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)

$x+ny$

Answer 2

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
$=y+(n-1) y=n y$
The total distance of the bird as estimated by fish is $x+n y$.