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# Two mirrors meet at right angles. A ray of light is incident on one at an angle of $30^{\circ}$ as shown in Fig. 16.19. Draw the reflected ray from the second mirror.

"

__Given__:

Two mirrors meet at right angles and a ray of light is incident on one at an angle of $30^{\circ}$ as shown in Fig. 16.19.

__To do__:

To draw the reflected ray from the second mirror.

__Solution__:

Before drawing the reflected ray from the second mirror, let us observe the given fig. 16.19 and draw the reflected ray from the first mirror.

__Incident ray on the first mirror and reflected ray from the first mirror__:

Let us assume that a ray PQ is an incident on the first mirror at an angle of $30^{\circ}$ to the normal. Here $30^{\circ}$ is the angle of incidence.

According to the law of reflection of a light ray from a plane surface-

Angle of reflection$=$Angle of incidence$=30^{\circ}$

QR is the reflected ray from the first mirror at an angle of $30^{\circ}$ as shown in the figure below:

__Incident ray on the second mirror and reflected ray from the second mirror__:

Reflected ray QR from the first mirror is the incident ray for the second mirror. Let us find the angle of incidence of the ray QR by drawing a normal on the second mirror at R:

Both normals intersect at O at the angle of $90^{\circ}$. Let's use the angle sum property of a triangle in $\triangle OQR$ to find the $\angle$ R:

$\angle$ Q$+\angle R+\angle$O$=180^{\circ}$

Or $30^{\circ}+\angle$ R$+90^{\circ}=180^{\circ}$

Or $\angle$ R$=180^{\circ}-(30^{\circ}+90^{\circ})$

Or $\angle$R$=60^{\circ}$

Therefore, the angle of incidence in the second mirror $=60^{\circ}$

As per the law of reflection, the angle of reflection from the second mirror$=60^{\circ}$

RS is the reflected ray from the second mirror as shown in the diagram below:

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