Question

In Young's double-slit experiment, the separation between two coherent sources $S_1$ and $S_2$ is $d$ and the distance between the source and screen is $D$. In the interference pattern, it is found that exactly in front of one slit, there occurs a minimum. Then the possible wavelengths used in the experiment are

• A. $\lambda =\frac{d^2}{D},\frac{d^2}{3D},\frac{d^2}{5D}$
• B. $\lambda =\frac{d^2}{D},\frac{d^2}{5D},\frac{d^2}{9D}$
• C. $\lambda =\frac{d^2}{D},\frac{d^2}{2D},\frac{d^2}{3D}$
• D. $\lambda =\frac{d^2}{3D},\frac{d^2}{7D},\frac{d^2}{11D}$

Answer 1

The correct option is A $\lambda =\frac{d^2}{D},\frac{d^2}{3D},\frac{d^2}{5D}$

Destructive interference has occurred in front of one slit, Its distance from centre is $x=\frac{d}{2}$

For dark fringes, $x=\frac{(2n-1)\lambda D}{2d}\Rightarrow \lambda =\frac{d^2}{(2n-1)D}$

For first dark fringes, $n=1,\lambda =\frac{d^2}{D}$

For second dark fringes, $n=2,\lambda =\frac{d^2}{3D}$

For third dark fringes, $n=3,\lambda =\frac{d^2}{5D}$