Question

prove that focal length of the concave mirror is half of the radius of curvature

Answer 1

Consider a Concave mirror as shown in the figure above.

A ray of light AB travelling parallel to the principal axis PC is incident on a concave mirror at B. After reflection, it goes through the focus F. P is the pole of the mirror. C is the centre of curvature.

The distance PF=focal length f.

The distance PC=radius of curvature R of the mirror.

BC is the normal to the mirror at the point of incidence B.

∠ABC=∠CBF (Law of reflection, ∠i=∠r)

∠ABC=∠BCF (alternate angles)

⇒∠BCF=∠CBF

∴ΔFBC is an isosceles triangle.

Hence, sides BF=FC

For a small aperture of the mirror, the point B is very close to the point P,

⇒BF=PF

∴PF=FC=1/2PC

⇒f=1/2R

Answer 2

to prove F=R/2

proof - Ray parallel to the principle axis strike at a point ‘A’ on the mirror.

To make a normal at any point on the spherical mirror a line passing through center of curvature is taken.

As per the rules of reflection: angle of incidence = angle of reflection

∠i = ∠r ........(1)

As OA and CP are parallel to each other, we get

∠ACF = i

And

∠FCA = i ............ from eq.1

∴ ∠ACF = ∠FCA = i

∴ CF = FA

When incident ray comes closer to principle axis FA = FP

∴ CF = FP

Then,

R = CP

R = CF + FP

R = f + f

∴ R = 2f