Question

With the help of a ray diagram, obtain the relation between its focal length and radius of curvature.

Answer 1

The distance between the centre of a lens or curved mirror and its focus.
The relationship between the focal length f and the radius of curvature R = 2f.

Consider a ray of light AB, parallel to the principal axis and incident on a spherical mirror at point B. The normal to the surface at point B is CB, and CP = CB = R is the radius of curvature. The ray AB, after reflection from a mirror, will pass through F (concave mirror) or will appear to diverge from F (convex mirror) and obeys the law of reflection, i.e., i = r.

From the geometry of the figure,

∠BCP = θ = i

In DCBF, θ = r

∴ BF = FC (because i = r)

If the aperture of the mirror is small, B lies close to P,

∴ BF = PF

Or, FC = FP = PF

Or, PC = PF + FC = PF + PF

Or, R = 2 PF = 2f

Or, f = R/2

A similar relation holds for a convex mirror also. In deriving this relation, we have assumed that the aperture of the mirror is small.

Answer 2

A ray of light BP' travelling parallel to the principal axis PC is incident on a spherical mirror PP'. It reflects along P'R.

For a concave mirror, it passes through the focus. For a convex mirror, extending the ray backward, it appears to pass through the focus.

P is the pole and Fis the focus of the mirror.

PF = f

C is the centre of curvature.

PC = radius of curvature = R

P'C is the normal to the mirror at the point of incidence P'.

For a concave mirror,

∠BP'C = ∠P'CF = θ    ...[alternate angles]

and ∠BP'C = ∠CP'F = θ    ...[law of reflection, ∠i = ∠r]

Hence ∠P'CF = ∠CP'F

∴ ΔFP'C is isosceles.

Hence, P'F = FC

If the aperture of the mirror is small, the point P' is very close to the point P, then:

P'F = PF

∴ PF = FC

= `1/2 PC`

∴ f = `1/2 R`