Question

Prove that
Refractive index =
real depth/
apparent depth​

Answer 1

Consider a ray of light incident normally along OA. It passes straight along

OAA’. Consider another ray from O (the object), incident at an angle i along OB. This ray gets refracted and passes along BC. On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO.

Since, AO and $\large\rm {BN^{I} }$ are parallel and OB is transerval,

$\large\rm { \angle AOB = \angle OBN^{I}}$ (alternate angles)

$\large\rm { \angle BIA^{I} = \angle CBN }$ (corresponding angles)

$\large\rm { In \ \triangle BAO , \sin \ i = \frac {BA}{OB}}$

$\large\rm { In \ \triangle IAB , \sin \ r = \frac {BA}{IB}}$

We know that refractive index of air w.r.t the medium

$\large\rm { _{a} \mu_{m} = \frac { \sin \ i}{ \sin \ r} = \frac {IB}{OB}}$

$\large \therefore$ Refracting index of medium w.r.t air is,

$\large\rm { _{a} \mu_{m} = \frac {1}{ _{m} \mu_{a}} = \frac {OB}{IB}}$

Since the point B is very close to point A, i.e. the object viewed from a point vertically above the object.

$\large\rm { \therefore IB=IA \ and \ OB=OA}$

Hence, Refractive index = real depth/apparent depth