Question

A beam of light is sent along the line $x-y=1$. It is incident on the x-axis as the refracting surface from under the x-axis. After refracting from the x-axis the line turns through ${30}^{\circ }$ away from the normal on the opposite side at the point of incidence on the x-axis. Then the equation of the refracted ray is

• A. $(2-\sqrt{3})x-y=2+\sqrt{3}$
• B. $(2+\sqrt{3})x-y=2+\sqrt{3}$
• C. $(2-\sqrt{3})x+y=(2+\sqrt{3})$
• D. $y=(2-\sqrt{3})(x-1)$

Answer 1

The correct option is D $y=(2-\sqrt{3})(x-1)$

Consider the ray
$y=x-1$
Therefore the angle it makes with the positive direction of $x$ axis is ${45}^0$
Now since the ray shifts by an angle of ${30}^0$ away from the normal, that is towards the x axis.
Therefore, the refracted ray makes an angle of ${45}^0-{30}^0={15}^0$ with the positive $x$ axis.
Since it passes through the point $(1,0)$
The equation of the ray would be
$(y-0)=\tan {15}^0(x-1)$
$y=(2-\sqrt{3})(x-1)$