edu
Diffraction from a Single Slit – Animations.physics.unsw.edu.au
Single-Slit Diffraction – Physics.nus.edu.sg
Single-Slit Diffraction – Openstax.org
Fraunhofer Single Slit – Hyperphysics.phy-astr.gsu.edu
Intensity of Single-Slit Diffraction – Scipp.ucsc.edu
Interference and Diffraction – Web.mit.edu

Article was last reviewed on Thursday, February 2, 2023

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Single Slit Diffraction

Single Slit Diffraction is a fundamental concept in wave optics that explains how light behaves as a wave when passing through a narrow slit. When coherent light (like a laser) goes through a single narrow slit, the waves spread out, and their interaction creates a pattern on a screen placed some distance away. This phenomenon, known as diffraction, leads to the formation of alternating bright and dark regions, showcasing the wave nature of light.

In this article, we’ll learn core concepts, types, and practical applications of Single Slit Diffraction, aiming to simplify and explore its patterns and formulas.

Table of Content

What is Single Slit Diffraction?

Central maximum, path difference, minima position, intensity distribution curve (pattern), single slit diffraction formula.

Diffraction is defined as the phenomenon in which light bends around the corners of an obstacle whose size is comparable to the wavelength of the light

Single Slit Diffraction occurs when light waves pass through a narrow aperture or slit. As the waves spread out, they interfere with each other, resulting in an interference pattern on a screen placed some distance away. This pattern includes a central maximum, secondary maxima, and minima, showing the bending and spreading of light waves around the edges of the slit.

The central maximum is the brightest region in a single slit diffraction pattern, appearing at the center of the screen. It’s formed due to constructive interference, where light waves traveling straight through the slit reinforce each other, resulting in a more intense central peak.

The formula for Central Maximum is:

sin θ = nλ/a where, θ is the Angle from the Center to Maxima n is the Order of the Maxima (for central maximum, it is 1) λ is the Wavelength of the incident light a is the Width of the slit

Path difference describes the variation in distances traveled by light waves from different points on the slit to a specific spot on the screen. It plays a critical role in determining how light waves interact—whether constructively or destructively—at that point. Understanding Single Slit Diffraction offers insights into the intriguing behavior of light waves as they pass through small openings, producing distinct patterns that reveal their wave-like nature. The formula for path difference in single slit diffraction is:

L = d⋅sin(θ) where, d represents the width of the slit θ is the angle of diffraction

The spots where we see darkness or dimness between the bright lines in a single slit diffraction pattern are called minima. They happen because some light waves meet at certain places on the screen and mix in a way that makes those spots look darker. This darkness occurs because these mixed light waves cancel each other out, resulting in reduced brightness or darkness in those particular areas.

The formula for Minima is:

a.sin θ= nλ where, θ is the Angle to Minima n is the Order of Minima λ is the Wavelength of the Incident light a is the Width of the Slit

Position of Secondary Maxima

Secondary maxima are smaller and less intense bright areas close to the brightest spot in a single slit diffraction pattern. They appear because some diffracted waves change direction around the edges of the slit, creating slightly brighter regions away from the brightest spot. This happens because these waves add up and reinforce each other at specific angles, making these areas a bit brighter than their surroundings.

In Diffraction by single slit intensity distribution curve shows how the brightness changes across the diffraction pattern. It explains how bright different angles or spots on the screen are. The curve usually has a highest point in the middle (which represents the central maximum) and then the brightness decreases gradually towards the secondary maxima and minima. This means that the brightest point in the middle is usually much brighter than the slightly less bright areas around it.

The diffraction single by single slit can be be best understood by the mathematical formula also called single slit diffraction formula.

The intensity distribution of diffracted light can be expressed by the single slit diffraction formula:

I(θ) = I 0 (sin(β)/ β) 2 where, I(θ) is the intensity at an angle θ I 0 is the intensity at the center of the diffraction pattern β is the phase difference between waves arriving at different points on the screen

Single Slit Diffraction vs Double Slit Diffraction

Diffraction of light can occur through single slit and double slit. However, the pattern observed in different in both the cases. The difference between single slit diffraction and double slit diffraction is tabulated below:


Involves a single narrow slit.	Involves two adjacent slits.
Results in an interference pattern with a central maximum, secondary maxima, and minima.	Displays multiple interference patterns with a series of bright and dark fringes.
Pattern created due to the diffraction of light waves passing through a single slit.	Pattern results from the interference of light waves from both slits.
Shows a central maximum and secondary maxima and minima.	Shows multiple interference patterns that incorporate characteristics of both single-slit and double-slit interference.
Exhibits a central maximum and decreasing intensity on either side with secondary maxima.	Shows alternating bright and dark fringes, with the central maximum being more pronounced and subsequent maxima decreasing in intensity.

Also, Check

Hugyen’s Wave Theory Difference between Diffraction and Interference Young’s Double Slit Diffraction

Single Slit Diffraction Solved Examples

Example 1: A single-slit diffraction experiment uses light of wavelength λ = 600 nm and a slit width of a = 0.1 mm. Calculate the angular position (in degrees) of the first minimum.

a = 0.1 mm = 0.1 × 10 (-3) m λ = 600 nm = 600 × 10 (-9) m For the first minimum, n = 1 Formula for the angular position of the first minimum: sin(θ) = n λ / a Substituting values: sin(θ) = (1) × (600 × 10 (-9) ) / (0.1 × 10 (-3) ) Calculating: sin(θ) ≈ 0.006 Hence, θ ≈ 0.6 degrees

Example 2: In a single-slit diffraction experiment, light of wavelength λ = 500 nm produces the first minimum at an angle x = 0.04 cm from the central maximum. Determine the slit width (in mm).

Given: λ = 500 nm = 500 × 10 -9 m x = 0.04 cm = 0.04 × 10 -2 m For the first minimum, n = 1 Formula for slit width: a = n × λ × x / sin(θ) sin(θ) = x / D (D is the distance between the slit and the screen) Substituting : a = (1) × (500 × 10 (-9) ) × (0.04 × 10 -2 )/ (sin(θ)) Sin(θ) ≈ θ (for small angles), therefore a ≈ 5 × 10 -3 m or 0.05 mm

Example 3. Light of wavelength λ = 500 nm passes through a slit of width a = 0.1 mm, creating a diffraction pattern on a screen. If the first minimum is observed at a distance x = 5 mm from the central maximum, what is the distance (in mm) between the screen and the slit?

Given: λ = 500 nm = 500 × 10 -9 m a = 0.1 mm = 0.1 × 10 -3 m x = 5 mm = 5 × 10 -3 m For the first minimum, n = 1 Formula for distance: D = n λ x / a Substituting values: D = (1) × (500 × 10 -9 ) × (5 × 10 -3 ) / (0.1 × 10 -3 ) D ≈ 6.4 mm

Example 4. For a single-slit diffraction pattern, light of wavelength λ = 500 nm produces the first minimum at an angular separation of Δθ = 60 degrees. Calculate the slit width (in mm).

λ = 500 nm = 500 × 10 -9 m Δθ = 60 degrees For the central maximum, n = 1 Formula for slit width: a = n × λ / sin(Δθ) Substituting values: a = (1) × (500 × 10 -9 ) / sin(60 degrees) Calculating: a ≈ 3.33 × 10 -3 m or 0.033 mm

Example 5: A monochromatic light of wavelength λ = 600 nm passes through a single slit and produces a diffraction pattern on a screen. If the angular width of the central maximum is 10 degrees, determine the width of the slit.

