Diffraction of Light

Wave Optics · Maharashtra HSC

Theory

What diffraction is

Definition: the bending of light near the edge of an obstacle or slit, and its spreading into the region of the geometrical shadow, is called diffraction of light.

Geometrical optics rests on rectilinear propagation and predicts a sharp shadow. In fact light is found in regions where a shadow is expected.
Strictly, light does not bend around the edge. Points on the wavefront at the edge of the obstacle act as secondary sources, and their wavelets reach the shadow region.
Diffraction is intimately related to interference: it is essentially the interference of MANY waves coming from different parts of the SAME wavefront, rather than of two waves from two different wavefronts.
Diffraction is noticeable only when the size of the obstacle or slit is of the order of the wavelength of the light. This is why the diffraction of sound, whose wavelength is large, is far easier to observe — we hear a person talking from behind a wall.

7.9.1 Fresnel and Fraunhofer diffraction

Fraunhofer diffraction	Fresnel diffraction
The distances between the source, the diffracting obstacle and the screen are very large	The distances are much smaller
The incident wavefront is a PLANE wavefront	The incident wavefront is cylindrical or spherical, depending on the source
Lenses are needed: one to make the wavefront plane, one to focus the pattern on the screen	No lens is required

7.9.2 Experimental set-up for Fraunhofer diffraction

A monochromatic source $S$ is placed at the focus of a converging lens, so that (ignoring aberrations) the emerging beam consists of plane parallel rays — plane wavefronts. These fall on the diffracting element: a slit, a circular aperture, a double slit, a grating. The emerging beam falls on a second converging lens which focuses it on a screen.

For a circular aperture, $S$ is a point source and the lenses are bi-convex.
For linear elements such as slits and gratings, the source is linear and the lenses are cylindrical, so that the focused image is also linear.
Either way a plane wavefront approaches and leaves the diffracting element, as Fraunhofer diffraction requires.

7.9.3 Fraunhofer diffraction at a single slit

The diagram to draw

A plane wavefront $YY'$ falls on a single slit $AB$ of width $a$ with its centre at $O$ ; the width lies in the plane of the paper and the length is perpendicular to it. A cylindrical lens next to the slit brings the emergent parallel beams to a focus on a screen at distance $D$ , effectively the focal length of the lens. Typically $a$ is $10^{-4}$ to $10^{-3}$ m and $D$ is of the order of 10 m, so $D\gg a$ . Draw the axial beam converging at the central point $P_{0}$ directly in front of $O$ , and a second set of parallel rays leaving at angle $\theta$ and converging at $P$ , with $AC$ drawn normal to the emergent direction at $A$ .

Derivation of the position of the minima

The slit is imagined to be divided into a large number of extremely thin slit elements. As the plane wavefront reaches the slit, each element becomes a secondary source of cylindrical wavelets, sending light in all directions.
Rays parallel to the axis from every point of $AB$ are focused at $P_{0}$ . All these rays have equal optical paths, so they arrive at $P_{0}$ in the same phase and interfere constructively — $P_{0}$ is the central maximum.
For a point $P$ at angular position $\theta$ : the optical paths from $AC$ onwards are equal, since $AC$ is drawn normal to the emergent rays. So the net path difference is produced only between $AB$ and $AC$ . The path difference between the extreme rays is

$BC=a\sin\theta$

Let $BC=\lambda$ . Take $K$ to be the mid-point of $AC$ ; then

$OK=\dfrac{1}{2}BC=\dfrac{\lambda}{2}$

so the ray from $A$ and the ray from $O$ arrive at $P$ with path difference $\dfrac{\lambda}{2}$ and interfere destructively.

Now take any pair of points such as $G$ and $H$ , equidistant from $A$ and from $O$ respectively, separated by $\dfrac{a}{2}$ along the slit. Every such pair has a path difference of $\dfrac{\lambda}{2}$ at $P$ . Every point of $AO$ is thus cancelled by a point of $OB$ , and $P$ is the FIRST MINIMUM.
Extending the argument to points on the screen for which the extreme rays differ by $2\lambda$ , $3\lambda$ , $\dots$ , $n\lambda$ shows that all these points are dark. The same holds on the other side of $P_{0}$ .

Hence at the $n^{\text{th}}$ minimum,

$\boxed{\ a\sin\theta=n\lambda\ },\qquad n=1,2,3,\dots$

Assuming the maxima to lie between consecutive minima, the path difference between the extreme rays at the $n^{\text{th}}$ secondary maximum is

$\boxed{\ a\sin\theta=\left(n+\dfrac{1}{2}\right)\lambda\ }$

(In reality the secondary maxima are nearly, but not exactly, midway between the dark fringes.)

