Question #89ca2

1 Answer
Mar 5, 2017

![images.tutorvista.com](useruploads.socratic.org)
Consider a Concave mirror as shown in the figure above.
A ray of light ABAB traveling parallel to the principal axis PCPC is incident on a convex mirror at BB. After reflection, it goes through the focus FF. PP is the pole of the mirror. CC is the center of curvature.

The distance PF=PF= focal length ff.
The distance PC=PC= radius of curvature RR of the mirror.
BCBC is the normal to the mirror at the point of incidence BB.

∠ABC = ∠CBFABC=CBF (Law of reflection, ∠i=∠ri=r)
∠ABC = ∠BCFABC=BCF (alternate angles)
=> ∠BCF = ∠CBFBCF=CBF
∴ Delta FBC 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= 1/2 PC
=> f = 1/2 R

Now consider a Convex mirror as shown in the figure below.

![tutorvista.com](useruploads.socratic.org)

A ray of light AB traveling parallel to the principal axis PC is incident on a convex mirror at B. After reflection, it goes to D and appear to be coming from the focus F.

The distance PF= focal length f.
The distance PC= radius of curvature R of the mirror.
Straight line NBC is the normal to the mirror at the point of incidence B.

∠ABN = ∠NBD (Law of reflection, ∠i=∠r)
∠CBF = ∠DBN (vertically opposite angles)
angleNBA=angleBCF (corresponding angles)
=> ∠BCF = ∠CBF
∴ Delta FBC 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= 1/2 PC
=> f = 1/2 R

Thus, for a spherical mirror (both for a concave and for convex), the focal length is half of radius of curvature.