Given: Wavelength (λ) = 600 nm = 600 × 10-9 m Angular width of the central maximum = 10 degrees For the central maximum, n = 1 Formula for the width of the slit: a = n × λ / sin(Δθ) Substituting values: a = (1) × (600 × 10-9) / sin(10 degrees) Calculating: a ≈ 3.47 × 10-5 m or 34.7 μm

Practice Problems on Single Slit Diffraction

Q1: A single-slit diffraction pattern is formed on a screen 2 meters away. If a light of wavelength 500 nm produces the first minimum at an angle of 30 degrees, what is the width of the slit?

Q2: Light with a wavelength of 600 nm passes through a single slit and produces a diffraction pattern. If the angular width of the central maximum is 20 degrees, calculate the width of the slit.

Q3: In a single-slit diffraction experiment, if the second minimum is observed at an angle of 45 degrees and the wavelength of light used is 450 nm, find the width of the slit.

Q4: A diffraction pattern is observed on a screen placed 1.5 meters away from a single slit. If light of wavelength 700 nm produces the first minimum at an angle of 45 degrees, determine the width of the slit.

Q5: When monochromatic light with a wavelength of 400 nm passes through a single slit, the first minimum is observed at an angle of 15 degrees. What is the width of the slit?

Single Slit Diffraction – FAQs

1. what is diffraction.

The bending of light around the corners of obstacle whose size is comparable to the wavelength of the light is called Diffraction

2. What causes Single Slit Diffraction?

Single Slit Diffraction is primarily caused by the wave nature of light. As light waves pass through a single slit, they interfere with each other, leading to the formation of a distinct diffraction pattern.

2. How does the Diffraction Pattern change with different Slit Widths?

The width of the slit directly influences the spacing and intensity of the diffraction maxima and minima. Wider slits result in broader and less intense patterns, while narrower slits yield sharper and more concentrated patterns.

4. Can Single Slit Diffraction occur with other forms of Waves?

Single Slit Diffraction is a universal wave phenomenon, and it can manifest with various types of waves, not limited to light waves.

5. What is the Difference between Fraunhofer and Fresnel Diffraction?

The key distinction lies in the distances involved. Fraunhofer Diffraction assumes an infinitely distant light source, whereas Fresnel Diffraction considers finite distances between the source, slit, and observing screen. This difference in perspective leads to variations in the observed diffraction patterns.

6. How does changing the Wavelength affect the Diffraction Pattern?

The wavelength of light directly influences the spacing of the diffraction maxima and minima. Shorter wavelengths result in narrower patterns, impacting the overall appearance of the diffraction pattern on the screen

7. What is Diffraction from One Slit?

Diffraction from one slit refers to the bending and spreading of light waves when they pass through a single narrow aperture or slit, causing the waves to spread out and interfere with each other, forming an interference pattern on a screen.

8. What are the two main Types of Diffraction?

The two main types of diffraction are: Single Slit Diffraction: Occurs when light passes through a single narrow aperture, resulting in an interference pattern with a central maximum and secondary maxima. Double Slit Diffraction: Involves two adjacent slits, producing multiple interference patterns on a screen, displaying a series of bright and dark fringes.

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Young's Experiment

Anatomy of a Two-Point Source Interference Pattern
The Path Difference
Young's Equation
Young's Experiment
Other Applications of Two-Point Source Interference

Today's classroom version of the same experiment is typically performed using a laser beam as the source. Rather than using a note card to split the single beam into two coherent beams, a carbon-coated glass slide with two closely spaced etched slits is used. The slide with its slits is most commonly purchased from a manufacturer who provides a measured value for the slit separation distance - the d value in Young's equation. Light from the laser beam diffracts through the slits and emerges as two separate coherent waves. The interference pattern is then projected onto a screen where reliable measurements can be made of L and y for a given bright spot with order value m . Knowing these four values allows a student to determine the value of the wavelength of the original light source.

To illustrate some typical results from this experiment and the subsequent analysis, consider the sample data provided below for d, y, L and m.


)
)
to AN ( )
)

(Note: AN 0 = central antinode and AN 4 = fourth antinode)

The determination of the wavelength demands that the above values for d, y, L and m be substituted into Young's equation.

Careful inspection of the units of measurement is always advisable. The sample data here reveal that each measured quantity is recorded with a different unit. Before substituting these measured values into the above equation, it is important to give some thought to the treatment of units. One means of resolving the issue of nonuniform units is to simply pick a unit of length and to convert all quantities to that unit. If doing so, one might want to pick a unit that one of the data values already has so that there is one less conversion. A wise choice is to choose the meter as the unit to which all other measured values are converted. Since there are 1000 millimeters in 1 meter, the 0.250 mm is equivalent to 0.000250 meter. And since there are 100 centimeters in 1 meter, the 10.2 cm is equivalent to 0.102 m. Thus, the new values of d, y and L are:

While the conversion of all the data to the same unit is not the only means of treating such measured values, it might be the most advisable - particularly for those students who are less at ease with such conversions.

Now that the issue regarding the units of measurement has been resolved, substitution of the measured values into Young's equation can be performed.

λ = 6.52 x 10 -7 m

As is evident here, the wavelength of visible light is rather small. For this reason wavelength is often expressed using the unit nanometer, where 1 meter is equivalent to 10 9 nanometers. Multiplying by 10 9 will convert the wavelength from meters to nanometers (abbreviated nm).

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Check Your Understanding

1. The diagram below depicts the results of Young's Experiment. The appropriate measurements are listed on the diagram. Use these measurements to determine the wavelength of light in nanometers. (GIVEN: 1 meter = 10 9 nanometers)

Answer: 657 nm

First, identify known values in terms of their corresponding variable symbol:

L = 10.2 m = 1020 cm y = 22.5 cm m = 10 d = 0.298 mm = 0.0298 cm

(Note: m was chosen as 10 since the y distance corresponds to the distance from the 5th bright band on one side of the central band and the 5th bright band on the other side of the central band.)

Then convert all known values to an identical unit. In this case, cm has been chosen as the unit to use. The converted values are listed in the table above.

Substitute all values into Young's equation and perform calculation of the wavelength. The unit of wavelength is cm.

λ = y • d / ( m • L) λ = ( 22.5 cm ) • ( 0.0298 cm ) / [ ( 10 ) • ( 1020 cm ) ] λ = 6.57 x 10 -5 cm

Finally convert to nanometers using a conversion factor. If there are 10 9 nm in 1 meter, then there must be 10 7 nm in the smaller centimeter.

λ = ( 6.57 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 657 nm

2. A student uses a laser and a double-slit apparatus to project a two-point source light interference pattern onto a whiteboard located 5.87 meters away. The distance measured between the central bright band and the fourth bright band is 8.21 cm. The slits are separated by a distance of 0.150 mm. What would be the measured wavelength of light?

