Question 69 of 191

In a Young’s double slit experiment, two slits are illuminated with a light of wavelength $800nm$. The line joining ${A}_{1}P$ is perpendicular to ${A}_{1}{A}_{2}$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation $a$ will be :

The distance of screen from slits $D=5cm$.

A$0.4mm$B$0.5mm$C$0.2mm$D$0.1mm$

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Answer: (C) $0.2mm$

The position of first minima is
$y=\dfrac{\lambda D}{2\times a}$($y$ is height of the point from the axis passing through the mid-point of ${A}_{1}\text{and}{A}_{2}$)
Since the first minima is in front of the slit as shown in the figure,
$\dfrac{\lambda D}{2\times a}=\dfrac{a}{2}$
$\Rightarrow a=\sqrt{\lambda D}=\sqrt{800\times {10}^{-9}\times 5\times {10}^{-2}}=2\times {10}^{-4}m=0.2mm$