Physics - Wave Optics Question with Solution | TestHub

$\frac{I_1}{I_2}=\frac{a^2}{b^2}=\beta \therefore \frac{a}{b}=\sqrt{\beta }$ Fringe visibility is given by $\begin{array}{rl}V=& \frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{(a+b{)}^2-(a-b{)}^2}{(a+b{)}^2+(a-b{)}^2}\\ & =\frac{4ab}{2\left(a^2+b^2\right)}=\frac{2(a/b)}{\left(\frac{a^2}{b^2}+1\right)}=\frac{2\sqrt{\beta }}{\beta +1}\end{array}$