2016 Phase II Wave OpticsInterference – maximum and minimum intensity Medium Question 27 of 40 in Wave Optics

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be:

A $\frac{\sqrt{n}}{(n + 1)^2}$
B $\frac{2\sqrt{n}}{(n + 1)^2}$
C $\frac{\sqrt{n}}{n + 1}$
D $\frac{2\sqrt{n}}{n + 1}$

The answer is D.

Solution

$I_{max} - I_{min} = 4\sqrt{I_1I_2}$ and $I_{max} + I_{min} = 2(I_1 + I_2)$.

Let $\frac{I_1}{I_2} = \frac{n}{1}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2} = \frac{4\sqrt{I_1I_2}}{2(I_1 + I_2)}$
Dividing numerator and denominator by $I_2$: ratio $= \frac{2\sqrt{I_1/I_2}}{\left(\frac{I_1}{I_2} + 1\right)} = \frac{2\sqrt{n}}{n + 1}$

Concept fringe-visibility-intensity-ratio

Wave Optics in past papers

40 questions from this chapter have appeared across 18 exam years.