Optics Ques 133
- In Young’s double slit experiment intensity at a point is $(1 / 4)$ of the maximum intensity. Angular position of this point is
(a) $\sin ^{-1} (\frac{\lambda}{d})$
(b) $\sin ^{-1} (\frac{\lambda}{2 d})$
(c) $\sin ^{-1} (\frac{\lambda}{3 d})$
(d) $\sin ^{-1} (\frac{\lambda}{4 d})$
$(2005,2 M)$
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Answer:
Correct Answer: 133.(c)
Solution:
Formula:
$ \begin{aligned} I & =I _{\max } \cos ^{2} (\frac{\phi}{2}) \\ \therefore \quad \frac{I _{\max }}{4} & =I _{\max } \cos ^{2} \frac{\phi}{2} \\ \cos \frac{\phi}{2} & =\frac{1}{2} \end{aligned} $
$ \begin{aligned} & \text { or } \quad \frac{\phi}{2}=\frac{\pi}{3} \\ & \therefore \quad \phi=\frac{2 \pi}{3}=(\frac{2 \pi}{\lambda}) \Delta x \quad …….(i) \\ & \text { where } \quad \Delta x=d \sin \theta \end{aligned} $
Substituting in Eq. (i), we get
$ \sin \theta=\frac{\lambda}{3 d} \text { or } \theta=\sin ^{-1} (\frac{\lambda}{3 d}) $
BETA
