Question

Figure shows two coherent sources $S_1$ and $S_2$ emitting wavelength $\lambda$. The separation $S_1{S}_2=1.5\lambda$ and $S_1$ is ahead in phase by $\pi /2$ relative to $S_2$. Then the maxima occur in direction $\theta$ given by ${\sin }^{-1}$ of
(i) $0$
(ii) $1/2$
(iii) $-1/6$
(iv) $-5/6$
Correct options are

• A. (ii), (iii), and (iv)
• B. (i), (ii), and (iii)
• C. (i), (iii), and (iv)
• D. All the above

Answer 1

Answer: (A)

$\delta =$ phase difference between the waves from $S_1$ and $S_2$ at
$P=\frac{\pi }{2}-\frac{2\pi }{2}(d\sin \theta )$
For maximum intensity at $P,\delta =n\pi$, where $n=0,\pm 1$, $\pm 2,\dots$
$\therefore \frac{2\pi }{2}(1.5\lambda \sin \theta )\frac{\pi }{2}=n\pi$
$\Rightarrow n-\frac{1}{2}=3\sin \theta$
$\Rightarrow \sin \theta =\left(\frac{n-\frac{1}{2}}{3}\right)$
For $n=0,\sin \theta =-\frac{1}{6}$
For $n=\pm 1,\sin \theta =\frac{1}{6},-\frac{1}{2}$
For $n=\pm 2,\sin \theta =\frac{1}{2},-\frac{5}{6}$