Physics - Wave Optics Question with Solution | TestHub

Given:

Slit width, $a=0.2\text{ mm}=2\times 10^{-4}\text{ m}$

Distance of screen, $D=2\text{ m}$

Distance between the first minimum on either side of the central maximum is the width of the central maximum, $W=1\text{ cm}=10^{-2}\text{ m}$

The width of the central maximum is given by:

$W=\frac{2\lambda D}{a}$

Substituting the given values:

$10^{-2}=\frac{2\times \lambda \times 2}{2\times 10^{-4}}$

$10^{-2}=\frac{2\lambda }{10^{-4}}$

$2\lambda =10^{-6}$

$\lambda =0.5\times 10^{-6}\text{ m}$

$\lambda =5000\times 10^{-10}\text{ m}=5000\text{ A˚}$

Answer: $5000\mathring{A}$