Question

Consider a usual set-up Young's double slit experiment with slits of equal intensity as shown in the figure. Take $O$ as origin and the $y$-axis as indicated. If average intensity of light at all points between $y_1=-\frac{\lambda D}{4d}$ and $y_2=+\frac{\lambda D}{4d}$ equals $n$ times the intensity of central maxima, then $n$ equals (take average over phase difference)

• A. $\frac{1}{2}(1+\frac{2}{\pi })$
• B. $2(1+\frac{2}{\pi })$
• C. $(1+\frac{2}{\pi })$
• D. $\frac{1}{2}(1-\frac{2}{\pi })$

Answer 1

The correct option is A $\frac{1}{2}(1+\frac{2}{\pi })$

Phase difference corresponding to $y_1$ is $-\pi /2$ and that for $y_2$ is $+\pi /2$
Intensity at a point on screes when phase difference is $\varphi$ is $I_P=I_{max}{\cos }^2\left(\varphi /2\right)$
$\therefore$ Average intensity between $y_1$ and $y_2$
$I_{Avg}=\frac{1}{\pi }\underset{-\pi /2}{\overset{\pi /2}{\int }}{I}_{max}{\cos }^2\left(\varphi /2\right)d\varphi$
$I_{Avg}=I_{max}(\pi +2)/2\pi$
$\therefore$ The required ratio $=\frac{1}{2}(1+\frac{2}{\pi })$