Answer: 524 nm

L = 5.87 m = 587 cm y = 8.21 cm m = 4 d = 0.150 mm = 0.0150 cm

λ = y • d / ( m • L) λ = ( 8.21 cm ) • ( 0.0150 cm ) / [ ( 4 ) • ( 587 cm ) ] λ = 5.24 x 10 -5 cm

λ = ( 5.24 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 524 nm

3. The analysis of any two-point source interference pattern and a successful determination of wavelength demands an ability to sort through the measured information and equating the values with the symbols in Young's equation. Apply your understanding by interpreting the following statements and identifying the values of y, d, m and L. Finally, perform some conversions of the given information such that all information share the same unit.

y =

d =

m =

L =

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit.

y = 12.8 cm	d = 0.250 mm	m = 4.5	L = 8.2 meters

y = 12.8 cm	d = 0.0250 cm	m = 4.5	L = 820 cm

(Note that m = 4.5 represents the fifth nodal position or dark band from the central bright band. Also note that the given values have been converted to cm.)

b. An interference pattern is produced when light is incident upon two slits that are 50.0 micrometers apart. The perpendicular distance from the midpoint between the slits to the screen is 7.65 m. The distance between the two third-order antinodes on opposite sides of the pattern is 32.9 cm.

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit.

y = 32.9 cm	d = 50.0 µm	m = 6	L = 7.65 m

y = 32.9 cm	d = 0.00500 cm	m = 6	L = 765 cm

(Note that m = 6 corresponds to six spacings. There are three spacings between the central antinode and the third antinode. The stated distance is twice as far so the m value must be doubled. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

c. The fourth nodal line on an interference pattern is 8.4 cm from the first antinodal line when the screen is placed 235 cm from the slits. The slits are separated by 0.25 mm.

y = 8.4 cm	d = 0.25 mm	m = 2.5	L = 235 cm

y = 8.4 cm	d = 0.025 cm	m = 2.5	L = 235 cm

( Note that the fourth nodal line is assigned the order value of 3.5. Also note that the given values have been converted to cm.)

d. Two sources separated by 0.500 mm produce an interference pattern 525 cm away. The fifth and the second antinodal line on the same side of the pattern are separated by 98 mm.

y = 98 mm	d = 0.500 mm	m = 3	L = 525 cm

y = 9.8 cm	d = 0.0500 cm	m = 3	L = 525 cm

( Note that there are three spacings between the second and the fifth bright bands. Since all spacings are the same distance apart, the distance between the second and the fifth bright bands would be the same as the distance between the central and the third bright bands. Thus, m = 3. Also note that the given values have been converted to cm.)

e. Two slits that are 0.200 mm apart produce an interference pattern on a screen such that the central maximum and the 10th bright band are distanced by an amount equal to one-tenth the distance from the slits to the screen.

y = 0.1 • L	d = 0.200 mm	m = 10	L - not stated

y = 0.1 • L	d = 0.200 mm	m = 10	L - not stated

( Note that there are 10 spacings between the central anti-node and the tenth bright band or tenth anti-node. And observe that they do not state the actual values of L and y; the value of y is expressed in terms of L. )

f. The fifth antinodal line and the second nodal line on the opposite side of an interference pattern are separated by a distance of 32.1 cm when the slits are 6.5 m from the screen. The slits are separated by 25.0 micrometers.

y = 32.1 cm	d = 25.0 µm	m = 6.5	L = 6.5 m

y = 32.1 cm	d = 0.00250 cm	m = 6.5	L = 650 cm

( Note that there are five spacings between the central anti-node and the fifth anti-node. And there are 1.5 spacings from the central anti-node in the opposite direction out to the second nodal line. Thus, m = 6.5. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

g. If two slits 0.100 mm apart are separated from a screen by a distance of 300 mm, then the first-order minimum will be 1 cm from the central maximum.

y = 1 cm	d = 0.100 mm	m = 0.5	L = 300 mm

y = 1 cm	d = 0.0100 cm	m = 0.5	L = 30.0 cm

( Note that a the first-order minimum is a point of minimum brightness or a nodal position. The first-order minimum is the first nodal position and is thus the m = 0.5 node. Also note that the given values have been converted to cm. )

h. Consecutive bright bands on an interference pattern are 3.5 cm apart when the slide containing the slits is 10.0 m from the screen. The slit separation distance is 0.050 mm.

y = 3.5 cm	d = 0.050 mm	m = 1	L = 10.0 m

y = 3.5 cm	d = 0.0050 cm	m = 1	L = 1000 cm

( Note that the spacing between adjacent bands is given. This distance is equivalent with the distance from the central bright band to the first antinode. Thus, m = 1. Also note that the given values have been converted to cm. )

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Single Slit Diffraction: Young’s Single Slit Experiment and Fringe Width

Jasmine Grover

Content Strategy Manager

Single slit diffraction takes place when light is incident on a slit with a size comparable to the wavelength of light and an alternating dark and bright pattern can be observed. Diffraction implies the bending of light around the sharp corner of an obstacle. Diffraction can be seen when the sources are small enough that they are fairly the size of light’s wavelength.

Key Terms: Diffraction, Single Slit Diffraction, Young’s Single Slit Experiment, Fringe Width, Huygens' Guideline , Wavelength, Light, Wave, Obstacles

What is Diffraction?

[Click Here for Previous Year Questions]

Diffraction is the inclination of a wave produced from a limited source or passing through a limited gap to spread out around obstacles . Diffraction results from the impedance of an endless number of waves transmitted by constant circulation of source focus in a few measurements.

$Diffraction$

For example, the silver lining occurring in the sky is caused by the diffraction of light. Thus, Diffraction is the slight bending of light as it passes around the edge of an object. The measure of the bending of light relies upon the overall size of the frequency of light to the size of the opening. However, if the opening is bigger than the light's frequency, the bend might be unnoticeable.

Relatable Links:

What is Single Slit Diffraction?

[Click Here for Sample Questions]

Single-slit diffraction can be defined as an observation where the bending of light (or, diffraction) causes the coherent source of light to interfere among themselves in order to produce a certain pattern on the screen, known as the Diffraction Pattern.

When the light goes through a single slit with a width w ( frequency of the light), then a single slit diffraction forms on a screen with distance L >> w away from the slit. Huygens' guideline reveals that each piece of the slit can be considered a producer of waves. This load of waves meddles to deliver the diffraction pattern.

$Single Slit Diffraction$

Single Slit Diffraction

In other words, if the light is incident on a slit with a width that is comparable to the wavelength of light, then an alternating shadow and bright pattern forms on a screen erected in front of the slit. This phenomenon of light is known as Single Slit Diffraction.

Young's Single Slit Experiment

According to Thomas Young’s double-slit experiment , performed in 1801, the wave nature of light was demonstrated. In this experiment, monochromatic light was illustrated on two narrow slits. After passing through each slit, the waves superimpose to give an alternate bright and dark distribution on a distant screen. And every bright fringe has the same intensity and width.

Monochromatic light in a single-slit experiment is transferred through one slit of finite width and an identical pattern is observed on the screen. As we move away from the central maximum, unlike the double-slit diffraction pattern, the width and intensity in the single-slit diffraction pattern reduce.

Single Slit Diffraction Formula

In the event that a monochromatic light of frequency λ falls on a slit of width a, the force on a screen a ways off L from the slit can be communicated as a component of θ. Here, θ is the point made with the first course of light. It is given by,

α/α … (1)

Here, α = π*λ

Sin θ and I0 are the intensity of the central bright fringe, situated at θ = 0.

$Single Slit Diffraction Formula$

Diffraction Maxima and Minima: Bright edges show up at points,

θ = 0, θ = sin-1 (±3λ/2), θ = sin-1 (±5λ/2)

θ = 0 is the focal most extreme.

… (2)

Dark fringes compared to the condition,

(3)

In a double-slit arrangement, diffraction through a single slit shows up as an envelope over the obstruction design between the two slits.