Why the first minimum needs $\lambda$ and not $\dfrac{\lambda}{2}$

For two waves, a dark point requires a path difference of $\dfrac{\lambda}{2}$ . In a single slit a point on the screen receives waves from every point of the slit. For that point to be dark, EVERY pair of waves must cancel — one wave from the upper half and its partner from the lower half, as with $A$ and $O$ , $G$ and $H$ , $O$ and $B$ . That requires $\dfrac{\lambda}{2}$ between each pair separated by $\dfrac{a}{2}$ , and hence a minimum path difference of $\dfrac{\lambda}{2}+\dfrac{\lambda}{2}=\lambda$ between the extreme points $A$ and $B$ .

Distances on the screen

Since $D\gg a$ , the angle $\theta$ is very small, so in radian $\sin\theta\approx\tan\theta\approx\theta=\dfrac{y}{D}$ .

At the $n^{\text{th}}$ dark point, $\theta_{nd}=\dfrac{y_{nd}}{D}=\dfrac{n\lambda}{a}$ , giving

$y_{nd}=\dfrac{n\lambda D}{a}=nW$

At the $n^{\text{th}}$ bright point, $\theta_{nb}=\dfrac{y_{nb}}{D}\approx\left(n+\dfrac{1}{2}\right)\dfrac{\lambda}{a}$ , giving

$y_{nb}\approx\left(n+\dfrac{1}{2}\right)\dfrac{\lambda D}{a}=\left(n+\dfrac{1}{2}\right)W$

where

$W=\dfrac{\lambda D}{a}$

is analogous to the fringe width in interference: the distance between consecutive bright fringes or consecutive dark fringes, EXCEPT for the central (zeroth) bright fringe.

Width of the central bright fringe

The central bright fringe extends between the first dark fringes on either side, so

$W_{c}=2y_{1d}=\dfrac{2\lambda D}{a}=2W$

The central maximum is twice as wide as every other bright fringe, and much the brightest.

7.9.4 Comparison of the two patterns

In a laboratory Young's set-up the slits are much thinner than their separation (a few mm), usually obtained with a biprism or Lloyd's mirror, and one sees about 30 to 40 equally spaced fringes of nearly the same brightness. A single slit used for diffraction is usually less than 1 mm wide, and one sees about 20 to 30 fringes on either side, with the central fringe the brightest and twice as wide as the rest.

Note: the central bright fringe is the ZEROTH bright fringe ( $n=0$ ), not the first. Here $d$ is the slit separation, $a$ the slit width, $D$ the slit-to-screen distance and $W$ the separation of consecutive bright or dark fringes.

Physical quantity	Young's double slit interference	Single slit diffraction
Fringe width	$W=\dfrac{\lambda D}{d}$	$W=\dfrac{\lambda D}{a}$ , except for the central bright fringe
$n^{\text{th}}$ bright: phase difference between extreme rays	$2n\pi$	$\left(n+\dfrac{1}{2}\right)2\pi$
$n^{\text{th}}$ bright: angular position $\theta$	$\dfrac{n\lambda}{d}$	$\left(n+\dfrac{1}{2}\right)\dfrac{\lambda}{a}$
$n^{\text{th}}$ bright: path difference $\Delta x$	$n\lambda$	$\left(n+\dfrac{1}{2}\right)\lambda$
$n^{\text{th}}$ bright: distance $y$ from centre	$\dfrac{n\lambda D}{d}=nW$	$\left(n+\dfrac{1}{2}\right)\dfrac{\lambda D}{a}=\left(n+\dfrac{1}{2}\right)W$
$n^{\text{th}}$ dark: phase difference between extreme rays	$\left(n-\dfrac{1}{2}\right)2\pi$	$2n\pi$
$n^{\text{th}}$ dark: angular position $\theta$	$\left(n-\dfrac{1}{2}\right)\dfrac{\lambda}{d}$	$\dfrac{n\lambda}{a}$
$n^{\text{th}}$ dark: path difference $\Delta x$	$\left(n-\dfrac{1}{2}\right)\lambda$	$n\lambda$
$n^{\text{th}}$ dark: distance $y$ from centre	$\left(n-\dfrac{1}{2}\right)\dfrac{\lambda D}{d}=\left(n-\dfrac{1}{2}\right)W$	$\dfrac{n\lambda D}{a}=nW$

Distinguishing interference from diffraction

Interference	Diffraction
Due to superposition of waves from two different wavefronts	Due to many waves coming from different parts of the same wavefront
All bright fringes are of equal intensity	Intensity decreases with the order of the bright band
Minimum intensity may be zero	Minimum intensity is not zero
The central maximum has the same width as the other bright fringes	The central maximum is broader than the others — twice the fringe width
Waves from the two coherent sources travel in straight lines	Light waves are bent at the corners and displaced from the straight line path
The resolving power of an optical instrument does not depend on interference	The resolving power of an optical instrument depends on diffraction

Double slit diffraction

With two slits rather than one, the pattern is decided both by the diffraction pattern of the individual slits and by the interference between them. One sees narrow interference fringes like those of Young's experiment, but of varying brightness, with the single slit diffraction pattern as the shape of their envelope.

