Question

Prove the relation among Object Distance, Image Distance & Focal length of Concave Mirror

Answer 1

Answer:

All of them are of equal distance from the mirror

Answer 2

Answer:

From the above diagram we can write that triangle PCQ and pCq are similar.

∴ PQ/pq = CQ/Cq .........(1)

Again, ΔABF and Δpdf are similar.

∴AB/pq = BF/Fq

or PQ/pq = BF/Fq [ since AB = PQ ] ........(2)

From (1) and (2) we get,

CQ/Cq = BF/Fq

Since the mirror is of small aperture, it can be assumed,

BF=OF.

∴ CQ/Cq = OF/Fq ........(3)

Object distance, OQ = -u ,

Image distance, Oq = -v,

Focal length, OF = -f ,

Radius of curvature, OC = -r = -2f

∴ CQ = OQ - OC = -u + 2f

Cq = OC - Oq = -2f + v

Fq = Oq - OF = -v + f

From (3) we get,

$\frac{ - u + 2f}{ - 2f + v} = \frac{ - f}{ - v + f}$

or uv - uf - 2fv + 2f² = 2f² - fv

or uv = uf + vf

Dividing both sides by uvf we get,

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

or $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{r}$

$\huge{ \bold{ \underline{ \green{ \sf{HENCE \: PROVED}}}}}$