Question

Two coherent sources of intensity ratio ${}^{\prime }{\alpha }^{\prime }$ interfere. In interference pattern $\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=$.

• A. $\frac{2\alpha }{1+\alpha }$
• B. $\frac{2\sqrt{\alpha }}{1+\alpha }$
• C. $\frac{2\alpha }{1+\sqrt{a}}$
• D. $\frac{1+\alpha }{2\alpha }$

Answer 1

Answer: (B)

$\frac{2\sqrt{\alpha }}{1+\alpha }$

Let the intensity of the second source be $I_2$.

Thus intensity of first source $I_1=\alpha I_2$

Maximum intensity $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

$\therefore$ $I_{max}=(\sqrt{\alpha I_2}+\sqrt{I_2}{)}^2=(\sqrt{\alpha }+1{)}^2{I}_2$

$⟹$ $I_{max}=(\alpha +1+2\sqrt{\alpha })I_2$

Minimum intensity $I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

$\therefore$ $I_{min}=(\sqrt{\alpha I_2}-\sqrt{I_2}{)}^2=(\sqrt{\alpha }-1{)}^2{I}_2$

$⟹$ $I_{min}=(\alpha +1-2\sqrt{\alpha })I_2$

From the two relations we get $I_{max}=\left(4\sqrt{\alpha }\right)I_2$

Also $I_{min}=2(\sqrt{\alpha }+1)I_2$

$⟹$ $\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=\frac{4\sqrt{\alpha }{I}_2}{2(\sqrt{\alpha }+1)I_2}=\frac{2\sqrt{\alpha }}{\sqrt{\alpha }+1}$