Real and Apparent depth Numerical problem – solved

Let’s solve a numerical based on real depth, apparent depth, and refractive index. Also, you will get to know about the RI formula to be used to solve this.

Real and Apparent depth Numerical problem – with solution

Question 1] A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water’s surface. How deep should the crane immerse its beak to pick that fish?
For the fish, how much above the water’s surface does the crane appear? The refractive index of water = 4/3.

Solution: For the crane, the apparent depth of the fish is 6 cm and the real depth is to be determined.
For fish, the real depth (height, in this case) of the crane is 6 m and the apparent depth (height) is to
be determined.

Formula to be used: n = R/A = Real Depth / Apparent depth
[Read our post on: How to prove that RI = real depth/apparent depth]

For crane, it is water with respect to air as real depth is in water and apparent depth is as seen from air. Here, apparent depth of fish is 6 cm.

n = 4/3
=> R/A = 4/3
=> R/6 = 4/3
=> R = 8 cm

So, real depth of the fish is 8 cm

For fish, it is air with respect to water as the real height is in air and seen from the water.

In this case, n will be 3/4. Real height of crane = 6 m.

n = 3/4
=> R/A = 3/4
=> 6/A = 3/4
=> A = 8 m

So, Apparent height of the crane to the fish is 8 m