A light ray is incident at $30^{\circ}$ on one of the two mirrors that are placed at right angle to each other. Find the angle of reflection for the ray reflected from mirror B.

20^{\circ}

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30^{\circ}

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90^{\circ}

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60^{\circ}

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Solution

The correct option is D $60^{\circ}$
Let the angle of incidence and angle of reflection on mirror A be $i_{1}$ and $r_{1}$ and on mirror B be $i_{2}$ and $r_{2}$ respectively. We know that the law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
$\Rightarrow ∠ i_{1} = ∠ r_{1} = 30^{\circ}$

On Mirror B
Angle between surface B and the incident ray

$∴ ∠ x + 90^{\circ} + 60^{\circ} = 180^{\circ}$
$\Rightarrow ∠ x = 180^{\circ} - 150^{\circ} = 30^{\circ}$

Angle between normal of surface B and the incident ray is $60^{\circ}$ .

From first law of reflection
$\Rightarrow ∠ i_{2} = ∠ r_{2} = 60^{\circ}$

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Question 6

Two mirrors A and B are placed at right angles to each other as shown in the figure.

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