# How do they derive the apparent-real depth formula?

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? $$^an_w=\frac{\mathrm{real\ depth}}{\mathrm{apparent\ depth}}=\frac{4}{3}$$ $$\mathrm{apparent\ depth}=\frac{3}{4}\mathrm{real\ depth}$$

• Can you post a picture of the page? – Alec Teal Oct 7 '15 at 19:50
• @AlecTeal added – Aaryan Dewan Oct 7 '15 at 19:55
• They draw some triangles. Almost everything involved in ray optics is done with a combination of triangles and algebra. Because rays move in straight lines except under specific circumstances and there are lots of theorems abut triangles. – dmckee Oct 7 '15 at 20:11

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, $$^a\mu_w=\frac{MA}{LA} =\frac{real~depth}{apparent~depth}$$

• Can someone help me add an image The image tab is not showing the option for adding image due to poor browser quality. – Sikander May 17 '16 at 6:05

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