Question

In a Young's double-slit experiment, let $\beta$ be the fringe width and let $I_0$ be the intensity at the central bright fringe. At a distance x from the central bright fringe, the intensity will be:

• A. $I_0$ cos$\left(\frac{x}{\beta }\right)$
• B. $I_0$ $co{s}^2\left(\frac{x}{\beta }\right)$
• C. $I_0$ $co{s}^2\left(\frac{\pi x}{\beta }\right)$
• D. $\left(\frac{I_0}{4}\right)$ $co{s}^2\left(\frac{\pi x}{\beta }\right)$

Answer 1

Answer: (C)

$I_0$ $co{s}^2\left(\frac{\pi x}{\beta }\right)$

$△$ = $x\frac{d}{D}$, where $△$ is path difference between two waves.
$\therefore$ Phase difference = $\varphi$ = $\frac{2\pi }{\lambda }△$.
Let a = amplitude at the screen due to each slit.
$\therefore$ I${}_0$ = k (2a)${}^2$ = 4ka${}^2$, where k is a constant.
For phase difference $\varphi$,
amplitude = A = $2acos(\varphi /2$).
[Since, a${}^2$ = a${}_1^2+a_2^2+2a_1{a}_{2}cos\pi$, here $a_1$ = a${}_2$]
Intensity $I$,
$I$ = KA${}^2$ = $k\left(4a^2\right)co{s}^2\left(\varphi /2\right)=I_{0}co{s}^2\left(\frac{\pi x}{\beta }△\right)$
= $I_{0}co{s}^2(\frac{\pi }{\lambda }.\frac{xd}{D})$ = $I_{0}co{s}^2\left(\frac{\pi x}{\beta }\right)$