Question

A plane wave front of wave length $6000A$ is incident upon a slit of $0.2mm$ width, which enables fraunhofer's diffraction pattern to be obtained on a screen $2m$ away. Width of the central maxima in mm will be

• A. $10$
• B. $12$
• C. $8$
• D. $2$

Answer 1

Answer: (B)

$12$

Given,

$\lambda =6000\times {10}^{-10}m$

$d=0.2mm$

$D=2m$

Fringe width, $\beta =\frac{\lambda D}{d}$

$\beta =\frac{6000\times {10}^{-10}\times 2}{0.2\times {10}^{-3}}=6mm$

Width of the central maxima,

$W=2\beta =2\times 6=12mm$

The correct option is B.