Question

In a Young's double slit experiment with light of wavelength $\lambda$ the separation of slits is $d$ and distance of screen is $D$ such that $D > > d > > \lambda$. If the Fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

• A. $\frac{\beta}{2}$
• B. $\frac{\beta}{4}$
• C. $\frac{\beta}{3}$
• D. $\frac{\beta}{6}$

Answer 1

Answer: (B)

For two identical slits, we can write the intensity as follows:
$I=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right)$
$I_{max}=4 I_{0}$
Therefore, now, the half of maximum intensity is
$\frac{I_{\max }}{2}=2 I_{0}=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right)$
$\Rightarrow \cos \left(\frac{\phi}{2}\right)=\frac{1}{\sqrt{2}}$
Therefore, $\frac{\phi}{2}=\frac{\pi}{2}$
$\Rightarrow \phi=\frac{\pi}{2}$
That is, $\frac{2 \pi}{\lambda} \Delta x=\frac{\pi}{2}$
$\Rightarrow \Delta x=\frac{\pi}{4}$
Now $y \frac{d}{D}=\frac{\lambda}{4}$
$\Rightarrow y=\frac{\lambda D}{4 d}= \frac{\beta}{4}$