Question

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

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

Answer 1

Answer: (C)

$I_o{\cos }^2\left(\frac{\pi x}{\beta }\right)$

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