Standard results

Minima: $a\sin\theta=n\lambda$ ; secondary maxima: $a\sin\theta=\left(n+\dfrac{1}{2}\right)\lambda$ .
$y_{nd}=\dfrac{n\lambda D}{a}$ ; $y_{nb}=\left(n+\dfrac{1}{2}\right)\dfrac{\lambda D}{a}$ ; $W=\dfrac{\lambda D}{a}$ .
Width of the central maximum $=\dfrac{2\lambda D}{a}=2W$ .
Increasing the slit width (or the diameter of the obstacle) DECREASES the fringe width in the diffraction pattern.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
What is 'diffraction of light'? Explain its two types.
Maharashtra HSC board question — asked in March 2016Answer 1 ↓
2
Distinguish between the phenomenon of interference and diffraction of light.
Maharashtra HSC board question — asked in March 2017Answer 2 ↓
3
What happens to the fringe width in diffraction pattern if the diameter of wire is increased?
Maharashtra HSC board question — asked in July 2022Answer 3 ↓
4
What are Fraunhofer diffraction and Fresnel diffraction?
Maharashtra HSC board question — asked in July 2023Answer 4 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

5
In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is 4 mm. The screen on which the pattern is displaced, is 2 m from the slit and wavelength of light used is 6000 Å. Calculate width of the slit and width of the central maximum.
Maharashtra HSC board question — asked in October 2015Answer 5 ↓
6
Monochromatic light of wavelength 4300 Å falls on a slit of width 'a'. For what value of 'a' the first maximum falls at 30°?
Maharashtra HSC board question — asked in March 2019Answer 6 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The bending of light near the edge of an obstacle or slit and spreading into the region of geometrical shadow is called diffraction of light. Two types of diffraction are: i. Fraunhofer diffraction: a. If the distances between the primary source of light, the obstacle/slit causing diffraction and the screen for viewing the diffraction pattern are very large, the diffraction is called Fraunhofer diffraction. b. In this case, the wavefront incident on the obstacle can be considered to be a plane wavefront. ii. Fresnel diffraction: In this case, the distances are much smaller and the incident wavefront is either cylindrical or spherical depending on the source.

2.↑ back to the question

Difference between interference and diffraction:

No.	Interference	Diffraction
i.	Interference is due to superposition of waves from different wavefronts.	Diffraction is due to many waves coming from different parts of the same wavefront.
ii.	All bright fringes are of equal intensity.	Intensity decreases with the order of bright band.
iii.	Minimum intensity may be zero.	Minimum intensity is not zero.
iv.	Width of the central maximum is same as that of other bright fringes i.e fringe width is same for all fringes including central maxima.	Width of central maximum is broader than other maxima and it is double the fringe width.
v.	The waves emitted by two coherent sources travel in straight line.	The light waves are bent at the corners and displaced from their straight line path.
vi.	The resolving power of an optical instrument does not depend on the phenomenon of interference.	The resolving power of an optical instrument depends on the phenomenon of diffraction.

3.↑ back to the question
If the diameter of wire is increased, the fringe width in diffraction pattern decreases.
4.↑ back to the question
Refer Subtopic: 7.9 Q. No. 1 (Types of diffraction only)
5.↑ back to the question
Given: λ = 6000 Å = $6\times10^{-7}$ m, D = 2 m, $2y_{1d}=4$ mm = $4\times10^{-3}$ m To find: Width of slit (a), Width of central maximum (W) Formulae: i. $2y_{1d}=\dfrac{2\lambda D}{a}$ ii. $W=2y_{1d}$ Calculation: From formula (i), $4\times10^{-3}=\dfrac{2\times6\times10^{-7}\times2}{a}$ $\therefore a=6\times10^{-4}$ m From formula (ii), Width of central maximum = $4\times10^{-3}$ m Ans: i. The width of the slit is $6\times10^{-4}$ m. ii. The width of the central maximum is $4\times10^{-3}$ m.
6.↑ back to the question
Given: λ = 4300 Å = $4300\times10^{-10}$ m, $\theta_1=30°$ , n = 1 To find: Slit width (a) Formula: For secondary maxima, $\sin\theta_n=(2n+1)\dfrac{\lambda}{2a}$ Calculation: From formula, for first secondary maximum, $a=(2n+1)\dfrac{\lambda}{2\sin\theta_n}=(2\times1+1)\dfrac{4300\times10^{-10}}{2\sin(30°)}=\dfrac{3\times4300\times10^{-10}}{2\times\dfrac{1}{2}}=1.29\times10^{-6}$ m = 1.29 μm Ans: For slit of width 1.29 μm, the first maximum falls at 30°.