Question

When viewed vertically a fish appears to be 4 meter below the surface of the lake. If the index of refraction of water is 1.33, then the true dept of the fish is

• A. 5.32 metres
• B. 3.32 metres
• C. 4.32 metres
• D. 6.32 metres

Answer 1

The correct option is A 5.32 metres

Here the apparent depth of fish is $d_{apparent}=4m$

and we know that

$refractiveindex=\frac{actualdepth}{apparentdepth}\mu =\frac{d_{actual}}{d_{apparent}}$

thus by putting the values we get,

$1.33=\frac{d_{actual}}{4}$

$d_{actual}=5.32m$

the actual depth of fish is 5.32 m

Option A is correct.