Prove that,

$\text {Refractive index} = \dfrac {\text {Real depth}}{\text {Apparent depth}}$

Consider a point object O kept at the bottom of a transparent medium ( such as water or glass ) separated from air by the surface PQ.

A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA'. Another ray OB, starting along the object O, strikes the boundary surface PQ at B and suffers refraction.

Since, the ray travels from denser to rarer medium so it bends away from normal N'BN drawn at the point of incidence B on the surface PQ and travels along BC in air.

When viewed by the eye , the ray BC appears to be coming from point I which is the virtual image of O, obtained on producing A'A and CB backwards.

Thus, any point seen from air will appear to be at I, which is lesser depth = AI than its actual depth AO.

Angle of incidence = ∠OBN'

Angle of refraction = ∠CBN.

Since, AO and BN’ are parallel and OB is transversal line, so

∠AOB = ∠OBN' = i

Similarly, IA’ and BN are parallel and IC is the transversal line, so

∠BIA’ = ∠CBN = r

In right-angle triangle BAO,

$\text {sin i} = \dfrac{BA}{OB}$

and,

In right-angle triangle IAB,

$\text {sin r} = \dfrac{BA}{IB}$

For refraction from medium to air, by Snell’s law

$_mμ _a = \dfrac{\text {sin i}}{\text {sin r}} = \dfrac {\dfrac {BA}{OB}}{\dfrac {BA}{IB}} \\[0.5em] \Rightarrow _mμ _a = \dfrac{IB}{OB} \\[0.5em]$ Hence, refractive index of medium with respect to air is,

$_aμ _m = \dfrac{1}{$mμa} = \dfrac{OB}{IB} \\[0.5em]

The object is viewed from a point vertically above the object O, since point B is very close to the point A. ∴ IB = IA and OB = OA

Hence,

$_aμ _m = \dfrac{1}{$mμa} = \dfrac{OA}{IA} \\[0.5em] \Rightarrow \dfrac{\text {Real depth}}{\text {Apparent depth}} \\[0.5em]