Expression for Real depth and apparent depth formula refractive index

Show that the bottom of a water tank appears to be raised. Hence find an expression for the normal shift in the position of an object placed in a denser medium.

Consider a tank filled with water upto the level PQ. Let an object O (say a coin) lies at the bottom of the water tank. The depth AO = $t$ of the object is known as the real depth. the object appears at position I instead of O when viewed obliquely.

The depth AI is known as apparent depth of the object O. 

According to Snell's law,

${ }^{w}n_{a} = \frac{\sin i}{\sin r} \quad \dots (i)$

From $\Delta AOC$, 

$\sin i = \frac{AC}{OC}$ and 

from $\Delta AIC$, 

$\sin r = \frac{AC}{IC}$

Substituting the values of $\sin i$ and $\sin r$ in equation (i), we get

${ }^{w}n_{a} = \frac{\frac{AC}{OC}}{\frac{AC}{IC}}$

${ }^{w}n_{a} = \frac{AC}{OC} \times \frac{IC}{AC} = \frac{IC}{OC}$

Since point C lies very close to A, so IC $\approx$ AI and OC $\approx$ AO

${ }^{w}n_{a} = \frac{AI}{AO}$

${ }^{a}n_{w} = \frac{1}{{ }^{w}n_{a}}$

Since

${ }^{a}n_{w} = \frac{AO}{AI} = \frac{\text{Real depth}}{\text{Apparent depth}} \quad \dots (iii)$

In general,

${ }^{1}n_{2} = \frac{\text{Real depth}}{\text{Apparent depth}} \quad \dots (iv)$

where ${ }^{1}n_{2}$ is the refractive index of the denser medium w.r.t. (with respect to) rarer medium.

Since refractive index of water w.r.t. air is greater than 1 ($^{a}n_{w} > 1$) therefore, apparent depth is less than the real depth.

Hence, the bottom of water tank appears to be raised}.

Normal Shift in the position of the object : 

normal shift is given by is given by,

$x = AO - AI = AO \left(1 - \frac{AI}{AO}\right)$

$x = AO \left(1 - \frac{1}{{ }^{a}n_{w}}\right)

$x = t \left(1 - \frac{1}{{ }^{a}n_{w}}\right)$

$x = t \left(1 - \frac{1}{{ }^{1}n_{2}}\right)$

In general,

$x = t \left(1 - \frac{1}{n}\right) \quad \dots (v)$

where $t$ is the real depth and $n$ is the absolute refractive index of the medium.