Question

In $\text{YDSE}$, having slits of equal width, let $\beta$ be the fringe width and $I_o$ be the maximum intensity. At a distance $x$ from the central bright fringe, the intensity will be

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

Answer 1

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


Intensity at a point on a screen is given by,

$I=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

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, $\beta =\frac{\lambda D}{d}$

Given that, $I_{max}=I_0$

So, the intensity at point $P$,

$I=I_0{\cos }^2\left(\frac{\pi x}{\beta }\right)$

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