Question

In Young's double slit experiment, having slits of equal width, $\beta$ is the fringe width and $I_0$ is the maximum intensity. At a distance $x$ from the central bright fringe, the intensity will be,

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

Answer 1

The correct option is C $I_0{\cos }^2\left(\frac{\pi x}{\beta }\right)$

At point $P$,
$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$
and, $\Delta x=\frac{xd}{D}$



$\Rightarrow \Delta \varphi =\frac{2\pi }{\lambda }\times \frac{xd}{D}=\frac{2\pi x}{\beta }$
Where, $\beta =\frac{\lambda D}{d}$ is the fringe width.

The resultant intensity at point $P$ is,
$I_p=I_1+I_2+2\sqrt{I_1{I}_2}\cos \Delta \varphi \because I_1=I_2=I$
$I_p=4I{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$
Now, $I_{max}=I_0=4I$

$\therefore I_p=I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)=I_0{\cos }^2\left(\frac{\pi x}{\beta }\right)$

Hence, $\left(C\right)$ is the correct answer.