Derive a relation between the focal length and radius of curvature stating the assumptions made in the case of a spherical mirror.
Assumptions made :
(a) Small aperture approximation:} Aperture of the spherical mirror is assumed to be small.
(b) Small angle approximation: Incident ray makes a very small angle with the principal axis and strikes the reflecting surface close to the pole of the mirror. Such a ray is called paraxial ray.
(c) Mirror is made of thin refracting material to avoid multiple reflecting.
Derivation:
Let a ray OA, travelling parallel to the principal axis, incident on a concave mirror at point A (Figure 11). After reflection, the ray passes through the focus (F) of the mirror. CA is the normal to the mirror at A.
According to the law of reflection,
$i = r = \theta$
Also,
$\angle AFP$ is the external angle of $\triangle ACF$, so
$\angle AFP = \angle ACF + \angle CAF$
$\angle AFP= \theta + \theta = 2\theta$
Now draw AN perpendicular to the principal axis.
From right angled $\triangle ANC$,
$\tan \theta = \frac{AN}{NC}$
$\theta$ is small, $ \tan \theta = \theta$
$\theta = \frac{AN}{NC}$
Also From right angled $\triangle ANF$,
$\tan 2\theta = \frac{AN}{NF}$
As $\theta$ is small,
so $\tan 2\theta = 2\theta$
$2\theta = \frac{AN}{NF}$
From equations (i) and (ii), we get,
$2\frac{AN}{NC} = \frac{AN}{NF}$
NC = 2NF
As aperture of the mirror is small, so point N lies very close to P.
$NF = PF$ and $NC = PC$
$PC = 2PF$...(iii)
Using New Cartesian Sign convention we find
$PC = -R$ and $PF = - f$
Hence , equation (iii) becomes
$R = 2f$
$f = \frac{R}{2}$