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# Trigonometry Snell’s Law Help (page 2)

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By — McGraw-Hill Professional
Updated on Oct 3, 2011

## Snell's Law Practice Problems

#### Practice 1

Suppose a laser is placed beneath the surface of a freshwater pond. The refractive index of fresh water is approximately 1.33, while that of air is close to 1.00. Imagine that the surface is perfectly smooth. If the laser is aimed upwards so it strikes the surface at an angle of 30.0° relative to the normal, at what angle, also relative to the normal, will the beam emerge from the surface into the air?

#### Solution 1

Envision the situation in Fig. 10-5 “upside down.”

Fig. 10–5. Snell’s law for a light ray that strikes a boundary where the index of refraction decreases.

Then M r is the water and M s is the air. The indices of refraction are r = 1.33 and s = 1.00. The measure of angle θ is 30.0°. The unknown is the measure of angle ø Use the equation for Snell’s law, plug in the numbers, and solve for ø . You’ll need a calculator. Here’s how it goes:

#### Practice 2

What is the critical angle for light rays shining upwards from beneath a freshwater pond?

#### Solution 2

Use the formula for critical angle, and envision the scenario of Problem 10-2, where the laser angle of incidence, θ, can be varied. Plug in the numbers to the equation for critical angle, θ c :

Remember that the angles in all these situations are defined with respect to the normal to the surface, not with respect to the plane of the surface.

#### Practice 3

Suppose a laser is placed above the surface of a smooth, freshwater pool that is of uniform depth everywhere, and aimed downwards so the light ray strikes the surface at an angle of 28° relative to the plane of the surface. At what angle, relative to the plane of the pool bottom, will the light beam strike the bottom?

#### Solution 3

This situation is illustrated in Fig. 10-6. The angle of incidence, θ, is equal to 90° minus 28°, the angle at which the laser enters the water relative to the surface. That means θ = 62°. We know that r, the index of refraction of the air, is 1.00, and also that s, the index of refraction of the water, is 1.33. We can therefore solve for the angle ø , relative to the normal N to the surface, at which the ray travels under the water:

We’re justified to go to two significant figures here, because that is the extent of the accuracy of the angular data we’re given. The angle at which the laser travels under the water, relative to the water surface, is 90° – 42°, or 48°.

Fig. 10–6. Illustration for Solution 3.

Because the pool has a uniform depth, the bottom is parallel to the water surface. Therefore, by invoking the geometric rule for alternate interior angles, we can conclude that the light beam strikes the pool bottom at an angle of 48° with respect to the plane of the bottom.

Practice problems for these concepts can be found at: Reflection and Refracion Practice Test

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