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 ‘$d$’ and the distance between source and screen is ‘$D$’. 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

Answer: (C)

As the second minima is formed just in front of one of the slit i.e. $y=\frac{d}{2}$
For second minima $(n=1),y=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light used.
Or $\frac{d}{2}=\left(1+\frac{1}{2}\right)\frac{\lambda D}{d}$
Or $\frac{d}{2}=\frac{3\lambda D}{2d}$
$\Rightarrow \lambda =\frac{d^2}{3D}$