Question

Derive the expression f = R\2 case of a concave mirror.​

Answer 1

I hope it will help you....

Answer 2

To Prove:

Focal length = (Radius of curvature)/2

Proof:

From the diagram, we can see that :

AB || CP and CB is transversal.

$\therefore \: \theta = i = r$

Again $\angle BFP = i + r = 2\theta$

Now, we can say that in ∆BPC (approximately a triangle):

$\therefore \: \tan( \theta) \approx \theta = \dfrac{BP}{CP} \: \: \: ......(1)$

Again , in ∆BFP (approximately a triangle):

$\therefore \: \tan( 2\theta) \approx 2\theta = \dfrac{BP}{FP} \: \: \: ......(2)$

Dividing eq.(2) by eq.(1):

$\therefore \: 2 = \dfrac{CP}{FP}$

$= > \: CP = 2 \times FP$

$= > \: c = 2 \times f$

[Hence Proved]