Question

A reflecting surface is represented by the equation $Y=\frac{2L}{\pi }\sin \left(\frac{\pi x}{L}\right),0\le x\le L$. A ray travelling horizontally becomes vertical after reflection. The coordinates of the point (s) dwhere this ray is incident is

• 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. $(\frac{2L}{4},\frac{\sqrt{3}L}{\pi })$

Answer 1

Answer: (B)

$(\frac{L}{3},\frac{\sqrt{3}L}{\pi })$

From the figure,

The slope of tangent can be given as

$slope=\tan {45}^{\circ }=1$

Assume the coordinate of point of incident be $(x,y)$

Also slope can be found by differentiating the given equation of curve

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2L}{\pi }\sin \left(\frac{\pi x}{L}\right)\right)=1$

$2\cos \left(\frac{\pi x}{L}\right)=1$

$x=\frac{L}{\pi }{\cos }^{-1}\left(\frac{1}{2}\right)$

$x=\frac{L}{3}$

Hence substituting the value of $x$ in equation of curve given

$y=\frac{2L}{\pi }\sin \left(\frac{\pi }{3}\right)$

$y=\frac{2L}{\pi }\times \frac{\sqrt{3}}{2}$

Therefore,

$y=\frac{\sqrt{3}L}{\pi }$

Hence the coordinate of point of incidence is $(x,y)=\left(\frac{L}{3},\frac{\sqrt{3}L}{\pi }\right)$ .