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Question: Answered & Verified by Expert
In Young's double slit experiment the slits are horizontal. The intensity at a point ' $\mathrm{P}$ ' on the screen shown in the figure is $\frac{\mathrm{I}_0}{4}$ where $\mathrm{I}_0$ is maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $2 \lambda$, then the value of ' $\theta$ ' is

PhysicsWave OpticsAP EAMCETAP EAMCET 2017 (25 Apr Shift 1)
Options:
  • A $\cos ^{-1}\left(\frac{1}{6}\right)$
  • B $\sin ^{-1}\left(\frac{1}{12}\right)$
  • C $\tan ^{-1}\left(\frac{1}{6}\right)$
  • D $\sin ^{-1}\left(\frac{3}{4}\right)$
Solution:
2612 Upvotes Verified Answer
The correct answer is: $\cos ^{-1}\left(\frac{1}{6}\right)$
No solution. Refer to answer key.

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