Proove $Refractive index=real depthapparent depth$ .

Open in App

Solution

The refractive index of medium a with respect to medium b is given by,
$_{b} μ_{a} = \frac{s i n i}{s i n r} \Rightarrow_{a} μ_{b} = \frac{s i n r}{s i n i}$
$I n Δ O A B, s i n i = \frac{A B}{O B}; I n Δ O^{'} A B, s i n r = \frac{A B}{O^{'} B}$
$∴_{a} μ_{b} = \frac{\frac{A B}{O^{'} B}}{\frac{A B}{O B}} = \frac{A B}{O^{'} B} \times \frac{O B}{A B} = \frac{O B}{O^{'} B}$
As the point A and B are very close, $O B \approx O A$ (real depth) and $O^{'} B \approx O^{'} A$ (apparent depth).
$∴_{a} μ_{b} = \frac{R e a l d e p t h}{A p p a r e n t d e p t h}$

Suggest Corrections

Similar questions

Prove that

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

\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} =

Join BYJU'S Learning Program