Question

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. $y+x\left(1-\frac{1}{n}\right)$

Answer 1

Answer: (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 $x+ny$.