Physics - Wave Optics Question with Solution | TestHub

Let the intensities of the two coherent sources be $I_1$ and $I_2$.

Given $\frac{I_1}{I_2}=b$.

The maximum and minimum intensities in the interference pattern are given by:

$I_{\text{max}}=(\sqrt{I_1}+\sqrt{I_2}{)}^2=I_1+I_2+2\sqrt{I_1{I}_2}$

$I_{\text{min}}=(\sqrt{I_1}-\sqrt{I_2}{)}^2=I_1+I_2-2\sqrt{I_1{I}_2}$

Adding these two equations, we get:

$I_{\text{max}}+I_{\text{min}}=2(I_1+I_2)$

Subtracting the two equations, we get:

$I_{\text{max}}-I_{\text{min}}=4\sqrt{I_1{I}_2}$

The required ratio is:

$\frac{I_{\text{max}}+I_{\text{min}}}{I_{\text{max}}-I_{\text{min}}}=\frac{2(I_1+I_2)}{4\sqrt{I_1{I}_2}}=\frac{I_1+I_2}{2\sqrt{I_1{I}_2}}$

Dividing the numerator and the denominator by $I_2$, we get:

$\frac{\frac{I_1}{I_2}+1}{2\sqrt{\frac{I_1}{I_2}}}=\frac{b+1}{2\sqrt{b}}$

Answer: $\frac{1+b}{2\sqrt{b}}$