Question

In the Young's double slit experiment a monochromatic source of wavelength $\lambda$ is used. The intensity of light passing through each slit is $I_0$. The intensity of light reaching the screen $S_C$ at a point $P$, a distance $X$ from $O$ is given by (Take $d<<D$ )

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

Answer 1

Answer: (B)

$\varphi =\left(\frac{2\pi }{\lambda }\right)$ path difference
Path difference $=d\sin \theta$
$=d\left[\frac{x}{D}\right]$
$\varphi =\frac{2\pi }{\lambda }\left[\frac{dx}{D}\right]$
$I=4I_o{\cos }^2\frac{\varphi }{2}$
$=4I_o{\cos }^2\frac{\pi d}{\lambda D}\cdot x$