Question

Directions: The following question has four choices, out of which one or more are correct.A ray of light is incident on a reflecting surface. The ray moves in horizontal direction and is reflected vertically after striking the surface. If the surface is denoted by $y=\frac{2L}{\pi }\sin \left(\frac{\pi }{L}x\right),0\le x\le L,$ what are the coordinates of the point(s) where the ray is incident?

• (A) $$(\frac{L}{4},\frac{\sqrt{2}L}{\pi })$$
• (B) $$(\frac{L}{3},\frac{\sqrt{3}L}{\pi })$$
• (C) $$(\frac{3L}{4},\frac{\sqrt{2}L}{\pi })$$
• (D) $$(\frac{2L}{3},\frac{\sqrt{3}L}{\pi })$$

Answer 1

The Correct Option is A, D

Explanation:
Consider the diagram for reflection at general point A shown below.

Here, we have.$$\begin{array}{rl}2i& ={90}^{\circ }\\ i& ={45}^{\circ }\end{array}$$So, $dy/dx$ at $A$ should be $\tan {45}^{\circ }.$ Therefore$$y=\frac{2L}{\pi }\sin \left(\frac{\pi }{L}x\right)$$$$\begin{array}{rl}\frac{dy}{dx}& =\frac{2L}{\pi }\cos \left(\frac{\pi }{L}x\right)\times \frac{\pi }{L}\\ 1& =2\cos \left(\frac{\pi }{L}x\right)\\ \frac{\pi x}{L}& =\frac{\pi }{3}\\ x& =\frac{L}{3}\end{array}$$Therefore,$$\begin{array}{rl}y& =\frac{2L}{\pi }\sin (\frac{\pi }{L}\times \frac{L}{3})\\ & =\frac{2L}{\pi }\times \frac{\sqrt{3}}{2}\\ & =\frac{\sqrt{3}L}{\pi }\end{array}$$Therefore, the point is $(\frac{L}{3},\frac{\sqrt{3}L}{\pi }).$ Another point, just opposite on the surface will also act same and its coordinates are $(\frac{2L}{3},\frac{\sqrt{3}L}{\pi })$