Question

When a light ray is incident on one of the two mirrors that are placed at right angle to each other, what can be said about the final reflected ray? It is given that the plane of incident ray and the the two normals is same.

• A. It is parallel to the incident ray and is in the same direction as that of the incident ray.
• B. It is parallel to the incident ray and is in the opposite direction as that of the incident ray.
• C. It is perpendicular to the incident ray and is in the same direction as that of the incident ray.
• D. It is perpendicular to the incident ray and is in the opposite direction as that of the incident ray.

Answer 1

The correct option is B

It is parallel to the incident ray and is in the opposite direction as that of the incident ray.

Let the light ray be incident at the point A with an incident angle equal to ${\theta }_1$.
From the figure, we see that the angle of reflection at point A is ${\theta }_1$ since the angles of incidence and reflection are equal.
Similarly, let the angle of incidence at point B be ${\theta }_2$. The angle of reflection there is ${\theta }_2$.

$\angle BAO={90}^{\circ }-{\theta }_1$
$\angle ABO={90}^{\circ }-{\theta }_2$

Sum of all the angles in a triangle is ${180}^{\circ }$.

$\angle BAO+\angle ABO+\angle AOB={180}^{\circ }$
${90}^{\circ }–{\theta }_1+{90}^{\circ }–{\theta }_2+{90}^{\circ }={180}^{\circ }$
${\theta }_1+{\theta }_2={90}^{\circ }$

Now the initial and the final rays can be considered as two lines, intercepted by the transversal which is the intermediate ray (that goes from A to B).

The sum of adjacent interior angles here is ${180}^{\circ }$ as shown below.
$2\times ({\theta }_1+{\theta }_2)=2\times {90}^{\circ }={180}^{\circ }$

Since the adjacent interior angles are supplementary, the initial and the final rays are parallel to each other.

Also, the direction of the final ray is opposite that of the initial incident ray.