Question

A beam of light from a flashlight at the bottom of a swimming pool is incident on the surface of the pool at an angle of 25 degrees to the normal. find the reflected angle in the water and refracted angle for the light entering air above the pool. n=1 for air and n=1.33 for water

Answer 1

The reflected angle is 25°.

The angle of refraction = 34.05°.

Step-by-step explanation:

For reflection,

Angle of incidence = Angle of the reflection.

Thus,

Given that the angle of incidence is 25°

So, the reflected angle is 25°.

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 water which is 1.33

n₂ is the refractive index of air which is 1

So,

`\frac {sin\theta_2}{sin25}=\frac {1.33}{1}`

`{sin\theta_2}=0.56`

Angle of refraction = sin⁻¹ 0.56 = 34.05°.