Question

In YDSE, having slits of equal width, let $\beta$ be the fringe width and $I_0$ be 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. $I_0{\cos }^2\frac{2\pi x}{\beta }$
• C. $I_0{\cos }^2\frac{\pi x}{\beta }$
• D. $\frac{I_0}{4}{\cos }^2\frac{\pi x}{\beta }$

Answer 1

Answer: (C)

$I_0{\cos }^2\frac{\pi x}{\beta }$

Intensity at a point P , $I_p=I_{o}co{s}^2(\frac{\delta }{2})$ -(1)

$\delta =\frac{2\pi }{\lambda }\Delta x$ where $\Delta x$ is the path difference.

Also $\Delta x=x\frac{d}{D}and\beta =\lambda \frac{D}{d}$

Thus $\Delta x=\frac{\lambda x}{\beta }$

$⟹\delta =\frac{2\pi x}{\beta }$

From equation (1), $I_p=I_{o}co{s}^2(\frac{\pi x}{\beta })$