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You observe a fish at an angle of = 20 to the water surface which is at an unknown depth Tfish below the water surface. However, you know that the fish is at a hori-zontal distance of 60cm from the incident light beam on the water surface. Due to the different refractive indices of air (1 = 1.0003) and water (2 = 1.333), the light beam is refracted at the surface of the water, causing you to underestimate the depthat which the fish is located. Calculate the depth T at which you assume the fish to be due to the refraction of light and the depth Tfish at which the fish is actually located.



Answer :

subhamjrp

Explanation:

A classic problem in optics!

Let's break it down step by step:

1. The angle of observation is 20° (θ1).

2. The horizontal distance from the incident light beam to the fish is 60 cm (d).

3. The refractive indices of air (n1) and water (n2) are given.

We can use Snell's law to relate the angles and refractive indices:

n1 sin(θ1) = n2 sin(θ2)

where θ2 is the angle of refraction in water.

First, we need to find the angle of refraction θ2:

θ2 = arcsin((n1 sin(θ1)) / n2)

θ2 ≈ 14.46° (using the given values)

Now, we can use the tangent function to relate the angles and distances:

tan(θ1) = d / T

tan(θ2) = d / Tfish

where T is the apparent depth (due to refraction) and Tfish is the actual depth.

Rearranging these equations, we get:

T = d / tan(θ1)

Tfish = d / tan(θ2)

Plugging in the values, we get:

T ≈ 34.64 cm (apparent depth)

Tfish ≈ 51.31 cm (actual depth)

So, due to refraction, you would underestimate the depth of the fish by about 16.67 cm (51.31 - 34.64).

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