Question

Write the relation between focal length and radius of curvature of a spherical mirror?

Answer 1

For a concave mirror:

In figure (a),

$\angle B{P}^{\prime }C=\angle P^{\prime }CF$ (alternate angles) and

$\angle B{P}^{\prime }C=\angle P^{\prime }F$ (law of reflection,$\angle i=\angle r$)

Hence $\angle P^{\prime }CF=\angle C{P}^{\prime }F$

∴ $F{P}^{\prime }C$ is isosceles.

Hence, $P^{\prime }F=FC$

If the aperture of the mirror is small, the point $P^{\prime }$ is very close to the point $P$,

then $P^{\prime }F=PF$

∴ $PF=FC=\frac{PC}{2}$

or $f=\frac{R}{2}$

For a convex mirror:

In figure (b),

$\angle B{P}^{\prime }N=\angle FC{P}^{\prime }$ (corresponding angles)

$\angle B{P}^{\prime }N=\angle N{P}^{\prime }R$ (law of reflection, $\angle i=\angle r$) and

$\angle N{P}^{\prime }R=\angle C{P}^{\prime }F$ (vertically opposite angles)

Hence $\angle FC{P}^{\prime }=\angle C{P}^{\prime }F$

∴ $F{P}^{\prime }C$ is isosceles.

Hence, $P^{\prime }F=FC$

If the aperture of the mirror is small, the point $P^{\prime }$ is very close to the point $P$.

Then $P^{\prime }F=PF$

∴$PF=FC=\frac{PC}{2}$

or $f=\frac{R}{2}$

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