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. $4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$
• C. $I_{o} \sin ^{2}\left(\frac{\pi d}{2 \lambda D} x\right)$
• D. $4 I _{ o } \cos \left(\frac{\pi d }{2 \lambda D } x \right)$

Answer 1

Answer: (B)

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