In Young's double slit experiment, one of the slit is wider than other,

Question:

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If $\mathrm{I}_{\mathrm{m}}$ be the maximum intensity, the resultant intensity I when they interfere at phase difference $\phi$ is given by :

  1. $\frac{I_{m}}{9}\left(1+8 \cos ^{2} \frac{\phi}{2}\right)$

  2. $\frac{\mathrm{I}_{\mathrm{m}}}{9}(4+5 \cos \phi)$

  3. $\frac{\mathrm{I}_{\mathrm{m}}}{3}\left(1+2 \cos ^{2} \frac{\phi}{2}\right)$

  4. $\frac{\mathrm{I}_{\mathrm{m}}}{5}\left(1+4 \cos ^{2} \frac{\phi}{2}\right)$

Correct Option: 1

Solution:

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