An interference pattern is obtained with two coherent sources of intensity ratio n:1. The ratio \(\dfrac{I_{Max} - I_{Min}}{I_{Max} + I_{Min}}\) will be maximum if
n = 1
For two coherent waves of intensities I1 and I2, \(I_{\text{Max}} = (\sqrt{I_1} + \sqrt{I_2})^2\) and \(I_{\text{Min}} = (\sqrt{I_1} - \sqrt{I_2})^2\). The given ratio is the fringe visibility, a measure of contrast.
Step 1 — \(I_{\text{Max}} - I_{\text{Min}} = 4\sqrt{I_1 I_2}\) and \(I_{\text{Max}} + I_{\text{Min}} = 2(I_1 + I_2)\).
Step 2 — With \(I_1 = nI_2\), the ratio is \(\dfrac{2\sqrt{I_1 I_2}}{I_1 + I_2} = \dfrac{2\sqrt{n}}{n + 1}\).
Step 3 — By the AM–GM inequality, \(n + 1 \ge 2\sqrt{n}\), so the ratio is at most 1, with equality only when n = 1.
For n = 2 the ratio is \(\dfrac{2\sqrt{2}}{3} \approx 0.943\), for n = 3 it is \(\dfrac{2\sqrt{3}}{4} \approx 0.866\) and for n = 4 it is \(\dfrac{4}{5} = 0.8\); all are less than the value 1 reached for equal intensities.
n = 1 option is correct.
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