Question

In the Young's double slit experiment the distance between two slits $S_1$ and $S_2$ is $d$. Interference pattern is obtained by these slits on a screen distant $D$ from the slits. A dark fringe is produced at point $P$ just in front of $S_1$, The wavelength of light used is :

• A. $\lambda =\frac{D}{d^2}$
• B. $\lambda =\frac{d^2}{D}$
• C. $\lambda =\frac{D}{d}$
• D. $\lambda =\frac{d}{D}$

Answer 1

The correct option is B $\lambda =\frac{d^2}{D}$

Path difference is $\sqrt{D^2+d^2}-D=\frac{\lambda }{2}$ (as dark fringe is formed )
so ,
$D(1+\frac{d}{D}{)}^{\frac{1}{2}}-D=\frac{\lambda }{2}$
$D+\frac{d^2}{2D}-D=\frac{\lambda }{2}$
$\lambda =\frac{d^2}{D}$
option $B$ is correct