Two coherent sources of light having intensity ratio ‘\( \alpha \)’ interfere to produce an interference pattern. From the intensity variation in the pattern, the ratio \( \displaystyle\frac{I_{\mathrm{max}}-I_{\mathrm{min}}}{I_{\mathrm{max}}+I_{\mathrm{min}}} \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{2\sqrt{\alpha}}{1+\alpha} \)
In interference the amplitudes add, not the intensities. With \( I \propto a^{2} \):
\( \displaystyle I_{\mathrm{max}} = (\sqrt{I_1} + \sqrt{I_2})^{2}, \qquad I_{\mathrm{min}} = (\sqrt{I_1} - \sqrt{I_2})^{2} \)
\( I_{\mathrm{max}} - I_{\mathrm{min}} = 4\sqrt{I_1I_2} \)
\( I_{\mathrm{max}} + I_{\mathrm{min}} = 2(I_1 + I_2) \)
\( \displaystyle \frac{I_{\mathrm{max}} - I_{\mathrm{min}}}{I_{\mathrm{max}} + I_{\mathrm{min}}} = \frac{2\sqrt{I_1I_2}}{I_1 + I_2} \)
Dividing numerator and denominator by \( I_2 \) and putting \( I_1/I_2 = \alpha \):
\( \displaystyle \frac{I_{\mathrm{max}} - I_{\mathrm{min}}}{I_{\mathrm{max}} + I_{\mathrm{min}}} = \frac{2\sqrt{\alpha}}{1 + \alpha} \)
Shortcut: Limiting case: for equal sources (\( \alpha = 1 \)) \( I_{\mathrm{min}} = 0 \), so the ratio must be 1. Only \( 2\sqrt{\alpha}/(1 + \alpha) \) gives 1; Options (1) and (3) become infinite and Option (2) gives \( \sqrt{2}/2 \).