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Wave Optics
Interference – maximum and minimum intensity
1 question from 1 paper, each with its answer.
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2016 Phase II 1 question
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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:
$I_{max} - I_{min} = 4\sqrt{I_1I_2}$ 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 $= \frac{2\sqrt{I_1/I_2}}{\left(\frac{I_1}{I_2} + 1\right)} = \frac{2\sqrt{n}}{n + 1}$