Q. A fish in water (refractive index$\ n$) looks at a bird vertically above in the air.If $y$ is the height of the bird and $x$ is the depth of the fish from the surface, then the distance of the bird as estimated by the fish is
A) $x\ + \ y\ \left( 1\ –\ \frac{1}{n} \right)$
B) $x\ + \ ny$
C) $x\ + \ y\ \left( 1 + \ \frac{1}{n} \right)$
D) $y\ + \ x\ \left( 1\ - \ \frac{1}{n} \right)$
Explanation
(b) When object is in rarer medium and observer is in denser medium.
Normal shift, $d = (n - 1)h$
Where $h = \ $real depth
Here, $h = y$
Now, apparent depth or the apparent height of the bird from the surface of the water $= y + (n - 1)y = ny$
The total distance of the bird as estimated by fish is$x + ny$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.