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 h1 and that of bird as estimated by the fish is h2, then the refractive index of liquid is :-

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 h1 and that of bird as estimated by the fish is h2, then the refractive index of liquid is :- (A) (h2/ h1) (B) (h1/h2) (C) (h1 + h2/ h1 - h2) (D) (h1 - h2/ h1 + h2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-l3r1r2kyt5l12aqo.jpg" /> Distance of fish seen by bird h1=y1+(y2/μ ) Distance of bird seen by fish h2=y2+μ y1 μ h1=μ y1+y2 h2=y2+μ y1 μ h1-h2=0 μ =(h2/ h1)

Q. 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 :-

$\frac{h _{2}}{h _{1}}$

$\frac{h _{1}}{h _{2}}$

$\frac{h _{1} + h _{2}}{h _{1} - h _{2}}$

$\frac{h _{1} - h _{2}}{h _{1} + h _{2}}$

Distance of fish seen by bird
$h_{1} = y_{1} + \frac{y _{2}}{μ}$
Distance of bird seen by fish
$h_{2} = y_{2} + μ y_{1}$
$μ h_{1} = μ y_{1} + y_{2}$
$h_{2} = y_{2} + μ y_{1}$
$μ h_{1} - h_{2} = 0$
$μ = \frac{h _{2}}{h _{1}}$

Q. 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 h1​ and that of bird as estimated by the fish is h2​, then the refractive index of liquid is :-

h1​h2​​

h2​h1​​

h1​−h2​h1​+h2​​

h1​+h2​h1​−h2​​