Fringe Width

The angular distance between the two first-order minima (on one or the other side of the centre) is known as the angular width of the central maximum, given by:

The straight width is as per the following,

Δ = L $\times$ 2θ= 2Lλa … (4)

The width of the central maximum in the diffraction formula is in inverse proportion with the slit width. Therefore, if the slit width decreases, the central maximum widens, and if the slit width increases, it narrows down. Hence, it can be concluded from this behaviour that light bends more as the dimension of the aperture becomes smaller.

Central Maximum

The maxima placed between the minima and the width of the central maximum can be expressed as the distance between the 1st order minima from the centre of the screen on either sides of the centre.

Position of minima is expressed by y (which is evaluated from the centre of the screen):

Now, for small θ, it can be said:

sin θ ≈ θ

⇒ λ = a sin θ ≈ aθ

⇒ θ = y/D = λ/a

⇒ y = λD/a

Width of central maximum is basically twice this value:

⇒ Therefore, Width of central maximum = 2λD/a

⇒ And, Angular width of central maximum = 2θ = 2λ/a

Types of Diffraction

There are two types of Diffraction:

Fresnel Diffraction : The light source and the screen both are at limited good ways from the slit. The episode waves are not equal.
Fraunhofer Diffraction : The light source and the screen both are vastly away from the slit with the end goal that the occurrence of light beams are equal.

$Types of Diffraction$

The constructive interference condition is that the path difference must be equivalent to the integral multiple of the wavelength.

Compton effect can be expressed as the effect observed when x-rays or gamma rays are scattered on a material with increase in wavelength.

The destructive interference condition is that the path difference must be equivalent to the odd integral multiple of half wavelength.

Temporal coherence can be defined as the correlation between field at one point and the field at the same point later some time.

Previous Year Questions

For a crystal, the angle of diffraction … [BITSAT 2008]
When light is incident on a diffraction grating the zero order … [KCET 2004]
The width of the diffraction band varies … [KCET 2006]
A parallel monochromatic beam of light is incident normally on a narrow slit … [JEE Advance 1998]
Calculate the wavelength of light used in an interference experiment … [BITSAT 2015]
Two beams of light having intensities … [JEE Advance 2001]
Angular width of the central maxima in the Fraunhofer diffraction … [NEET 2019]
For a parallel beam of monochromatic light of wavelength λ … [NEET 2015]
The angular resolution of a 10 cm diameter telescope at a wavelength of … [NEET 2005]
The ratio of resolving powers of an optical microscope … [NEET 2017]
A beam of light of wavelength 600nm from a distant source falls on a single slit … [KCET 2004, UPSEE 2006]
The aperture diameter of a telescope is 5 m … [JEE Mains 2020]
The angle of minimum deviation for an incident light ray on an equilateral prism … [KCET 2009]
In refraction, light waves are bent on passing from one medium … [KCET 2021]
Two plane wavefronts of light, one incident on a thin convex lens … [KEAM]

Things to Remember

Diffraction is the inclination of a wave produced from a limited source or going through a limited gap to fan out as it spreads. Diffraction results from the impedance of an endless number of waves transmitted by constant circulation of source focus in a few measurements.
For monochromatic light, I(θ) = Io Sin 2 α/α 2
For Bright Fringes, θ = sin-1 ± (2n+1λ)/2
For Dark Fringes, asinθ = m λ

Sample Questions

Ques: What do you mean by the Compton effect? (1 mark)

Ans: Compton effect, discovered by Arthur Holly Compton, is the increase in the wavelength of the X-rays and the gamma rays which occurs when they are scattered.

Ques: What is fringe width? (2 marks)

Ans: Fringe width is defined as the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes.

Ques: What is meant by phase difference? (2 marks)

Ans: Phase difference is referred to as the difference between any two waves or the particles having the same frequency and starting from the same point. The phase difference is expressed in degrees or radians.

Ques: What is the condition for constructive and destructive interference? (2 marks)

Ans: The condition for constructive interference is that the way contrast ought to be equivalent to an essential variable of the frequency. The condition for destructive interference is that the way contrast ought to be equivalent to an odd fundamental numerous of half frequency.

Ques: Consider a solitary slit diffraction design for a slit width w. It is seen that for the light of frequency 400 nm the point between the primary least and the focal greatest is 4*10 -3 radians. What is the worth of w? (3 marks)

Ans: Dark edges in the diffraction example of the single slit are found at points θ for which w*sinθ = mλ, where m is a number, m = 1, 2, 3, ... . For the principal dim periphery we have w*sinθ = λ. Here we are approached to address this condition for w.

Subtleties of the estimation:

First least: w*sinθ = λ, w = (400 nm)/sin(4*10 -3 radians) = 1*10 -4 m.

Ques: When a monochromatic light source radiates through a 0.2 mm wide slit onto a screen 3.5 m away, the main dim band in the example seems 9.1 mm from the focal point of the brilliant band. What is the frequency of light? (3 marks)

Ans: Dark edges in the diffraction example of a single slit are found at points θ for which w*sinθ = mλ, where m is a whole number, m = 1, 2, 3, ... . For the primary dim periphery we have w*sinθ = λ. Here we are approached to settle this condition for λ.

z = 9.1 mm = 9.1*10 -3 m.

w = 0.2 mm = 2*10 -4 m.

L >> z, thus sinθ ~ z/L and λ = zw/(mL).

λ = (9.1*10 -3 m)(2*10 -4 m)/(3.5 m).

λ = 5.2*10 -7 m = 520 nm.

Ques: What is the distinction between the Fresnel and Fraunhofer class of diffraction? (2 marks)

Ans: The light source and the screen both are at limited good ways from the slit for Fresnel diffraction while the distances are endless for Fraunhofer diffraction. The occurrence of light beams is equal (plane wavefront) for the last mentioned. For Fresnel diffraction, the occurrence light can have a circular or barrel-shaped wavefront.

Ques: What is stage contrast? (2 marks)

Ans: The stage distinction is characterized as the contrast between any two waves or the particles having a similar recurrence and beginning from a similar point. It is communicated in degrees or radians.

Ques: What is single-slit diffraction? (2 marks)

Ans: If monochromatic light falls on a thin slit having a width practically identical to the frequency of the episode light, a trademark example of dull and splendid districts is acquired on a screen set before the slit. The waves from each place the slit begin to spread in stage however procure a stage contrast on the screen as they cross various distances. The noticed example is brought about by the connection among force and way contrast.

Ques: What are the conditions for diffraction? (2 marks)

Ans: The conditions for diffraction are as follows:

The episode light ought to be monochromatic.
The slit width ought to be practically identical to the frequency of episode light.

Ques: Fraunhofer diffraction at a solitary slit is performed utilizing a 700 nm light. On the off chance that the primary dull periphery shows up at a point of 300, discover the slit width. (3 marks)

Ans: Using the diffraction equation for a solitary slit of width a, the nth dull periphery happens for,

a sin θ= nλ

At point θ=300, the main dull periphery is found. Utilizing n=1 and λ= 700 nm=700 X 10 -9 m,

a sin 300=1 X 700 X 10 -9 m

a=14 X 10 -7 m

The slit width is 1400 nm.

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CBSE CLASS XII Related Questions

1. a capillary tube of radius r is dipped inside a large vessel of water. the mass of water raised above water level is m. if the radius of capillary is doubled, the mass of water inside capillary will be.

