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

Answer:

option A

Explanation:

given,

lamp position from pool wall = 2.5 m

height of the pool = 2.5 m

now,

$tan \theta = \dfrac{P}{B}$

$\theta =tan^{-1}(\dfrac{2.5}{2.5})$

$\theta =45^0$

from the triangle

θ = i = 45°

using Snell's law

n₁ sin i = n₂ sin r

n₁ =4/3     n₂ = 1  

now,

$\dfrac{sin r}{sin i}=\dfrac{n_1}{n_2}$

$\dfrac{sin r}{sin 45^0}=\dfrac{\dfrac{4}{3}}{1}$

$sin r=\dfrac{1}{\sqrt{2}}\times \dfrac{4}{3}$

r = sin⁻¹(0.9428)

r = 70.5°

hence, the correct answer is option A