Question

In Young's double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ${}^{\prime }{d}^{\prime }$ and the distance between source and screen is ${}^{\prime }{D}^{\prime }$. The wavelength of light source used is :

• A. $\frac{d^2}{D}$
• B. $\frac{d^2}{2D}$
• C. $\frac{d^2}{3D}$
• D. $\frac{d^2}{4D}$

Answer 1

The correct option is C $\frac{d^2}{3D}$

As the second minima is formed just in front of one of the slit i.e. $y=\frac{d}{2}$

For second minima $(n=1)$, $y=(n+\frac{1}{2})\frac{\lambda D}{d}$

where $\lambda$ is the wavelength of light used.

Or $\frac{d}{2}=(1+\frac{1}{2})\frac{\lambda D}{d}$

Or $\frac{d}{2}=\frac{3\lambda D}{2d}$

$⟹$ $\lambda =\frac{d^2}{3D}$