Physics - Wave Optics Question with Solution | TestHub

Hint: $\frac{I_1}{I_2}=n$ $\begin{array}{rl}& {\mathrm{I}}_1={\mathrm{nI}}_2\\ & {\mathrm{I}}_{\max }={\left(\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_2}\right)}^2={\left(\sqrt{\mathrm{nI}}2+\sqrt{{\mathrm{I}}_2}\right)}^2=(\sqrt{\mathrm{n}}+1{)}^2{\left(\sqrt{{\mathrm{I}}_2}\right)}^2\end{array}$ $\begin{array}{rl}& I_{\min }={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2={\left(\sqrt{n{I}_2}-\sqrt{I_2}\right)}^2=(\sqrt{n}-1{)}^2{\left(\sqrt{I_2}\right)}^2\\ & \frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{(\sqrt{n}+1{)}^2\left(I_2\right)-(\sqrt{n}-1{)}^2{I}_2}{(\sqrt{n}+1{)}^2{I}_2+(\sqrt{n}-1{)}^2{I}_2}=\frac{(\sqrt{n}+1{)}^2-(\sqrt{n}-1{)}^2}{(\sqrt{n}+1{)}^2+(\sqrt{n}-1{)}^2}\\ & =\frac{(n+1+2\sqrt{n})-(n+1-2\sqrt{n})}{n+1+2\sqrt{n}+n+1-2\sqrt{n}}=\frac{4\sqrt{n}}{2(n+1)}=\frac{2\sqrt{n}}{n+1}\end{array}$ $\therefore$ decreases with increasing $\mathrm{n}$. $\therefore$ It will be maximum if $n=1$