Derive Relation Between Focal Length and Radius of Curvature - Param Himalaya

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 

$R = 2f$

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