$\frac M4$

2. A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

3. (a) a circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 a is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 t. the field lines make an angle of 60° with the normal of the coil. calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area (all other particulars are also unaltered.), 4. a series lcr circuit connected to a variable frequency 230 v source. l = 5.0 h, c = 80mf, r = 40 ω. (a) determine the source frequency which drives the circuit in resonance. (b) obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) determine the rms potential drops across the three elements of the circuit. show that the potential drop across the lc combination is zero at the resonating frequency, 5. a boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. he runs to the other, end. the distance through which the centre of mass of the boat boy system moves is, 6. a convex lens of glass is immersed in water compared to its power in air, its power in water will.

decrease for red light increase for violet light

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( ) ( ) with the phasor sum

Diffraction pattern from a single slit

The sketch shows the view from above a single slit. Let's assume that the slit is constant width and very tall compared with that width, so that we can consider the system as two-dimensional. With light at normal incidence, the pattern is symmetrical about the axis of the slit. On a distant screen, the light arriving on the axis from all points in the slit has travelled an equal distance from the slit, so the centre of the pattern is a maximum. The next question is what determines its width.

Notice the broad central maximum, and the equally spaced, successively weaker maxima on either side.

First order minima

This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum. We call the slit width , and we imagine it divided into two equal halves. Using the , we consider a point at the very top of the slit, and another point a distance /2 below it, a point at the very top of the lower half of the slit. Consider parallel rays from both points, at angle to the axis of symmetry. (Why parallel? Because the screen is distant. Typically in diffraction experiments, the slit is ~ 10 µm wide, while the distance to the screen might be ~ 1 m.)

The ray from the distance /2 below has to travel an extra distance ( sin /2). If this distance is half a wavelength, . if sin =

then they are π/2 out of phase and they interfere destructively. Now, for every point in the top half of the slit, there is one in the bottom half a distance /2 below and, at the angle that satisfies sin = , they all interfere destructively.

So the first mimimum has sin = / . On the other side of the axis of symmetry, sin = / is also a minimum. These two minima limit the broad central maximum.

Higher order minima

An argument like the one applies if, in our imagination, we divide the slit into any even number of equal slices. THe diagram shows a division into four. Each half is divided into quarters, and light from a source in the first quarter cancels that from one in the second quarter. Similarly, sources in the third quarter are nullified by those in the fourth quarter. So this diagram represents the second order minima, where sin = / , or sin = / . For the nth order minima, we have

sin = , where n is an integer, but not zero.

Remember that, on the axis where = 0, there is a minimum, so the minima are equally spaced in sin , except either side of the central maximum.

We can note too that, for light diffracting the throught slits, the slit is usually much wider than a wavelength, so the pattern is usually very small, so the approximation that sin = is usually good.

Next we calculate how the intensity varies with sin .

Intensity I ( θ )

The slit of width a is divided into N slits, each of width δa . The path difference between rays from successive slices of the slit are equal, and so too are the angles between successive phasors . The slices of the slit all have equal width and length, so the lengths of all the phasors are equal. For a large number of slices, the phasors approximate the arc of a circle, as shown in red. We'll call the angle that this subtends 2 α . Now 2 α is also the phase difference between the first and the last phasor, which is 2π a sin θ / λ .

The phasor sum has magnitude A , which we can write as R .sin α , where R is the radius of the arc formed by the phasor sum. On the axis, where the phasors are all in phase, the phasor sum is the straight line shown in red at right. This is the amplitude of the diffraction patter at θ /= 0, which we call A 0 . By the definition of angle, 2 α = A 0 / R , which gives us R = A 0 /2 α . So the magnitude of the phasor sum is

Graphing ( ) with the phasor sum

( ) and

As we remarked when looking at the intenstiy of Young's experiement in Interference , the eye does not respond linearly to intensity. To my eyes, at least, the difference in brightness seems much less than the difference in the intensity graph.

Varying the slit width

DIffraction effects are most noticeable when the slit with a is not very much larger than the wavelength λ . This apparatus allows us to vary the slit width, but the pattern is still projected on a distant screen to make the diffraction effects cl

Young's experiment with finite slit width.

d a
a

In the Young's experiment in Interference , we didn't mention the effect of finite slit width. From the equations above, we can see that, if d is an integral number of times the slit width ( d = na ), then the n th interference fringe is absent: neither slit radiates power at this angle so there are no rays for constructive interference.

a
a d = 3a

Note that the envelope for the interference pattern has four times the magnitude of the diffraction pattern for a single slit. The integral of the combined interference pattern (the area under the red curve) must be twice the integral of diffraction pattern (the area under the black curve), because two slits admit twice as much light, so its maximum must be four times the maximum of the diffraction pattern. We show this experimentally in Interference with single photons. (In the photograph above, it's hard to see by eye that the factor is four. However, note that the interference peaks are white, because of extreme saturation of the camera.)

Compare Young’s Double Slit Interference Pattern and Single Slit Diffraction Pattern. - Physics

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Compare Young’s Double Slit Interference Pattern and Single Slit Diffraction Pattern.

Solution Show Solution


i.	For a common laboratory setup, the slits in Young’s double-slit experiment are much thinner than their separation. They are usually obtained by using a biprism or a Lloyd’s mirror. The separation between the slits is a few mm only.	The single slit used to obtain the diffraction pattern is usually of width less than 1 mm.
ii.	Size of the pattern obtained: With the best possible setup, the observer can usually see about 30 to 40 equally spaced bright and dark fringes of nearly the same brightness.	Size of the pattern obtained: Taken on either side, the observer can see around 20 to 30 fringes with the central fringe being the brightest.
iii.	W = `(λ "D")/"d"`	W = `(λ "D")/"a"` Except for the central bright fringe
iv.
a.	Phase difference, Φ between extreme rays: n(2π)	Phase difference, Φ between extreme rays: `("n" + 1/2)` (2π) OR (2n + 1)π
b.	Angular position, θ: n`(λ/"d")`	Angular position, θ: `("n" + 1/2)(λ/"a")` OR `((2"n" + 1)λ)/(2"a")`
c.	Path difference, ∆l between extreme rays: nλ	Path difference, ∆l between extreme rays: nλ
d.	Distance from the central bright spot, y: n`((λ"D")/"d")` = nW	Distance from the central bright spot, y: `("n" + 1/2)((λ"D")/"a") = ("n" + 1/2)`W
v.	dark fringe
a.	Phase difference, Φ between extreme rays: `("n" - 1/2) (2pi)` OR (2n - 1)π	Phase difference, Φ between extreme rays: n(2π)
b.	Angular position, θ: `("n" - 1/2)(λ/"d")` OR (2n - 1)`λ/(2"d")`	Angular position, θ: n`(λ/"a")`
c.	Path difference, ∆l between extreme rays: `("n" - 1/2)λ` OR `(2"n" - 1)λ/2`	Path difference, ∆l between extreme rays: nλ
d.	Distance from the central bright spot, y': `("n" - 1/2)((λ "D")/"d") = ("n" - 1/2)`W	Distance from the central bright spot, y': n`((λ"D")/"a")` = nW

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Interference: Young’s Double Slit Experiment
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Optics of Class 12

Interference: young’s double slit experiment .

It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light.

Two slits S 1 and S 2 are made in an opaque screen, parallel and very close to each other. These two are illuminated by another narrow slits S which in turn lit by a bright source. Light wave spread out from S and fall on both S 1 and S 2 which behave like coherent sources. Note that the coherent sources are derived from the same source. In this way, any phase change in S occurs in both S 1 and S 2 . The phase difference (φ 1 - φ 2 ) between S 1 and S 2 is unaffected and remain constant.

