Question

An insulated lamp bulb is on the bottom of a swimming pool at a point 2.5 m from a wall; the pool is 2.5 m deep and filled to the top with water (n = 4/3). At what angle is light refracted out of the water at the edge of the pool?A. 70.5° B. 25.1° C. 40.9° D. 30.4° E. 64.8°

Answer 1

option A

Step-by-step explanation:

given,

lamp position from pool wall = 2.5 m

height of the pool = 2.5 m

now,

`tan \theta = (P)/(B)`

`\theta =tan^(-1)((2.5)/(2.5))`

`\theta =45^0`

from the triangle

θ = i = 45°

using Snell's law

n₁ sin i = n₂ sin r

n₁ =4/3 n₂ = 1

now,

`(sin r)/(sin i)=(n_1)/(n_2)`

`(sin r)/(sin 45^0)=((4)/(3))/(1)`

`sin r=(1)/(√(2))* (4)/(3)`

r = sin⁻¹(0.9428)

r = 70.5°

hence, the correct answer is option A