Question

In Young's double'sIit experiment, Let $\beta$ be the fringe width and $I_0$ 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. $I_0{\cos }^2\left(\frac{x}{\beta }\right)$
• C. $I_0{\cos }^2\left(\frac{\pi x}{\beta }\right)$
• D. $\frac{I_0}{4}{\cos }^2\left(\frac{\pi x}{\beta }\right)K-3$

Answer 1

Answer: (C)

$\Delta =x\frac{d}{D}$
$\therefore$ phase difference $=\varphi =\frac{2\pi }{\lambda }\Delta$
Let $a=$ amplitude at the screen due to each slit.
$\therefore I_0=k(2a{)}^2=4k{a}^2$,
where $k$ is a constant.
For phase difference $\varphi$,
amplitude $=A=2a\cos (\varphi /2)$.
Intensity, I $=k{A}^2=k\left(4a^2\right)\cos (\varphi /2)=I_0{\cos }^2\left(\frac{\pi }{\lambda }\Delta \right)$
$I_0{\cos }^2\left(\frac{\pi }{\lambda }\cdot \frac{xd}{D}\right)=I_0{\cos }^2\left(\frac{\pi x}{\beta }\right)$