Question

A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0m behind the slit?

Answer 1

Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

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

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

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

where we have

$\lambda=500 nm = 5\cdot 10^{-7}m$ is the wavelength

D = 2.0 m is the distance of the screen

$d=0.50 mm=5\cdot 10^{-4}m$ is the width of the slit

Substituting, we find

$w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m$