Optics 6 Question 18
20. 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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Solution:
$$ \begin{aligned} I & =I _{\max } \cos ^{2} \frac{\varphi}{2} \\ \therefore \quad \frac{I _{\max }}{4} & =I _{\max } \cos ^{2} \frac{\varphi}{2} \\ \cos \frac{\varphi}{2} & =\frac{1}{2} \end{aligned} $$
$$ \begin{aligned} & \text { or } \quad \frac{\varphi}{2}=\frac{\pi}{3} \\ & \therefore \quad \varphi=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \Delta x \\ & \text { where } \quad \Delta x=d \sin \theta \end{aligned} $$
Substituting in Eq. (i), we get
$$ \sin \theta=\frac{\lambda}{3 d} \text { or } \theta=\sin ^{-1} \frac{\lambda}{3 d} $$
