The intensity ratio of the two interfering beams of light is \( \beta \). The value of \( \left[\displaystyle\frac{I_{\mathrm{max}} - I_{\mathrm{min}}}{I_{\mathrm{max}} + I_{\mathrm{min}}}\right] \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{2\sqrt{\beta}}{(1+\beta)} \)
In interference the amplitudes add at a maximum and subtract at a minimum, and intensity is proportional to the square of the amplitude:
\( \displaystyle I_{\mathrm{max}} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^{2}, \qquad I_{\mathrm{min}} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^{2} \)
\( \displaystyle I_{\mathrm{max}} - I_{\mathrm{min}} = 4\sqrt{I_1 I_2}, \qquad I_{\mathrm{max}} + I_{\mathrm{min}} = 2\left(I_1 + I_2\right) \)
\( \displaystyle \frac{I_{\mathrm{max}} - I_{\mathrm{min}}}{I_{\mathrm{max}} + I_{\mathrm{min}}} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2} \)
Dividing numerator and denominator by \( I_2 \) and putting the intensity ratio \( I_1/I_2 = \beta \):
\( \displaystyle \frac{I_{\mathrm{max}} - I_{\mathrm{min}}}{I_{\mathrm{max}} + I_{\mathrm{min}}} = \frac{2\sqrt{\beta}}{1 + \beta} \)