Light now spreads out from both S 1 and S 2 falls on a screen. It is essential that the waves from the two sources overlap on the same part of the screen. If one slit is covered up, the other produces a wide smoothly illuminated patch on the screen. But when both slits are open, the patch is seen to be crossed by dark and bright bands called interference fringes.

Condition for bright fringes or maxima

φ = 2nπ or path difference , p = nλ where n = 0, 1, 2, ……

Interference: Young’s Double Slit Experiment

Condition for dark fringes or minima

φ = (2n – 1)πorpath difference, p = (n - 1/2)λ

where n = 1, 2, 3, …………

The relation between phase difference (φ) and path difference (p) is given by

φ = 2π/λ p (15.32)

How to find the position of the nth maxima or minima on the screen?

Let P be the position of the nth maxima on the screen. The two waves arriving at P follow the path S 1 P and S 2 P, thus the path difference between the two waves is

p = S 1 P – S 1 P = d sin θ

From experimental conditions, we know that D >> d, therefore, the angle θ is small,

thus sinθ ≈ tanθ = Yn/D

∴p = d sinθ = d tan θ = d (Yn/D) ⇒ y n = pD/d

For nth maxima

p = nλ

∴y n = nλ D/d where n = 0, 1, 2, ……..(15.33)

For nth minima

p = (n - 1/2)λ

∴y n = (n - 1/2)λD/d where n = 1, 2, 3, ……(15.34)

Note that the nth minima comes before the nth maxima.

Fringe Width

It is defined as the distance between two successive maxima or minima.

Example 15.23

In a YDSE , if D = 2 m; d = 6 mm and λ = 6000 Å, then

(a)find the fringe width

(b)find the position of the 3 rd maxima

(c)find the position of the 2 nd minima

Solution

(a)Using equation (15.33),

ω = λD/d = (6000 x 10 -10 )(2)/6 x 10 -3 = 0.2 mm

(b)Position of 3 rd maxima is obtained by putting n = 3 in the equation (15.33),

y 3 = 3 λD = 3ω = 3(0.2) = 0.6 mm

(c)Position of 2 nd minima is obtained by putting n = 2 in the equation (15.34),

Example 15.24

What is the effect on the interference fringes in a YDSE due to each of the following operations.

(a)the screen is moved away from the plane of the slits

(b)the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength

(c)the separation between the two slits is increased

(d)the monochromatic source is replaced by source of white light

(e)the whole experiment is carried out in a medium of refractive index μ

(a)Angular separation (= λ/d ) of the fringes remains constant. But the linear separation or fringe width increases in proportion to the distance (D) from the screen.

(b)As λ decreases, fringe width (ω ∝ λ) decreases

(c)As d increases, fringe width (ω ∝ 1/d) decreases

(d)The interference pattern due to different component colours of white light overlap (in-coherently). The central bright fringes of different colours are at the same position. Therefore, the central fringe is white. Since blue colour has the lower λ, the fringe closest on either side of the central white fringe is blue; the farthest is red.

Thus, fringe width decreases by μ.

Example 15.25

Light from a source consists of two wavelength λ 1 = 6500Å and λ 2 = 5200Å. If D = 2m and d = 6.5 mm find the minimum value of y (≠ 0) where the maxima of both the wavelengths coincide.

Let n 1 th maxima of λ 1 coincides with n 2 th maxima of λ 2 .

Then yn1 = yn2

Thus, fourth maxima of λ 1 coincides with fifth maxima of λ 2 .

The minimum value of y (≠ 0) is given by

Optical Path

It is defined as distance travelled by light in vacuum in the same time in which it travels a given path length in a medium. If light travels a path length d in a medium at speed c, the time taken by it will be (d/c). So optical path length−

L = c o × [d/c] = μd(because μ = c0/c )(15.36)

Since for all media μ > 1, optical path length is always greater than geometrical path length. When two light waves arrive at a point by travelling different distances in different media, the phase difference between the two is related by their optical path difference instead of simply path difference.

Phase Difference = 2π/ λ (optical path difference) = 2π/ λ L

Fringe Shift

When a transparent film of thickness t and refractive index μ is introduced in front of one of the slits, the fringe pattern shifts in the direction where the film is placed.

How much is the fringe shift?

Consider the YDSE arrangement shown in the figure.

The optical path difference is given by

p = [(S2P – t) + μt] – S1P

orp = (S2P – S1P) + t(μ - 1)

SinceS2P – S1P = d sin θ

∴p = d sin θ + t (μ - 1)

As sin θ ≈ tan θ =

∴p =

In the absence of film, the position of the nth maxima is given by equation (15.34)

y n = n λD/d

Therefore, the fringe shift is given by

FS = y n - y′ n = (μ - 1) tD/d (15.37)

Note that the shift is in the direction where the film is introduced.

Example 15.26

Ιn a YDSE, λ = 6000Å, D = 2 m, d = 6 mm. When a film of refractive index 1.5 is introduced in front of the lower slit, the third maxima shifts to the origin.

(a)Find the thickness of the film

(b)Find the positions of the fourth maxima

(a)Since 3 rd minima shifts to the origin, therefore, the fringe shift is given by

FS = y 3 = 3 λD/d

From equation (15.37), we know

FS = (μ - 1) tD/d

∴(μ - 1) tD/d = 3 λD/d

or t = 3 λ/μ - 1

Here λ = 0.6 × 10 -6 m; μ = 1.5

∴t = 3(0.6 x 10 -6 )/1.5 -1 = 3.6 μm

(b)There are two positions of 4th maxima: one above and the other below the origin.

y4 = 1ω = λD/d = 0.2 mm

y′4 = -7ω = -7λD/d = 1.4 mm

Intensity Distribution

When two coherent light waves of intensity I 1 and I 2 with a constant phase difference φ superimpose, then the resultant intensity is given by

I = I 1 + I 2 + 2 √I 1 I 2 cosφ

In YDSE, usually the intensities I 1 and I 2 are equal

i.e.I 1 = I 2 = I o

Then I = 2I o ( 1+ cos φ)

orI = 4I o cos 2 (φ/2)(15.38)

For maxima:φ = 2nπ ⇒I max = 4I o

For minima:φ = (2n – 1)π⇒I min = 0

Example 15.27

The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.

SinceI 1 = 2I o and I 2 = I o , therefore

Example 15.28

In a YDSE, λ = 60 nm; D = 2m; λ = 6 mm. Find the positions of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen.

Using equation (15.32)

I = 4I o cos 2 φ/2

here I = 3/4 (4I o ) = 3I o

∴cos φ/2 = √3/2

Thus,φ/2 = nπ ± π/6

or φ = 2nπ ± π/3

For the point lying between third minima and third maxima,

or y 3 = 17/6 λD/d

Putting λ = 0.6 × 10 -6 m; D = 2m; d = 6 mm

Geometrical Optics
Reflection from a Plane-surface
Reflection from Curved Surfaces
Refraction Through Plane Surfaces
Refraction on Curved Surfaces
Wave Optics

Talk to Our counsellor

He placed a screen that had two slits cut into it in front of a monochromatic ( single color ) light.
The results of Young's Double Slit Experiment should be very different if light is a wave or a particle.
Let’s look at what the results would be in both situations, and then see how this experiment supports the wave model.

If light is a particle…

We set up our screen and shine a bunch of monochromatic light onto it.

Imagine it as being almost as though we are spraying paint from a spray can through the openings.
Since they are little particles they will make a pattern of two exact lines on the viewing screen ( Figure 1 ).

