Question

A single slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if (a) the wavelength is doubled, (b)the slit width is doubled, or (c) the distance from the slit to the screen is doubled?

Answer 1

The formula for the single slit diffraction (in the small angle approximation) is:

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the screen (so, for n=1 it corresponds to half the width of the central maximum)

n is the order of the minimum

D is the distance between the slit and the screen

d is the width of the slit

$\lambda$ is the wavelength

So, the width of the central bright spot is equal to 2y when n=1:

$w=2y = \frac{2\lambda D}{d}$

using this formula, let's analyze the different situations:

(a) The width doubles

The initial width of the central bright spot is

$w= \frac{2\lambda D}{d}$

Here the wavelength is doubled, so

$\lambda' = 2\lambda$

Therefore, the new width of the central spot will be

$w'=\frac{2 \lambda' D}{d}=\frac{2(2\lambda) D}{d}=2(\frac{2\lambda D}{d})=2w$

so, the width has doubled.

(b) The width halves

The initial width of the central bright spot is

$w= \frac{2\lambda D}{d}$

Here the slit width is doubled, so

$d'=2d$

Therefore, the new width of the central spot will be

$w'=\frac{2 \lambda D}{d'}=\frac{2\lambda D}{2d}=\frac{1}{2}(\frac{2\lambda D}{d})=\frac{w}{2}$

so, the width has halved.

(c) The width doubles

The initial width of the central bright spot is

$w= \frac{2\lambda D}{d}$

Here the distance from the slit to the screen is doubled, so

$D' = 2 D$

Therefore, the new width of the central spot will be

$w'=\frac{2 \lambda D'}{d}=\frac{2\lambda (2D)}{d}=2(\frac{2\lambda D}{d})=2w$

so, the width has doubled.