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(A) The resultant intensity $I$ of two superposed coherent light beams is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$, where $\phi$ is the ph

Vedclass · Accepted Answer

$\frac{9 L}{4}, \frac{L}{4}$

Vedclass · Answer

(A) The resultant intensity $I$ of two superposed coherent light beams is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$, where $\phi$ is the phase difference between the beams.For maximum intensity, $\cos \phi = 1$ (constructive interference):$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$Given $I_1 = L$ and $I_2 = \frac{L}{4}$:$I_{\max} = (\sqrt{L} + \sqrt{\frac{L}{4}})^2 = (\sqrt{L} + \frac{\sqrt{L}}{2})^2 = (\frac{3\sqrt{L}}{2})^2 = \frac{9L}{4}$For minimum intensity, $\cos \phi = -1$ (destructive interference):$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2$$I_{\min} = (\sqrt{L} - \frac{\sqrt{L}}{2})^2 = (\frac{\sqrt{L}}{2})^2 = \frac{L}{4}$Thus, the maximum and minimum intensities are $\frac{9L}{4}$ and $\frac{L}{4}$ respectively.

Two monochromatic coherent light beams $A$ and $B$ have intensities $L$ and $\frac{L}{4},$ respectively. If these beams are superposed, the maximum and minimum intensities will be

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