If light is a wave…

If light is a wave, everything starts the same way, but results we get are very different.

Remember, diffraction is when light passes through a small opening and starts to spread out. This will happen from both openings ( Figure 2 ).
Where crest meets crest , there will be constructive interference and the waves will make it to the viewing screen as a bright spot .
Where crest meets trough there will be destructive interference that cancel each other out… a black spot will appear on the screen.
When this experiment is performed we actually see this, as shown in Figure 3 .

We must conclude that light is made up of waves, since particles can not diffract.

Calculations

When you set up this sort of an apparatus, there is actually a way for you to calculate where the bright lines (called fringes ) will appear.

There is always a middle line, which is the brightest. We call it the central fringe .
The central fringe is n = 0 .
The fringe to either side of the central fringe has an order of n = 1 (the first order fringe ).
The order of the next fringe out on either side is n = 2 (the second order fringe ).
And so on, as shown in Figure 4 .

The formula that we will use to figure out problems involving double slit experiments is easy to mix up, so make sure you study it carefully.

λ = wavelength of light used (m) x = distance from central fringe (m) d = distance between the slits (m) n = the order of the fringe L = length from the screen with slits to the viewing screen (m)

It is very easy to mix up the measurements of x, d, and L.

Make sure to look at Figure 5 and see the different things each is measuring.
If you mix up x and d it's not so bad, since they are both on top in the formula. If you were to mix them up with L, you would get the wrong answer.
Almost all questions that you will see for this formula just involve sorting out what each variable is... you might find it helpful to write out a list of givens.

Example 1 : A pair of screens are placed 13.7m apart. A third order fringe is seen on the screen 2.50cm from the central fringe. If the slits were cut 0.0960 cm apart, determine the wavelength of this light. Roughly what colour is it?

Just to make sure you’ve got all the numbers from the question matched with the correct variables… L = 13.7 m n = 3 x = 2.50cm = 0.0250 m d = 0.0960cm = 9.60e-4 m It’s probably a yellow light being used given the wavelength we've measured.

If a white light is used in the double slit experiment, the different colours will be split up on the viewing screen according to their wavelengths.

The violet end of the spectrum (with the shortest wavelengths) is closer to the central fringe, with the other colours being further away in order.

There is also a version of the formula where you measure the angle between the central fringe and whatever fringe you are measuring.

The formula works the same way, with the only difference being that we measure the angle instead of x and L.
Make sure that your calculator is in degree mode before using this version of the formula.

Example 2 : If a yellow light with a wavelength of 540 nm shines on a double slit with the slits cut 0.0100 mm apart, determine what angle you should look away from the central fringe to see the second order fringe?

Do not forget to: Change the wavelength into metres. Change the slit separation into metres. "Second order" is a perfect number and has an infinite number of sig digs.

The Single Slit

A surprising experiment is that you can get the same effect from using a single slit instead of a double slit.

In Figure 7 , the blue path has to travel further than the red path... if this difference is equal to half a wavelength, they will meet each other out of sync .
If they meet crest to crest or trough to trough they will be in sync , but if they meet crest to trough they will be out of sync .
Being in sync will result in constructive interference , while meeting out of sync will result in destructive interference .
After the first couple of fringes (n = 1 and 2), the edges start getting really fuzzy, so you have a hard time measuring anything.
The only real difference in calculations is that "d" is now the width of the single opening.
If two slits work better than one, would more than two slits work better? This is a question that we will answer in the next section.

Example 3 : For a single slit experiment apparatus like the one described above, determine how far from the central fringe the first order violet (λ = 350nm) and red (λ = 700nm) colours will appear if the screen is 10 m away and the slit is 0.050 cm wide.

We need to solve the formula for “x”, the distance from the central fringe. For the violet light… For the red light…

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Chapter 12 – Wave Optics

12.3 Young’s Double Slit Experiment

Explain the phenomena of interference.
Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for colour, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1 ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Why do we not ordinarily observe wave behaviour for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single colour, single wavelength, λ ) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4 . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4 (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths ( 1/2λ, 3/2 λ, 5/2 λ … etc.) then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths (1 λ, 2λ, 3λ… , etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapour lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 5 below shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be dsinθ, where d is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where λ is the wavelength of the light, d is the distance between slits, and θ is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, m=4 is fourth-order interference.

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the centre. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed λ and m , the smaller d is, the larger θ must be, since dsinθ= mλ.

This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large θ , hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Example 1: Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10.95 o relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m=3 . We are given d = 0.0100 mm and θ = 10.95 o . The wavelength can thus be found using the equation d sinθ = m λ for constructive interference.

The equation is d sinθ = m λ. Solving for the wavelength λ gives

Substituting known values yields

λ = (d sinθ) / m = (0.0100 x 10 -3 m) (sin 10.95 o ) / 3 = 6.33 x 10 -7 m or

λ = 633 nm

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red colour is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2: Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation d sinθ = mλ for m = 0, 1, -1, 2, -2, 3, -3, 4, -4, … describes constructive interference. For fixed values of d and λ , the larger m is, the larger sinθ is. However, the maximum value that sinθ can have is 1, for an angle of 90 o . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m corresponds to this maximum diffraction angle.

Solving the equation d sinθ = mλ for m gives

Taking sinθ =sin90 o = 1 and substituting the values of d and λ, and remembering the values for the metric prefixes, from the preceding example gives

Therefore, the largest integer m can be is 15, or

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the centre. Consequently, not all 15 fringes may be observable.

Section Summary

Young’s double slit experiment gave definitive proof of the wave character of light.
An interference pattern is obtained by the superposition of light from two slits.
There is constructive interference when d sinθ = mλ for m = 0, 1, -1, 2, -2, 3, -3, 4, -4, ….(constructive) , where d is the distance between the slits, θ is the angle relative to the incident direction, and m is the order of the interference.
There is destructive interference when d sinθ = ( m + 1 / 2 )λ for m = 0, 1, -1, 2, -2, 3, -3, 4, -4, ….(constructive)

Conceptual Questions

1: Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

2: Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the colour of the light change? Explain.

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4: Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the centre. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Problems & Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm ?
Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30.0 degrees?
Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0 degrees.
Calculate the wavelength of light that has its third minimum at an angle of 30.0 o when falling on double slits separated by 3.00 μm. Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .
What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0 o ?
At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?
What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm ?
Find the largest wavelength of light falling on double slits separated by 1.20 μm for which there is a first-order maximum. Is this in the visible part of the spectrum?
What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light.
(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10.0o, at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?
Figure 8 shows a double slit located a distance x from a screen, with the distance from the centre of the screen given by y. When the distance d between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where sinθ≈tanθ, with θ in radians), the distance between fringes is given by Δy = x λ /d.

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

1: 0.516 o

3: 1.22 x10 -6 m

4: 0.290 μm

9: 1200 nm (not visible)

10 : 1.44 μm

11: (a) 760 nm (b) 1520 nm

13: For small angles sinθ ≈ tanθ≈θ when the angle is in radians

For two adjacent fringes we have d sinθ m = m λ

and d sinθ m +1 = (m+1) λ

Subtracting these equations gives

$\begin{array}{r @{{}={}}l} \boldsymbol{d (\textbf{sin} \; \theta _{\textbf{m} + 1} - \textbf{sin} \; \theta _{\textbf{m}})} & \boldsymbol{[(m + 1) - m] \lambda} \\[1em] \boldsymbol{d(\theta _{{\textbf{m}} + 1} - \theta _{\textbf{m}})} & \boldsymbol{\lambda} \end{array}$

14: 2.37 cm

Derivation of Physics Formula
Youngs Double Slit Experiment Derivation

Young’s Double Slits Experiment Derivation

Introduction to young’s double slits experiment.

