Question

A fish in water (refractive index n) looks at a bird vertically above in the air If y is the height of the brid and x is the depth of the fish from the surface,then distance of thee bird as estimated by the fish is

• A. $x+y(1+\frac{1}{n})$
• B. $y+x(1-\frac{1}{n})$
• C. $x+y(1-\frac{1}{n})$
• D. $x+ny$

Answer 1

The correct option is D $x+ny$

If the object is in rarer medium and the fish is looked from

the denser medium, we know Normal shift $d=(n-1)h$,

where h represents the real depth.

Again I the question it is given that $h=y$.

Therefore apparent depth or the height of the bird from

the water surface is $y+(n-1)y=NY.$

Hence the total distance is $X+ny.$

The option $D$ is correct.