physicsa.com
 class code: SPS22 teacher: Mr. Elert classroom: A314 office: A214 test day: Friday phone: (718) 724–8500 ext. 2141 email: elert@midwoodscience.org

## Physics A: Problem Set 18: Refraction

 Barron's Let's Review: 11.6 Refraction, 12.6 Refraction physics.info: Refraction Wikipedia: Refraction, Snell's law, Total internal reflection HyperPhysics: Refraction of Light Khan Academy: Reflection and refraction YouTube: Upside Down Glasses Challenge

### classwork

1. inverting-prism.pdf
The diagram below shows 11 parallel rays of light incident on a trapezoidal, crown glass prism. Choose one of the numbered rays at random. On the diagram, trace the transmitted portion of your ray until it emerges from the prism. Just as each incident ray was assigned a number, each emergent ray can be assigned a number. Determine this number.

inverting-prism-key.pdf
The diagram below shows the solutions. Basically, each numbered ray is mapped onto its inversion from 12. 1 goes through 11, 2 goes through 10, 3 goes through 9, etc.

2. refraction-drill.pdf
The diagrams on the accompanying pdf show a ray of light incident upon an interface. In some cases the ray is traveling through air and entering glass. In other cases it is traveling through glass and entering air. On each diagram, sketch the approximate path of the light in the second medium. Ignore any reflected light. It is not necessary to calculate any angles, but do clearly show the change in direction of the rays, if any, at each surface.

refraction-drill-key.pdf
The solutions are on a separate document.

### homework

1. Identify a likely material if the speed of light (v) in the material is the following fraction of the speed of light (c) in a vacuum…
1. v = ⅔c
2. v = ¾c
3. v = ⅗c

Use the index of refraction equation.

 n = c v

Replace the denominator with a fraction of the speed of light. The speed of light cancels out and the original fraction inverts. Compute this fraction as a decimal. Pick the substance with the closest index of refraction. Do it three times.

1. This is probably lucite.

 n = c = 3 = 1.50 ⅔c 2
2. This is probably water.

 n = c = 4 = 1.33 ¾c 3
3. This is probably flint glass.

 n = c = 5 = 1.66 ⅗c 3
2. Three related questions.
1. Clean air is effectively invisible and so is clean glass, which is why people sometimes walk into freshly washed glass doors. How is it possible to see a clean glass object in clean air if both materials are effectively invisible? (Think about an empty drinking glass or an empty glass jar instead of a thin, flat window.)
2. Pyrex is a brand name for the type of glass used for kitchenware (like glass baking pans) and laboratory glassware (like beakers and stirring rods). Glycerol is a clear syrup used by the food industry to keep foods moist and add sweetness. Pyrex is transparent and so is glycerol, but Pyrex in glycerol is nearly invisible. Why does this happen?
3. Find another pair of materials that have the same optical relationship as Pyrex and glycerol.

1. Windows, drinking glasses, and jars are made of crown glass. Crown glass and air have two very different indexes of refraction (1.52 and 1.00, respectively). When light travels from one medium to another, it's partly reflected and partly transmitted. The reflected part obeys the law of reflection and the transmitted part refracts according to Snell's law. We see glass in air because it reflects a little bit of what's in front of it and distorts what's behind it. Our brains interpret these reflections and distortions as something with substance.

2. Pyrex and glycerol have nearly the same index of refraction (about 1.47). When light goes from glycerol to Pyrex (or Pyrex to glycerol) the light travels across each interface with almost no change in direction. With no change in direction, there's no way for the light to know where one medium ends and another begins. All of the light is transmitted straight through and none is reflected back. Pyrex is effectively invisible in glycerol.

3. Vegetable oil (especially corn oil) has an index of refraction that is close to both Pyrex and glycerol (about 1.47) — so Pyrex and vegetable oil would make a good combo. Vegetable oil and glycerol would also work, but there's an obvious color difference that would make the interface between the two media visible. Microscopics sometimes use a special immersion oil that matches the index of refraction of crown glass, but this is a specialized material. Other combinations that are reasonably close are acrylic and mineral oil (1.495 and 1.470) and ice and water (1.309 and 1.333). This second pair doesn't seem to agree with most people's personal experience, however. Ice cubes in water are pretty easy to see because of air bubbles. Ice and air have very different indexes of refraction (1.000 and 1.309). Clear ice cubes with no bubbles, however, are hard to see in clean water.

3. 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

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.

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°

### project 3

1. 1The diagram below shows 18 rays of light from the sun incident on an equilateral, triangular ice crystal. (The width of the crystal is much larger than the wavelength of the light incident on it.)

Choose one of the rays at random. On the diagram, trace the transmitted portion of your ray until it emerges from the crystal. Determine the deviation angle between the incident ray and the emergent ray.

2. 2The diagram to the below shows 20 rays of light from the sun incident on a spherical drop of water. (The diameter of the drop is much larger than the wavelength of the light incident on it.)

Choose one of the rays at random. On the diagram, trace the portion of your ray that is transmitted at the interface, reflects once off the back of the drop, and then emerges. Determine the angle between the incident ray and the emergent ray.

3. 3Repeat the problem above, but this time trace the portion of your ray that is transmitted at the interface, reflects twice off the inside of the drop and then emerges.