Question

You shine a flashlight on a pond. The light hits the pond at an angle of 70° to the normal of the pond. What is angle between the refracted ray entering the pond and the surface normal of the pond? The index of refraction of water is 4/3 Select One of the following: (a) 86.99 (b) 23.4° (c) 35 (d) 44.8 (e) Toby

Answer 1

(d) 44.8°

Step-by-step explanation:

For refraction,

Using Snell's law as:

`\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}`

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index of air which is 1

n₂ is the refractive index of water which is 4/3

Given that angle of incidence = 70°

So,

`\frac {sin\theta_2}{sin70}=\frac {1}{\frac {4}{3}}`

`{sin\theta_2}=0.7048`

Angle of refraction = sin⁻¹ 0.7048 = 44.8°.