Question

The reflective surface is given by \[y = 2\sin x\]. The reflective surface is facing a positive axis. What are the least values of co-ordinate of the point where a ray parallel to the positive x axis becomes parallel to the positive y axis after reflection?
A. \[\left( {\dfrac{\pi }{3},\sqrt 3 } \right)\]
B. \[\left( {\dfrac{\pi }{2},\sqrt 2 } \right)\]
C. \[\left( {\dfrac{\pi }{3},\sqrt 2 } \right)\]
D. \[\left( {\dfrac{\pi }{4},\sqrt 3 } \right)\]

Answer 1

Hint: In this case the angle of refraction plus the angle of reflection would be equal to 90 degrees. We need to find the slope at that point, and equate it to the expression of the slope of the equation.

Formula used: In this solution we will be using the following formulae;
\[m = \dfrac{{dy}}{{dx}}\] where \[m\] is the gradient of a line at a point, \[\dfrac{{dy}}{{dx}}\] is the differentiation of the equation of the line with respect to x.
\[m = \tan \theta \] where \[\theta \] is the angle the line makes with the horizontal.

Complete step by step answer:
To solve, we note that since the incident light is parallel to the x axis, and the reflected light is parallel to the y axis, then the incident light and parallel light are perpendicular to each other. This means that the angle of incidence (angle incident light makes with a normal perpendicular to reflective surface at the point of contact) is 45 degrees. The gradient line is perpendicular to the normal line, hence, the angle the gradient makes with the horizontal is 45 degrees.
The gradient of a line can be given as
\[m = \tan \theta \] where \[\theta \] is the angle the line makes with the horizontal.
Hence,
\[m = \tan \left( {45} \right) = 1\]
Also, we are given the expression,
\[y = 2\sin x\]
The gradient of this is given as
\[m = \dfrac{{dy}}{{dx}}\] where \[m\] is the gradient of a line at a point, \[\dfrac{{dy}}{{dx}}\] is the differentiation of the equation of the line with respect to x, which is
\[m = \dfrac{{dy}}{{dy}} = 2\cos x\]
Hence,
\[2\cos x = 1\]
\[ \Rightarrow \cos x = \dfrac{1}{2}\]
By finding the inverse, we have
\[x = {\cos ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{3}rad\]’
Hence,
\[y = 2\sin x = 2\sin \dfrac{\pi }{3} = \sqrt 3 \]
Thus, the coordinates are
\[\left( {\dfrac{\pi }{3},\sqrt 3 } \right)\]
Hence, the correct option is A.

Note: For clarity, note that the line is drawn which is used to calculate the gradient at a point is the tangent to the curve at that point. This is why it is perpendicular to the normal, as, from geometry, the tangent to a point is always perpendicular to the tangent at the point.