Question #89ca2

1 Answer
Mar 5, 2017

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Consider a Concave mirror as shown in the figure above.
A ray of light AB traveling parallel to the principal axis PC is incident on a convex mirror at B. After reflection, it goes through the focus F. P is the pole of the mirror. C is the center 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
∴ 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.

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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.