Hence, phase difference between $S_{1}$ and $S_{2}, \phi=\frac{\lambda}{6} \times \frac{2 \pi}{\lambda}=\frac{\pi}{3}$
So intensity between $S_{1}$ and $S_{2}$ is,
$I^{\prime}=I \cos ^{2}\left(\frac{\pi}{6}\right)^{2}=\frac{3 I}{4}$ $............(i)$
Given, $S_{1} P-S_{3} P=\frac{2 \lambda}{3}$
Hence, phase difference between $S_{1}$ and $S_{3}, \phi=\frac{2 \lambda}{3} \times \frac{2 \pi}{\lambda}=\frac{4 \pi}{3}$
So intensity between $S_{1}$ and $S_{2}$ is,
$I^{\prime \prime}=I \cos ^{2}\left(\frac{2 \pi}{3}\right)^{2}=\frac{I}{4} \quad \ldots \ldots \ldots(i i)$
The intensity at P when all the three slits are open is, $I \frac{I^{\prime}}{I^{\prime \prime}}=3 I$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $-I$ | List $-II$ |
| $(A)$ Moment of inertia of solid sphere of radius $(R)$ about any tangent | $(I)$ $\frac{5}{3} MR ^{2}$ |
| $(B)$ Moment of inertia of hollow sphere of radius $(R)$ about any tangent | $(II)$ $\frac{7}{5} MR ^{2}$ |
| $(C)$ Moment of inertia of circular ring of radius $(R)$ about its diameter. | $(III)$ $\frac{1}{4} MR ^{2}$ |
| $(D)$ Moment of inertia of circular disc of radius $(R)$ about any diameter. | $(IV)$ $\frac{1}{2} MR ^{2}$ |
Question: Choose the correct answer from the options given below
