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Expression for Apparent Depth

Prove that Re...

Question

Prove that

Refractive index = $\frac{r e a l d e p t h}{a p p a r e n t d e p t h}$

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Solution

This can be proved by the following picture:

The refractive index of the medium a with respect to medium b is given by,

$_{b} μ_{a} = \frac{s i n i}{s i n r} \Rightarrow_{a} μ_{b} = \frac{s i n r}{s i n i}$

$I n △ O A B, s i n i = \frac{A B}{O B} I n △ O^{'} A B, s i n r = \frac{A B}{O^{'} B}$

$_{a} μ_{b} = \frac{\frac{A B}{O^{'} B}}{\frac{A B}{O B}} = \frac{A B}{O^{'} B} \times \frac{O B}{A B} = \frac{O B}{O^{'} B}$

As the points, A and B are very close, $O B \approx O A$ (real depth) and $O^{'} B \approx O^{'} A$ (apparent depth).

$_{a} μ_{b} = \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$

Hence proved

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Refractive index=real depthapparent depth

Q. Refractive Index of a medium =

\frac{R e a l d e p t h}{A p p a r e n t d e p t h}

\frac{Real Depth}{Apparent Depth} =

n_{12} \times n_{23} \times n_{31} = 1

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is the refractive index of medium-2 with respect to medium-1,

n_{23}

is the refractive index of medium-3 with respect to medium-2,

n_{31}

is the refractive index of medium-3 with respect to medium-1.

Q. Define critical angle. Prove that the refractive index of the denser medium is reciprocal of the sine of the critical angle .

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Expression for Apparent Depth

Standard X Physics

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