# Question #c5820

Jan 29, 2017

It will be ${30}^{\circ}$ when incident ray is above the horizontal.
Again it will be ${60}^{\circ}$ when incident ray is below the horizontal.

#### Explanation:

It maybe caused in two ways depending on how the incident ray is made to fall on the mirror.

Case-I When the ray is incident at an angle ${30}^{\circ} \text{ above}$ the horizontal. Let the angle be ${\theta}^{\circ}$with which the mirror to be placed with the horizontal so as to reflect vertically the ray incident on the mirror at ${30}^{\circ}$ with the horizontal.

Incident ray is inclined at angle ${30}^{\circ}$ (above) with the horizontal and reflected ray is vertical. So the angle between reflected ray and incident ray may be $\left({90}^{\circ} - {30}^{\circ}\right) = {60}^{\circ}$ Hence angle of incidence will be $= \frac{1}{2} \times {60}^{\circ} = {30}^{\circ}$

So it is obvious that

$\theta + {30}^{\circ} + {30}^{\circ} = {90}^{\circ}$

$\implies \theta = {30}^{\circ}$

Case-II When the ray is incident at an angle ${30}^{\circ} \text{ below}$ the horizontal. In this case if mirror makes an angle $\theta$ to reflect the incident ray in vertical direction, then the vertical reflected ray will make ${\left(90 - \theta\right)}^{\circ}$ with the mirror.
So the angle of reflection becomes $\left({90}^{\circ} - {\left(90 - \theta\right)}^{\circ}\right) = {\theta}^{\circ}$

But the incident ray makes an angle ${30}^{\circ}$ below the horizontal. So it makes an angle $\left({90}^{\circ} + {30}^{\circ}\right) = {120}^{\circ}$with the vertical reflected ray and as a result the angle of incidence or the angle of reflection becomes $\frac{1}{2} \times {120}^{\circ} = {60}^{\circ}$

So we have $\theta = {60}^{\circ}$