Two coherent light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are :

5 I

and

I

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5 I

and

3 I

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9 I

and

I

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9 I

and

3 I

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Solution

The correct option is C $9 I$ and $I$
The net intensity due to interference of two beams of intensities $I_{1}, I_{2}$ with a phase difference of $ϕ$ is
$I_{n e t} = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} c o s ϕ$
Thus, $I_{n e t_{m a x}}$ occurs for $c o s ϕ = 1$
$\Rightarrow I_{n e t_{m a x}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$ $= 9 I$
Similarly, $I_{n e t_{m i n}}$ occurs for $c o s ϕ = - 1$
$\Rightarrow I_{n e t_{m i n}} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$ $= I$

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