Question

A bird in air looks at fish vertically below it and inside water in a tank. If the distances of the fish as estimated by the bird is $h_{1}$ and that of bird as estimated by the fish is $h_{2},$ then the refractive index of liquid is :-

• A. $\frac{h_{2}}{ h_{1}}$
• B. $\frac{h_{1}}{h_{2}}$
• C. $\frac{h_{1} + h_{2}}{ h_{1} - h_{2}}$
• D. $\frac{h_{1} - h_{2}}{ h_{1} + h_{2}}$

Answer 1

Answer: (A)

Distance of fish seen by bird
$h_{1}=y_{1}+\frac{y_{2}}{\mu }$
Distance of bird seen by fish
$h_{2}=y_{2}+\mu y_{1}$
$\mu h_{1}=\mu y_{1}+y_{2}$
$h_{2}=y_{2}+\mu y_{1}$
$\mu h_{1}-h_{2}=0$
$\mu =\frac{h_{2}}{ h_{1}}$