Physics - Wave Optics Question with Solution | TestHub

$\text{I}={\text{I}}_{\text{max}}{\text{cos}}^2\left(\frac{\varphi }{2}\right)$

$\therefore \; \frac{{\text{3I}}_{\text{max}}}{4}={\text{I}}_{\text{max}}{\text{ cos }}^2\frac{\varphi }{2}$

$\text{cos}\frac{\varphi }{2}=\frac{\sqrt{3}}{2}$

or,$\frac{\varphi }{2}=\frac{\pi }{6}$

$\therefore \; \; \mathrm{\varphi }=\frac{\mathrm{\pi }}{3}=\left(\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\right)\Delta \mathrm{x}\text{ .....(i) }$

where,$\Delta \mathrm{x}=\text{d sin}\mathrm{\theta }$

Substituting in Eq.(i) we get

$\text{sin}\theta =\frac{\lambda }{\text{6d}}$

or,$\theta ={\text{sin}}^{-1}\left(\frac{\lambda }{\text{6d}}\right)$

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