The Physics
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Opus in profectus

Refraction

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Practice

practice problem 1

Waves travel in all directions in the open ocean, but they always approach the land nearly perpendicular to the shore. Why does this happen?

solution

Ocean waves normally form when the wind grips the surface of the water and tries to drag it along. The friction at the interface gives the water a little tug and piles it up into a wave. Short burst of wind make little ripples and strong, steady winds make larger waves or swells. Regardless of size, waves generated by this means generally propagate in the direction that the wind is blowing. Since the wind can and does blow in every direction, waves can and do travel in any direction when they are formed.

Overhead cartoon representation of ocean waves approaching the shore

The speed of an ocean wave is affected by the depth of the water through which it is propagating. As the sea floor approaches the shore it rises and depth decreases. The shallower the water, the slower the wave speed. (The relationship is a complex one, but near shore speed is approximately proportional to the square root of depth.) Waves entering a medium with slower wave speed are refracted towards the normal. Where the sea floor rises suddenly, the refraction is abrupt. Where it rises gradually, the refraction is gradual. The closer a wave gets, the more perpendicular its propagation. The result is that most waves near the shore will eventually wind up heading nearly perpendicular to the shore no matter what direction they were traveling in initially.

practice problem 2

A ray of light is traveling from air to crown glass. The angle that this ray makes with the surface of the glass is 30°. Determine each of the following angles…
  1. the angle of incidence
  2. the angle of reflection
  3. the angle of refraction
  4. the angle between the reflected and refracted rays
  5. the angle between the incident and refracted rays
  6. the angle between the incident and reflected rays

solution

Start with a sketch. Let's make the surface horizontal, because we gotta pick something. Label one side air. Label the other side crown glass. Add an incident ray 30° above the surface. Add a normal to the surface at the point where the ray strikes the surface, since angles in geometric optics are measured from the normal. Sketch a reflected ray that looks symmetric to the incident ray, since it's obeying the law of reflection. Sketch in a transmitted ray that's refracted toward the normal, since the ray is entering a medium whe the speed of light is slower.

The set up

  1. The angle of incidence is measured from the normal, not the surface. Use the compliment of 30°

    90° − 30° = 60°

  2. The angle of reflection equals the angle of incidence. That's the law of reflection.

    60° = 60°

  3. Use Snell's law to compute the angle of refraction.

    n1 sin θ1 =  n2 sin θ2
    (1.00)(sin 60°) =  1.52 sin θ2
    θ2 =  35°
  4. Subtract the two angles on the right side of the normal from 180° to determine the angle between the reflected and refracted rays.

    180° − 60° − 35° = 85°

  5. Add the two angles adjacent to the right angle in the lower left corner to determine the angle between the incident and refracted rays.

    30° + 90° + 35° = 155°

  6. Double the angle of incidence to determine the angle between the incident and reflected rays.

    60° + 60° = 120°

The solution

practice problem 3

The diagram below shows a ray of light (A) incident upon an air-water interface. Several possible rays (B–K) are also shown.

Rays of light going into and out of a point on an interface with a protractor overlay

  1. Using the protractor in the diagram, measure the angle of incidence.
  2. Using the law of reflection, determine the angle of reflection, then select the lettered ray that best represents the reflected ray.
  3. Using Snell's law, determine the angle of refraction, then select the lettered ray that best represents the transmitted ray.

solution

  1. The angle of incidence is the angle between a line normal to the interface and the incident ray. Using the protractor supplied we get 53°. You can get this number by just counting off the tick marks, or by doing a little math. The incident ray is at 143° and the normal to the interface is at 90°.

    143° − 90° = 53°

  2. According to the law of reflection, the angle of reflection equals the angle of incidence. So once again the answer is 53°. Ray C best represents this direction.

  3. Snell's law is usually written like this.

    n1 sin θ1 = n2 sin θ2

    For this problem…

    n1 =  1.00 (air)
    θ1 =  53°
    n2 =  1.33 (water)
    θ2 =  ?

    Substituting these numbers…

    (1.00)(sin 53°) = 1.33 sin θ2

    gives us this answer…

    θ2 = 37°

    Ray G best represents this direction.

Rays of light going into and out of a point on an interface with a protractor overlay

practice problem 4

Write something completely different.

solution

Answer it.