A ray of light parallel to the principal axis falls on a concave mirror of radius of curvature 50 cm. It meets the mirror at a point 1 cm above the principal axis. The angle between the reflected ray and the principal axis is about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.04 rad
Let the ray meet the mirror at M, and let MD be its height above the axis. The normal at M is the line CM through the centre of curvature. So the angle of incidence is θ = ∠MCP. The reflected ray then crosses the axis at F, and it makes an angle 2θ with the axis. The ray is paraxial, so θ is small and \( \tan\theta\approx\theta \).
\( MD=1\ \mathrm{cm} \)
\( R=50\ \mathrm{cm} \)
For a paraxial ray, D is very close to P, so \( CD\approx R \).
\( \begin{aligned} \theta &\approx \frac{MD}{CD} \\ &= \frac{1}{50} \\ &= 0.02\ \mathrm{rad} \end{aligned} \)
\( \begin{aligned} \angle MFP &= 2\theta \\ &= 2\times0.02 \\ &= 0.04\ \mathrm{rad} \end{aligned} \)
As a check, the ray crosses the axis about 1/0.04 = 25 cm from the mirror, which is \( R/2 \). Hence, the reflected ray makes an angle of about 0.04 rad with the principal axis.
NCERT (2026-27):
Focal length of spherical mirrors – “Let C be the centre of curvature of the mirror. Consider a ray parallel to the principal axis striking the mirror at M. Then CM will be perpendicular to the mirror at M. Let θ be the angle of incidence, and MD … be the perpendicular from M on the principal axis. Then, ∠MCP = θ and ∠MFP = 2θ”
“Now, \( \tan\theta=\frac{MD}{CD} \) and \( \tan 2\theta=\frac{MD}{FD} \qquad (9.1) \)”
“For small θ, which is true for paraxial rays, tanθ ≈ θ, tan 2θ ≈ 2θ.”