Question

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width 'a'. If the distance between the slit and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is 1.5 mm, calculate the width of the slit.

Answer 1

Given $\lambda =600nm=600\times {10}^{-9}m=6.0\times {10}^{-7}m,D=0.8m$,

$y_2=1.5mm=1.5\times {10}^{-3}m,n=2,a=?$

Position of $n^{th}$ maximum in diffraction of a single slit

$y_n=(n+\frac{1}{2})\frac{\lambda D}{a}\Rightarrow a=(n+\frac{1}{2})\frac{\lambda D}{y_n}$

Substituting given values $a=(2+\frac{1}{2})\frac{6.0\times {10}^{-7}\times 0.8}{1.5\times {10}^{-3}}$

$a=\frac{5}{2}\times 4.0\times 0.8\times {10}^{-4}m=0.8\times {10}^{-3}m=0.8mm$