Question

A long narrow horizontal slit lies $1mm$ above a plane mirror. The interference pattern produced by the slit and its image is viewed on a screen at a distance $1m$ from the slit. The wavelength of light is $700nm$. Then the distance of the first maxima above the mirror is equal to (d<<D);

• A. $0.30mm$
• B. $0.35mm$
• C. $60mm$
• D. $7.5mm$

Answer 1

The correct option is B $0.35mm$

Given that,

$D=1m$

$\lambda =700nm=700\times {10}^{-9}m$

Since,

$a=2mm$, $d=2a=2mm=2\times {10}^{-3}m$

Fringe width $=\frac{\lambda D}{d}=\frac{700\times {10}^{-9}m\times 1m}{2\times {10}^{-3}m}=0.35mm$