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

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  • $\begingroup$ Can you post a picture of the page? $\endgroup$ – Alec Teal Oct 7 '15 at 19:50
  • $\begingroup$ @AlecTeal added $\endgroup$ – Aaryan Dewan Oct 7 '15 at 19:55
  • $\begingroup$ 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. $\endgroup$ – dmckee Oct 7 '15 at 20:11
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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, $$^a\mu_w=\frac{MA}{LA} =\frac{real~depth}{apparent~depth}$$

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

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