Physics - Wave Optics Question with Solution | TestHub

For possible interference maxima on the screen, the condition is $d\sin \theta =n\lambda \dots (i)$ Given : $d=$ slit - width $=2\lambda$ $\begin{array}{rlrl}& \therefore & 2\lambda \sin \theta & =n\lambda \\ \Rightarrow & & 2\sin \theta & =n\end{array}$ The maximum value of $\sin \theta$ is 1 , hence, $n=2\times 1=2$ Thus, Eq. (i) must be satisfied by 5 integer values $ie,-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is 5 .