Question 89ca2

Mar 5, 2017 Consider a Concave mirror as shown in the figure above.
A ray of light $A B$ traveling parallel to the principal axis $P C$ 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 $P F =$ focal length $f$.
The distance $P C =$ radius of curvature $R$ of the mirror.
$B C$ 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 $B F = F C$

For a small aperture of the mirror, the point $B$ is very close to the point $P$,
$\implies B F = P F$
∴ PF = FC= 1/2 PC
$\implies f = \frac{1}{2} R$

Now consider a Convex mirror as shown in the figure below. A ray of light $A B$ traveling parallel to the principal axis $P C$ 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 $P F =$ focal length $f$.
The distance $P C =$ radius of curvature $R$ of the mirror.
Straight line $N B C$ is the normal to the mirror at the point of incidence $B$.

∠ABN = ∠NBD (Law of reflection, ∠i=∠r)
∠CBF = ∠DBN (vertically opposite angles)
$\angle N B A = \angle B C F$ (corresponding angles)
=> ∠BCF = ∠CBF
∴ Delta FBC is an isosceles triangle.
Hence, sides $B F = F C$

For a small aperture of the mirror, the point $B$ is very close to the point $P$,
$\implies B F = P F$
∴ PF = FC= 1/2 PC#
$\implies f = \frac{1}{2} R$

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