Chapter 15 Pretest Light and Refraction 1 Refraction

Chapter 15 Pretest Light and Refraction

1. Refraction is the term for the bending of a wave disturbance as it passes at an angle from one _____ into another. A. glass B. medium C. area D. boundary

1. Refraction is the term for the bending of a wave disturbance as it passes at an angle from one _____ into another. A. glass B. medium C. area D. boundary

2. When a light ray passes from water (n = 1. 333) into diamond (n = 2. 419) at an angle of 45°, its path is: A. bent toward the normal. B. bent away from the normal. C. parallel to the normal. D. not bent.

2. When a light ray passes from water (n = 1. 333) into diamond (n = 2. 419) at an angle of 45°, its path is: A. bent toward the normal. B. bent away from the normal. C. parallel to the normal. D. not bent.

3. A beam of light in air is incident at an angle of 35° to the surface of a rectangular block of clear plastic (n = 1. 49). What is the angle of refraction? A. 42° B. 23° C. 55° D. 59°

3. A beam of light in air is incident at an angle of 35° to the surface of a rectangular block of clear plastic (n = 1. 49). What is the angle of refraction? A. 42° B. 23° C. 55° D. 59°

4. Which of the following describes what will happen to a light ray incident on a glass-to-air boundary at greater than the critical angle? A. total reflection B. total transmission C. partial reflection, partial transmission D. partial reflection, total transmission

4. Which of the following describes what will happen to a light ray incident on a glass-to-air boundary at greater than the critical angle? A. total reflection B. total transmission C. partial reflection, partial transmission D. partial reflection, total transmission

1. How is the index of refraction calculated? How is light refracted as it speeds up? How is light refracted as it slows down?

• Index of refraction = speed of light in a vacuum divided by speed of light in the substance ‘c’ in a vacuum n = -------------‘c’ in the substance

• When light speeds up it bends away from the normal. • When light slows down it bends toward the normal.

2. Why do you see ‘wet spots’ on the road on a hot day?

2. Why do you see ‘wet spots’ on the road on a hot day? • If the air close to the ground is warmer than the air at higher altitudes, light from the sky is refracted upward into the observer’s eyes. The blue sky appears to be on the ground and looks like it is reflected in water.

3. Explain why total internal reflection occurs. Why are prisms used as optical reflectors? Why are diamonds so bright?

• As light moves into a medium in which it moves faster, it bends away from the normal. As the angle of incidence increases, the angle of refraction reaches 90 degrees before the angle of incidence does.

• At this angle of incidence, no light is transmitted, 100% of the light is reflected. This is total internal reflection.

• Prisms are used as reflectors because total internal reflection is 100%. No mirrored surface is as efficient.

• Diamonds are bright because the critical angle for total internal reflection in diamond is so small that most of the light that enters the diamond is reflected back out. The critical angle is small for diamond because the speed of light in diamond is so much slower than it is in air.

4. Explain how a prism disperses light.

4. Explain how a prism disperses light. • Different colors of light are refracted different amounts. A prism refracts light twice in the same direction. Each bend splits the colors up a little more, producing a spectrum.

5. Why do stars twinkle?

5. Why do stars twinkle? • The atmosphere distorts the light from stars because of differences in the density of air. This distortion is seen as twinkling.

6. Why does the atmosphere make our days 4 minutes longer?

6. Why does the atmosphere make our days 4 minutes longer? • The atmosphere refracts sunlight toward the surface of the earth. This allows the sun to be seen after it has passed below the horizon and before it moves above the horizon. This adds about 4 minutes to each day.

7. A 3 cm object is placed 10 cm in front of a convex lens with a focal length of 5 cm. Draw a ray diagram and calculate the location, magnification , and size of the image formed. What is the type and orientation of the image?

• First draw a line parallel to the principle axis which refracts through the focal point.

• Then draw a line through the focal point which refracts parallel. Where they cross is the image.

• This image is real and inverted (case 3). We use the equations to find the actual distance and size of the image.

1/f = 1/ do + 1/di 1/5 = 1/ 10 + 1/di di = 10 cm

h i / ho = di / do hi / 3 cm = 10 cm / 10 cm hi = 3 cm, mag = -1

8. A 4 cm object is placed 7 cm in front of a concave lens with a focal length of -4 cm. Draw a ray diagram and calculate the location, magnification , and size of the image formed. What is the type and orientation of the image?

• First draw lines from each end of the object parallel to the principle axis which refract through the focal point.

• First draw lines from each end of the object parallel to the principle axis which refract through the focal point.

• Then draw lines from each end through the optical center of the lens. Where they cross forms the ends of the image.

• This image is virtual and upright. We use the equations to find the actual distance and size of the image.

1/f = 1/ do + 1/di 1/-4 = 1/ 7 + 1/di di = -2. 54 cm

h i / ho = di / do hi / 4 cm = 2. 54 cm / 7 cm hi = 1. 45 cm, mag = 0. 36

9. A 5 cm object is placed 3 cm in front of a convex lens with a focal length of 8 cm. Draw a ray diagram and calculate the location, magnification, and size of the image formed. What is the type and orientation of the image? diagram

• First draw a line parallel to the principle axis which refracts through the focal point.

• First draw a line parallel to the principle axis which refracts through the focal point.

• Then draw a line through the optical center of the lens. Where they cross is the image.

• This image is virtual and upright (case 6). We use the equations to find the actual distance and size of the image.

1/f = 1/ do + 1/di 1/8 = 1/ 3 + 1/di di = -4. 8 cm

h i / ho = di / do hi / 5 cm = 4. 8 cm / 3 cm hi = 8 cm, mag = 1. 6

10. In what two ways can a convex lens be used to produce an image that is larger than the object?

10. In what two ways can a convex lens be used to produce an image that is larger than the object? • Case 4 and case 6.

11. How does the production of images with mirrors compare with the production of images with lenses?

11. How does the production of images with mirrors compare with the production of images with lenses? • Convex mirrors produce images like concave lenses. • Concave mirrors produce images like convex lenses.

12. An object is placed along the principle axis of a thin converging lens that has a focal length of 39 cm. If the distance from the object to the lens is 51 cm, what is the distance from the image to the lens?

1/f = 1/ do + 1/di 1/39 = 1/51 + 1/di di = 166 cm

M = d i / do M = 166 cm / 51 cm mag = -3. 25

• First draw a line parallel to the principle axis which refracts through the focal point.

• First draw a line parallel to the principle axis which refracts through the focal point.

• Then draw a line through the optical center of the lens. Where they cross is the image.

• Or, you could draw a line through the other focal point which refracts parallel. All lines cross at the image.