Physics - Wave Optics Question with Solution | TestHub

If a is the amplitude of the wave then

$\frac{I_{\max }}{4}=a^2=a^2+a^2+2aa\cos \varphi$

or$\cos \varphi =-\frac{1}{2}$

or$\varphi =\frac{2\pi }{3}$

Corresponding path difference,

$\mathrm{\Delta }x=\frac{\varphi \times \lambda }{2\pi }=\frac{\left(\frac{2\pi }{3}\right)\times \lambda }{2\pi }=\frac{\lambda }{3}$

So$d\sin \theta =\frac{\lambda }{3}$

Or$\theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right).$