Physics - Wave Optics Question with Solution | TestHub

The condition of interference maxima is $\begin{array}{rl}d\sin \theta & =n\lambda \\ \sin \theta & =\frac{n\lambda }{d}\end{array}$ Given, $\begin{array}{c}d=2\lambda \\ \sin \theta =\frac{n\lambda }{2\lambda }=n/2\end{array}$ The magnitude of $\sin \theta$ lies between 0 and 1 When $n=0,\sin \theta =0\Rightarrow \theta =0$ When $n=1,\sin \theta =1/2\Rightarrow \theta =30^{\circ }$ When $n=2,\sin \theta =1\Rightarrow \theta =90^{\circ }$ Thus, there is central maximum $\left(\theta =0^{\circ }\right)$, on other side of it maxima lie at $\theta =30^{\circ }$ and $\theta =90^{\circ }$, so maximum number of possible interference maxima is 5.