In a Young's double slit experiment, the path difference at a point on the screen is $\Delta x=\frac{\lambda }{4}$. What is the ratio of intensity at this point to the maximum intensity on the screen?

Solution

1. The phase difference $\varphi$ is related to the path difference $\Delta x$ by the formula $\varphi =\frac{2\pi }{\lambda }\Delta x$.
2. Substituting $\Delta x=\frac{\lambda }{4}$, we get $$\begin{array}{rl}\varphi & =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}\\ & =\frac{\pi }{2}\end{array}$$.
3. The intensity $I$ at any point is given by $I=I_{\text{max}}{\cos }^2(\frac{\varphi }{2})$.
4. Substituting $\varphi =\frac{\pi }{2}$, we find $$\begin{array}{rl}I& =I_{\text{max}}{\cos }^2(\frac{\pi }{4})\\ & =I_{\text{max}}(\frac{1}{\sqrt{2}}{)}^2\\ & =\frac{I_{\text{max}}}{2}\end{array}$$.
5. Hence the answer is (A).