Question

9. A ray of light LM is incident on a mirror as shown in the figure. The angle of incidence is half of the angle between ray LM and the line joining two other points in the figure. What are these two points? nav M Р F C A. M and F B. M and C C. P and F D. P and C​

Answer 1

The angle of incidence is half of the angle between ray LM and the line joining two other points in the figure. These points are A). M and F.

• The angle of incidence is the angle made by the incident ray with the normal to the point of incidence.
• The point of incidence, in this case, is M while C is the centre of the curvature of the concave mirror shown here. Hence, the perpendicular to point M is MC.
• Thus, the angle of incidence is ∠LMC₁. (Refer to figure attached).
• Here, it is asked that the angle of incidence is half of angle X. So, we need to find an angle that is twice the value of angle of incidence, i.e., ∠LMF₁.
• Let us consider the triangle Δ CMP.
• As we know F is at half of the radius of curvature (CP). Thus, MF is the bisector of side CP in the Δ CMP.

$Hence, \angle PMF = \angle FMC.$

$In \triangle\: MFC, MF = FC \:\:(radius\: of\: circle)$

$\therefore \angle FMC = \angle FCM$

$Now, \angle FMC = \angle F_{1}MC_{1}\:(Alternate\:\: angles)$

$Also, \angle FCM = \angle LMC_{1} \:(Corresponding\: angles)$

$Thus, \angle F_{1}MC_{1} = \angle LMC_{1}.$

Thus, ∠ NMC₁ (angle of incidence) = $\frac{1}{2}$ ∠ LMF₁ (angle formed between ray LM and the line joining MF).