Question

In a single slit diffraction pattern, the distance between the first maximum on the left and the first maximum on the right is $5mm$. The screen on which the diffraction pattern is displaced is at a distance of $80cm$ from the slit. The wavelength is $6000A$. The slit width in $mm$ is about

• A. $0.576$
• B. $0.348$
• C. $0.192$
• D. $0.096$

Answer 1

Answer: (C)

$0.192$

Given data :

The order of the fringe is m = 1,

Distance of first maximum (or minimum) from centeral fringe is y = 2.5 mm,

Distance between screen and slit is D = 80 cm = 800 mm,

The wavelength of incident wave is $\lambda ={6000}^{\circ }A$

In a single slit diffraction, the slit width is given as

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

$d=\frac{\left(1\right)\left(800mm\right)\left({6000}^{\circ }A\right)}{2.5mm}$

$d=(19.2\times {10}^5){10}^{-7}mm$

$d=0.192mm$

Hence, option C is correct.