Question

A flashlight at the bottom of a 4.00m deep swimming pool sends a ray upward and at an angle such that the ray strikes the surface of the water 1.50 meters from a point directly above the flashlight. What angle in air does the emerging ray make with the water

Answer 1

Answer:

62°

Explanation:

h = 4 m

x = 1.5 m

refractive index of water, n = 4/3

According to the diagram

tanθ = 4/1.5

θ = 69.44°

Angle of incidence = i = 90° - 69.44° = 20.56°

Let the angle of refraction is r.

Use Snell's law

$\frac{1}{n}=\frac{Sin i}{Sin r}$

$\frac{3}{4}=\frac{Sin 20.56}{Sin r}$

Sinr = 0.468

r = 28°

Thus, the emerging ray makes an angle with water = 90° - 28° = 62°.

Answer 2

The angle in air which the emerging ray make with the water is equal to 28°.

Given the following data:

• Height at bottom = 4.00 m.

• Height at surface of the water = 1.50 meters.

Note: The refractive index of water (n) = 4/3

What is Snell's law?

Snell's law gives the relationship between the angle of incidence and angle of refraction with respect to light passing through a media such as water, glass, or air.

From the ray diagram, we have:

tanθ = 4/1.5

θ = tan⁻¹(2.6667)

θ = 69.44°

Angle of incidence, i = 90° - 69.44°

Angle of incidence, i  = 20.56°

Mathematically, Snell's law is given by this formula:

1/n = sini/sinr

Substituting the given parameters into the formula, we have;

3/4 = sin20.56/sinr

sinr = 0.468

r = sin⁻¹(0.468)

r = 28°

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