wave-optics Question 14
Question: Q. 2. Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves have a path difference of (i) $\lambda / 6$, and (ii) $\lambda / 2$.
A [Foreign 2017]
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Solution:
Ans. Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference
Path difference $=\frac{\lambda}{6} \Rightarrow$ Phase difference $=\frac{\pi}{3} 1 / 2$
Path difference $\frac{\lambda}{2} \Rightarrow$ Phase difference $=\pi \quad 1 / 2$
$$ I=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) $$
(i)
$$ \begin{equation*} I_{1}=4 I_{0} \times \frac{3}{4}=3 I_{0} \tag{ii} \end{equation*} $$
Detailed Answer :
Phase difference between interfering waves
$$ \phi=\frac{2 \pi}{\lambda} \times \text { path difference } $$
(i) Path difference $=\frac{\lambda}{6}$
$\therefore$ Phase difference $=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}$
$$ \begin{aligned} I & =4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \ & =4 I_{0} \cos ^{2} \frac{\pi}{6} \ & =4 I_{0} \times \frac{3}{4}=3 I_{0} \end{aligned} $$
(ii) Path difference $=\frac{\lambda}{2}$
$\therefore$ Phase difference, $\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$
$$ \begin{aligned} I & =4 I_{0} \cos ^{2} \frac{\phi}{2} \ & =4 I_{0} \cos ^{2} \frac{\pi}{2}=0 \end{aligned} $$
