We know that the refractive index of water to air is 4/3 . So,
n=4/3
which should be n = 4/3 = n2/n1 , but my book says,
4/3= real depth/apparent depth.
How have they put the “depths” in the formula and Why? $$^an_w=\frac{\mathrm{real\ depth}}{\mathrm{apparent\ depth}}=\frac{4}{3}$$ $$\mathrm{apparent\ depth}=\frac{3}{4}\mathrm{real\ depth}$$
We can prove this if you dont understand the formula.
Suppose $M$ is a point object at an actual depth $MA$ below the free surface of water $XY$ in a tank.
A ray of light incident on $XY$ normally along $MA$ passes straight along $MAA'$.Another ray of light from $M$ incident at $\angle i$ on $XY$,along $MB$ gets deviated away from normal and is refracted at $\angle r$ along $BC$.If we produce $BC$ we will find that it meets $OA$ at $L$.Therefore $L$ is virtual image of $M$ which appears when we see from $C$.Now the apparent depth is $AL$.
$$\angle AMB=\angle MBN'$$ $$\angle ALB= \angle NBC$$ In ∆$AMB$, $$\sin i= \frac{AB}{MB}$$
In ∆$IAB$,$$\sin r=\frac{AB}{LB}$$
Now, $$^a\mu_w =\frac{AB}{LB}×\frac{MB}{AB}=\frac{MB}{LB}$$
Suppose that $\angle i\rightarrow0$ then B will near A
Therefore, $$^a\mu_w=\frac{MA}{LA} =\frac{real~depth}{apparent~depth}$$
The picture in this question may be helpful to visualize the situation:
Basically, if you observe an object immersed at some (real) depth $h$ in water, as a consequence of the refraction of light you will see it as if it were at a different (apparent) depth $h'$. The relation between $h$ and $h'$ can b easily found geometrically in terms of the angle of incidence ($\theta_{water}$) and angle of refraction ($\theta_{air}$). The Snell formula, namely $$ n_{water} \sin \theta_{water} = n_{air} \sin \theta_{air} $$ finally allows you to express the relation between $h$ and $h'$ in terms of the refractive indices of two mediums
