Question

A reflecting surface is represented by the equation $y = \dfrac{{2L}}{\pi }\sin (\dfrac{{\pi x}}{L}),0 \leqslant x \leqslant L$.
A ray traveling horizontal becomes vertical after reflection. The coordinates of the point(s) on which this ray is incident.
A.$(\dfrac{L}{4}:\dfrac{{\sqrt 2 L}}{\lambda })$
B.$(\dfrac{L}{3}:\dfrac{{\sqrt 3 L}}{\pi })$
C.$(\dfrac{{3L}}{4}:\dfrac{{\sqrt 2 L}}{\lambda })$
D.$(\dfrac{{4L}}{3}:\dfrac{{\sqrt 4 L}}{\pi })$

Answer 1

Hint: We know that the light travels in a straight path unless an obstacle comes along its path. According to the law of reflection, the angle of incidence and angle of reflection are equal. The slope of any line in the x-y plane is defined by the ratio of its vertical and horizontal components.

Complete answer:
The equation of reflecting surface given in the question is below.
$y = \dfrac{{2L}}{\pi }\sin (\dfrac{{\pi x}}{L}),0 \leqslant x \leqslant L$
To find the slope let us differentiate the above equation with respect to x.
$\dfrac{{dy}}{{dx}} = \dfrac{{2L}}{\pi }\cos \left( {\dfrac{{\pi x}}{L}} \right) \times \dfrac{\pi }{L} = 2\cos \left( {\dfrac{{\pi x}}{L}} \right)$ ……………….. (1)
Now, according to the question, the incident ray is horizontal which is zero-degree and the reflected ray is vertical which is 90-degree. Hence, we can say that the angle of incident and reflection will be \[90^\circ \] . So, we conclude that the normal to the point is at \[45^\circ \] .
Hence, the slope of the point of incidence is given by,
$\dfrac{{dy}}{{dx}} = \tan {45^o} = 1$ …………………..(2)
Equating equations (1) and (2) we get the following expression.
$2\cos \left( {\dfrac{{\pi x}}{L}} \right) = 1$
Let us further simplify it.
$\cos \left( {\dfrac{{\pi x}}{L}} \right) = \dfrac{1}{2}$
Now, the angle of cosine whose value is $\dfrac{\pi }{3}$.
$\cos \left( {\dfrac{{\pi x}}{L}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) \Rightarrow x = \dfrac{L}{3}$
Now, to find the value of \[y\] , let us substitute the value of \[x\] in the equation of the reflecting surface.
$y = \dfrac{{2L}}{\pi }\sin \left( {\dfrac{\pi }{L} \times \dfrac{L}{3}} \right) = \dfrac{{2L}}{\pi }\sin \dfrac{\pi }{3}$
Let us further simplify it.
$y = \dfrac{{2L}}{\pi } \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 L}}{\pi }$
Therefore, the coordinates of the incidence point are $\left( {\dfrac{L}{3},\dfrac{{\sqrt 3 L}}{\pi }} \right)$.

Hence, the correct option is (B) $\left( {\dfrac{L}{3},\dfrac{{\sqrt 3 L}}{\pi }} \right)$.

Note:
When light incidents on the interface of any two different optical media, their reflection and transmission are described by the Fresnel equations.
 When the light goes from one medium to another then reflection and refraction both can occur. The Fresnel equation gives the ratio of the reflected wave to incident wave and the ratio of the transmitted wave to incident wave.