During the year 1801, Thomas Young carried out an experiment where the wave and particle nature of light and matter were demonstrated. The schematic diagram of the experimental setup is shown below-

young double slits experiment derivation

A beam of monochromatic light is made incident on the first screen, which contains the slit S 0 . The emerging light then incident on the second screen which consists of two slits namely, S 1 , S 2 . These two slits serve as a source of coherent light. The emerging light waves from these slits interfere to produce an interference pattern on the screen. The interference pattern consists of consecutive bright and dark fringes. The dark fringes are the result of destructive interference and bright fringes are the result of constructive interference.

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young double slits experiment derivation-1

The light falls on the screen at the point P. which is at a distance y from the centre O.
The distance between the double-slit system and the screen is L
The two slits are separated by the distance d
Distance travelled by the light ray from slit 1 to point P on the screen is r 1
Distance travelled by the light ray from slit 2 to point P on the screen is r 2
Thus, the light ray from slit 2 travels an extra distance of ẟ = r 2 -r 1 than light ray from slit 1.
This extra distance is termed as Path difference.

Refer to Figure(3) Applying laws of cosines; we can write –

∵ Cos(90° – θ) = Sinθ

∵ Cos(90° + θ) = -Sinθ

Subtracting equation (2) from (1) we get-

Let us impose the limit that the distance between the double slit system and the screen is much greater than the distance between the slits. That is L >> d. The sum of r 1 and r 2 can be approximated to r 1 + r 2 ≅2r. Thus, the path difference becomes –

In this limit, the two rays r 1 and r 2 are essentially treated as parallel. (See Figure(4))

young double slits experiment derivation-3

Figure(4): Assuming L >> d, The path difference between two rays.

To compare the phase of two waves, the value of path difference (ẟ) plays a crucial role.

Constructive interference is seen when path difference (𝛿) is zero or integral multiple of the wavelength (λ).

(Constructive interference) 𝛿 = dSin𝜃 = mλ —-(5), m = 0, ±⁤1, ±2, ±3, ±4, ±5, ……

Where m is order number. The Zeroth order maximum (m=0)corresponds to the central bright fringe, here 𝜃=0. The first order maxima(m=±⁤1)(bright fringe) are on either side the central fringe.

Similarly, when 𝛿 is an odd integral multiple of λ/2, the resultant fringes will be 180 0 out of phase, thus, forming a dark fringe.

(Destructive interference) 𝛿 = dSin𝜃 = λ—(6), m = 0, ±⁤1, ±2, ±3, ±4, ±5, ……

Assuming the distance between the slits are much greater than the wavelength of the incident light, we get-

Substituting it in the constructive and destructive interference condition we can get the position of bright and dark fringes, respectively. The equation is as follows-

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Chapter 27 Wave Optics

27.3 Young’s Double Slit Experiment

Explain the phenomena of interference.
Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1 ).

Take-Home Experiment: Using Fingers as Slits

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

$\boldsymbol{d \;\textbf{sin} \;\theta = m +}$

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

$\boldsymbol{d \;\textbf{sin} \;\theta = m \theta , \;\textbf{for} \; m = 0, \;1, \; -1, \; 2, \; -2, \; \dots}.$

Example 1: Finding a Wavelength from an Interference Pattern

$\boldsymbol{10.95 ^{\circ}}$

Substituting known values yields

$$\begin{array}{r @{{}={}}l} \boldsymbol{\lambda} & \boldsymbol{\frac{(0.0100 \;\textbf{mm})(\textbf{sin} 10.95^{\circ})}{3}} \\[1em] & \boldsymbol{6.33 \times 10^{-4} \;\textbf{mm} = 633 \;\textbf{nm}}. \end{array}$

Example 2: Calculating Highest Order Possible

Strategy and Concept

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \; (\textbf{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}$

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

Young’s double slit experiment gave definitive proof of the wave character of light.
An interference pattern is obtained by the superposition of light from two slits.

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}$

Conceptual Questions

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4: Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

Problems & Exercises

2: Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

$\boldsymbol{30.0 ^{\circ}}$

7: At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?

$\boldsymbol{25.0 \;\mu \textbf{m}}$

10: What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

11: (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

$\boldsymbol{10.0^{\circ}}$

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

$\boldsymbol{0.516 ^{\circ}}$

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

$\boldsymbol{\textbf{sin} \;\theta - \;\textbf{tan} \;\theta \approx \theta}$

For two adjacent fringes we have,

$\boldsymbol{d \;\textbf{sin} \;\theta _{\textbf{m}} = m \lambda}$

Subtracting these equations gives

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Higher order minima

Intensity I ( θ )

Varying the slit width

Young's experiment with finite slit width.

Further reading.

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The Single Slit

12.3 Young’s Double Slit Experiment

Take-Home Experiment: Using Fingers as Slits

Example 1: Finding a Wavelength from an Interference Pattern

Example 2: Calculating Highest Order Possible

Section Summary

Conceptual Questions

Young’s Double Slits Experiment Derivation

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27.3 Young’s Double Slit Experiment

Take-Home Experiment: Using Fingers as Slits

Example 1: Finding a Wavelength from an Interference Pattern

Example 2: Calculating Highest Order Possible

Section Summary

Conceptual Questions

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Diffraction of Light - Young’s Single Slit Experiment

Single Slit Diffraction of Light

What is Diffraction of Light?

Diffraction Due to Single Slit

Young’s Single Slit Experiment

Explanation of The Phenomenon and Diffraction Formula

Diffraction Maxima and Minima

Fringe width

Conditions for Diffraction

Types of Diffraction

Solved Examples

About Fraunhofer and Fresnel

Fraunhofer, The Scientist

Fresnel, The Scientist

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FAQs on Diffraction of Light - Young’s Single Slit Experiment

Single-Slit Diffraction

Huygens’ Principle

Single Slit Diffraction Pattern

Single Slit Diffraction Equation

Difference Between Single- and Double-Slits Diffraction

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Single Slit Diffraction

What is Single Slit Diffraction?

Position of Secondary Maxima

Single Slit Diffraction vs Double Slit Diffraction

Single Slit Diffraction Solved Examples

Practice Problems on Single Slit Diffraction

Single Slit Diffraction – FAQs

2. What causes Single Slit Diffraction?

2. How does the Diffraction Pattern change with different Slit Widths?

4. Can Single Slit Diffraction occur with other forms of Waves?

5. What is the Difference between Fraunhofer and Fresnel Diffraction?

6. How does changing the Wavelength affect the Diffraction Pattern?

7. What is Diffraction from One Slit?

8. What are the two main Types of Diffraction?

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Single Slit Diffraction: Young’s Single Slit Experiment and Fringe Width

What is Diffraction?

What is Single Slit Diffraction?

Young's Single Slit Experiment

Single Slit Diffraction Formula

Fringe Width

Central Maximum

Types of Diffraction

Previous Year Questions

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

2. A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

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First order minima

Higher order minima

Intensity I ( θ )

Varying the slit width

Young's experiment with finite slit width.

Further reading.

Compare Young’s Double Slit Interference Pattern and Single Slit Diffraction Pattern. - Physics

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Optical Path

Fringe Shift

Intensity Distribution

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If light is a particle…

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Calculations

The Single Slit