Question

$\left(a\right)$ Find the angle of incidence.
$\left(b\right)$ Find the angle between the incident and reflected rays.

Figure shows an incident ray $AO$ and the normal $ON$ on a plane mirror. The angle which the incident ray $AO$ makes with a mirror is ${30}^{\circ }$.

• A. $\left(a\right){60}^{\circ },\left(b\right){60}^{\circ }$
• B. $\left(a\right){120}^{\circ },\left(b\right){120}^{\circ }$
• C. $\left(a\right){60}^{\circ },\left(b\right){120}^{\circ }$
• D. $\left(a\right){120}^{\circ },\left(b\right){60}^{\circ }$

Answer 1

Answer: (C)

$\left(a\right){60}^{\circ },\left(b\right){120}^{\circ }$

We know that angle of incidence and angle of reflection both are measured from normal which is perpendicular to the plane.
Let,
Angle of incidence $=i$
Angle of reflection $=r$
Normal ON is perpendicular to plane.
So, angle of incidence, $i={90}^0-{30}^0={60}^0$
We know that angle of incidence$=$angle of reflection
So, angle of incidence$=$angle of reflection $=$${60}^0$
Hence, angle between incident ray(AO) and reflected ray(OB) is $=i+r={60}^0+{60}^0={120}^0$