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# Diffraction - single slit diffraction, intensity at general point, practice problems, FAQs

Have you seen the following phenomena around you i.e. a bright ring around the light source like sun and moon, a rainbow pattern on the CD/DVD disc, the speckle pattern when laser light falls on an optical rough surface. You can hear your friend calling even when hiding behind a tree etc? These all phenomena occur due to diffraction. In diffraction the wave is spread out as a result of passing via a narrow area or across an edge. Let's understand Diffraction in detail!

Table of content

• Introduction to diffraction
• Types of diffraction
• Single slit diffraction
• Intensity at any general point
• Practice problems
• FAQs

## Introduction to diffraction

When a wave is obstructed by an obstacle having a sharp edge, the rays bend around the corner. This phenomenon is known as diffraction.

Now the question is how bending occurs? To explain the bending of light, we have to consider light as a wave and using Huygens principle we can explain the bending of light as follows:

When there is a plane wavefront incident on a narrow slit having sharp edges, each point on the plain wavefront acts as a secondary source of light waves. The secondary sources emit spherical wavelets as shown in the figure and the envelope of the wavelets denote the secondary wavefront. This explains the reason behind the bending of light when light strikes on a narrow slit having sharp edges.

## Types of diffraction

The diffraction is mainly of two types:

1. Fraunhofer Diffraction : In this diffraction all the rays are parallel when passing through a narrow slit. It is achieved by placing the light source to an infinite distance from the screen. A convex lens is used to differentiate the light rays.
2. Fresnel Diffraction : In this diffraction the source of light used is placed at a finite distance from the screen where the diffraction pattern is to be obtained. The shape of the incident wavefront is spherical. The convex lens is not required in this diffraction.

## Single slit diffraction

Consider that parallel light rays are coming to the narrow slit of width b. Here, we have used a converging lens as the diffraction element.

When light rays fall on the single slit, each point on the slit acts as the source of secondary wavelets. Therefore, interference will take place between these wavelets, and we will see the interference pattern on the screen.

We want to find the intensity at any general point P on the screen. Point P makes an angle θ with the central line passing through point B.

From geometry, we can say that the path difference between two rays (shown by dotted line) coming from point A and B is

Since the source and the screen are very far away in comparison with the width of the slit, each light rays from the slit will make an angle θ with the horizontal straight line extending from the point from where light is emitting.

Therefore, any two points on the slit which are b2 distance away from each other will form a pair and they will constitute 1st dark fringe on the screen if the following condition holds:

The general condition for dark fringe on the screen in case of nth fringe is given by:

b sin⁡ θ=nλ

Where n is the order of dark fringe

If n=0 ⇒ b sin⁡ θ=0 ⇒ θ=0 ⇒ Path difference between the rays is zero ⇒ The constructive interference will occur at Po . Thus, Po is the central maxima.

Using the similar concept we can find the position of secondary maximum as,

Where n is the order of bright fringe.

## Intensity at any general point

Intensity at any general point is given by

Where Io= Intensity at central point Po

= Phase difference

If sin =0 Central maxima

If sin =nb minima

If b is large, b0 ; the fringe pattern is collapsed to the central maxima and hence, only a single bright spot is visible on the screen. Thus, when b is large, rectilinear propagation of light is valid.

## Practice problems

Q. Beam of light from a source far away falls on a single slit which has a width of 1 mm and a resulting diffraction pattern is observed on a screen which is 2 m away. If the wavelength of the beam is 600 nm, then what is the distance on both sides between the first dark fringes?

1. 1.2 cm
2. 1.2 mm
3. 2.4 cm
4. 2.4 mm

A.

We know that the first maxima is formed at λ/b distance away on both sides of the central maxima. Now, the distance between the first dark fringes on either side of the central bright fringe is nothing but the linear width of the central maxima which is denoted by ‘y’ in the figure below.

Since b≪D, θ is small. Thus,

sin

and $2\theta =\frac{2\lambda }{b}$

Let 2=

$\alpha =\frac{2\lambda }{b}$

$\frac{y}{D}=\frac{2\lambda }{b}$

$y=\frac{2\lambda }{b}D$

$y=\frac{2×600×{10}^{-9}}{{10}^{-3}}×2$

y=2.4 mm

Q. A parallel beam of monochromatic light of 4500 Å is allowed to be incident on a long narrow slit of width 0.2 mm. What is the angular divergence in which maximum light is diffracted on a screen far away?

A.

Since this is Fraunhofer diffraction, most of the light is diffracted in between the location of the 1st order dark fringe on either side of the central maxima.

Therefore, the angular divergence in which most of the light is diffracted is nothing but the angular width of the central maxima.

Since b≪D, θ is small. Thus,

$\alpha =\frac{2\lambda }{b}$

$\alpha =\frac{2×4500×{10}^{-10}}{0.2×{10}^{-3}}$

Q. Consider the Fraunhofer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first minima, what is the minimum phase difference (in radians) between the wavelets from the opposite edges of the slit?

1. 4
2. 2
3. 2
4.

A.

The path difference between ray from A and B at point P is:

For minima to occur at P, the condition to be satisfied is:

Therefore, phase difference between rays from A and B at point P is:

${\delta }_{A,B}=\pi$

Similarly, phase difference between ray from B and C at point P is:

${\delta }_{B,C}=\pi$

The phase difference (in radians) between the wavelets from the opposite edges of the slit is:

Q. Find the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm when light of wavelength 5890 Å is an incident.

A.

For bright fringe

First maximum n=1

When is small, sin

$\theta =\frac{3\lambda }{2b}$

$\theta =\frac{3\lambda }{2b}=\frac{3×5890×{10}^{-10}}{2×0.25×{10}^{-3}}$

## FAQs

Q. How do you increase diffraction?
A.
The amount of diffraction increases as wavelength increases and decreases as wavelength decreases.

Q. Where do we see diffraction in nature?
A.
Rings of light that are observed around the sun and other celestial bodies are the result of diffraction. This is caused by light wave diffraction by small particles present in the atmosphere.

Q. How many types of diffraction are there?
A.
Diffraction is of two types- 1. Fresnel diffraction , 2. Fraunhofer diffraction.

Q. What is fringe width ?
A.
Distance between any two consecutive bright or dark fringe is called fringe width.

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