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

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 $\frac{d}{2}=\frac{3 \lambda D}{2 d}$
$\Rightarrow \lambda=\frac{d^{2}}{3 D}$