Physics - Wave Optics Question with Solution | TestHub

${\text{I}}_1={\text{I}}_{\text{A}}+{\text{I}}_{\text{B}}+2\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}}}cos\frac{\pi }{3}$...(1)

${\text{I}}_2={\text{I}}_{\text{A}}+{\text{I}}_{\text{B}}+2\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}}}cos\frac{2\pi }{3}$...(2)

$\Rightarrow \;$${\text{I}}_1+{\text{I}}_2=2\left({\text{I}}_{\text{A}}+{\text{I}}_{\text{B}}\right)$...(3)

From (1)

$\Rightarrow$${\text{I}}_1=\frac{{\text{I}}_1+{\text{I}}_2}{2}+\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}}}$

${\text{I}}_1-\left(\frac{{\text{I}}_1+{\text{I}}_2}{2}\right)=\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}}}$

$\frac{{\text{I}}_1-{\text{I}}_2}{2}=\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}}}$

${\text{I}}_{\text{max}}={\left(\sqrt{{\text{I}}_{\text{A}}}+\sqrt{{\text{I}}_{\text{B}}}\right)}^2$

$={\text{I}}_{\text{A}}+{\text{I}}_{\text{B}}+2\sqrt{{\text{I}}_{\text{A}}{\text{I}}_{\text{B}} }$...(4)

${\text{I}}_{\text{max}}=\frac{{\text{I}}_1+{\text{I}}_2}{2}+\left({\text{I}}_1-{\text{I}}_2\right)$

$=\frac{{3\text{I}}_1-{\text{I}}_2}{2}$