Question

A ray of light parallel to $x-$axis is incident on a circle $x^2+y^2=R^2$ polished from inside as shown in the figure. Find the co-ordinate of $x\text{and}y$ of the point at which the incident ray should fall such that the reflected ray deviated by ${60}^{\circ }$ after reflection.

• A. $(x,y)=(\frac{\sqrt{3}R}{2},-\frac{R}{2})$
• B. $(x,y)=(\frac{R}{\sqrt{2}},-\frac{R}{2})$
• C. $(x,y)=(-\frac{R}{2},\frac{\sqrt{3}R}{2})$
• D. $(x,y)=(-\frac{R}{\sqrt{2}},\frac{R}{\sqrt{2}})$

Answer 1

The correct option is C $(x,y)=(-\frac{R}{2},\frac{\sqrt{3}R}{2})$

The given situation can be shown below as
So, here from the geometry we can say that the co-ordinates are
$(x,y)=(-R\cos {60}^{\circ },R\sin {60}^{\circ })$
$\Rightarrow (x,y)=(-\frac{R}{2},\frac{\sqrt{3}R}{2})$