Question
Prove that
Refractive index $=\frac{\text { Real depth }}{\text { Apparent depth }}$
Refractive index $=\frac{\text { Real depth }}{\text { Apparent depth }}$
✓
Answer
Consider a ray of light incident normally along OA. It passes straight along OAA'. Consider another ray from $O$ (the object) incident at an angle $i$ along $OB$. This ray gets refracted and passes along $BC$. On producing this ray BC backward, it appears to come from point I, and hence, Al represents the apparent depth, which is less than the real depth $A O$.
Since $AO$ and $BNI$ are parallel and $OB$ is the transversal
$\angle AOB =\angle OBNI$ (Alternate angles)
$\angle BIAI =\angle CBN$ (Corresponding angles)
In $\triangle B A O, \sin i=\frac{B A}{O B}$
In $\triangle IAB , \sin r=\frac{ BA }{ IB }$
We know that refractive index of air w.r.t the medium
$a \mu m=\frac{\sin i}{\sin r}=\frac{( BA / OB )}{( BA / IB )}=\frac{I B}{O B}$
$\therefore$ Refractive index of medium w.r.t. air is,
$a \mu m=\frac{1}{m \mu a}=\frac{O B}{I B}$
Since point $B$ is very close to point $A$, i.e the object is viewed from a point vertically above the object.
$\therefore I B=I A$ and $O B=O A$.
Hence, $a \mu m=\frac{O A}{I A}=\frac{\text { Real depth }}{\text { Apparent depth }}$.
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