Question

Two coherent sources of intensity ratio $\beta$ interfere. Then the value of $(I_{\max }-I_{\min })/(I_{\max }+I_{\min })$ is:

• A. $\frac{1+\beta }{\sqrt{\beta }}$
• B. $\sqrt{\left(\frac{1+\beta }{\beta }\right)}$
• C. $\frac{1+\beta }{2\beta }$
• D. $\frac{2\sqrt{\beta }}{1+\beta }$

Answer 1

Answer: (D)

We know $I\propto a^2$ or $a\propto \sqrt{I}$
$\therefore \frac{a_1}{a_2}=\sqrt{\left(\frac{I_1}{I_2}\right)}$
So, $\frac{I_{\max }}{I_{\min }}=\frac{{\left(a_1+a_2\right)}^2}{{\left(a_1-a_2\right)}^2}$
$=\frac{\left(a_1+a_2\right)}{{\left(a_2-a_1\right)}^2}=\frac{(1+\sqrt{\beta }{)}^2}{(1-\sqrt{\beta }{)}^2}$
Applying componendo and dividendo
$\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}=\frac{(1+\sqrt{\beta }{)}^2+(1-\sqrt{\beta }{)}^2}{(1+\sqrt{\beta }{)}^2-(1-\sqrt{\beta }{)}^2}$
or $=\frac{2+2\beta }{4\sqrt{\beta }}$
or $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2\sqrt{\beta }}{1+\beta }$