Question

Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit. The distance on the screen between the second order minimum and the central maximum is 1.05 cm . What is the width of the slit in micrometers (μm)?

Answer 1

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

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

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$, with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$