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Physics Question on wave interference

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

A

nn+1\frac{\sqrt{n}}{n + 1}

B

2nn+1\frac{2 \sqrt{n}}{n + 1}

C

n(n+1)2\frac{\sqrt{n}}{(n + 1)^2}

D

2n(n+1)2\frac{2 \sqrt{n}}{(n + 1)^2}

Answer

2nn+1\frac{2 \sqrt{n}}{n + 1}

Explanation

Solution

Imax⁡=(I+nI)2I_{\max} = \left(\sqrt{I} + \sqrt{nI}\right)^{2}
Imin⁡=(I−nI)2I_{\min} = \left(\sqrt{I} - \sqrt{nI}\right)^{2}
Imax⁡−Imin⁡Imax⁡+Imin⁡=(I+nI)2−(I−nI)2(I+nI)2+(I−nI)2\frac{I_{\max}- I_{\min}}{I_{\max} + I_{\min}} = \frac{\left(\sqrt{I} + \sqrt{nI}\right)^{2} - \left(\sqrt{I} - \sqrt{nI}\right)^{2}}{\left(\sqrt{I} + \sqrt{nI}\right)^{2} + \left(\sqrt{I} - \sqrt{nI}\right)^{2}}
=1+n+2n−1−n+2n1+n+2n+1+n−2n=4n2+2n=2n1+n= \frac{1+n+2\sqrt{n} - 1 - n+2\sqrt{n}}{1+n+2\sqrt{n} +1+n-2\sqrt{n}} = \frac{4\sqrt{n}}{2+2n} = \frac{2\sqrt{n}}{1+n}