Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
CHAPTER 33 THE NATURE AND PROPAGATION OF LIGHT (14TH ED.) 33 33.1. IDENTIFY: For reflection, r a . SET UP: The desired path of the ray is sketched in Figure 33.1. 140 cm EXECUTE: tan , so 506. r 90 394 and r a 394. 115 cm EVALUATE: The angle of incidence is measured from the normal to the surface. Figure 33.1 33.9. IDENTIFY and SET UP: Use Snell’s law to find the index of refraction of the plastic and then use c n to calculate the speed v of light in the plastic. v EXECUTE: na sina nb sinb . sin a sin 627 nb na 100 1194. sin sin 481 b c c n so v (300 108 m/s)/1194 251108 m/s. v n EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent toward the normal. 33.11. IDENTIFY: The figure shows the angle of incidence and angle of refraction for light going from the water into material X. Snell’s law applies at the air-water and water-X boundaries. SET UP: Snell’s law says na sin a nb sin b . Apply Snell’s law to the refraction from material X into the water and then from the water into the air. © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 33-1 33-2 Chapter 33 EXECUTE: (a) Material X to water: na nX , nb nw 1333. a 25 and b 48. sin b sin 48 na nb (1333) 234. sin sin 25 a (b) Water to air: As Figure 33.11 shows, a 48. na 1333 and nb 100. n sin b a sin a (1333)sin 48 82. nb EVALUATE: n 1 for material X, as it must be. Figure 33.11 33.15. IDENTIFY: The critical angle for total internal reflection is a that gives b 90 in Snell’s law. SET UP: In Figure 33.15 the angle of incidence a is related to angle by a 90. EXECUTE: (a) Calculate a that gives b 90. na 160, nb 100 so na sina nb sinb gives 100 and a 387. 90 a 513. 160 1333 (b) na 160, nb 1333. (160)sina (1333)sin90. sin a and a 564. 160 90 a 336. (160)sina (100)sin90. sina = EVALUATE: The critical angle increases when the ratio na decreases. nb © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The Nature and Propagation of Light 33-3 Figure 33.15 33.19. IDENTIFY and SET UP: For glass water, crit 487. Apply Snell’s law with a crit to calculate the index of refraction na of the glass. EXECUTE: na sincrit nb sin90, so na nb 1333 177 sincrit sin 487 EVALUATE: For total internal reflection to occur the light must be incident in the material of larger refractive index. Our results give nglass nwater , in agreement with this. 33.23. IDENTIFY: The index of refraction depends on the wavelength of light, so the light from the red and violet ends of the spectrum will be bent through different angles as it passes into the glass. Snell’s law applies at the surface. SET UP: na sina nb sinb . From the graph in Figure 33.17 in the textbook, for 400 nm (the violet end of the visible spectrum), n 167 and for 700 nm (the red end of the visible spectrum), n 162. The path of a ray with a single wavelength is sketched in Figure 33.23. Figure 33.23 EXECUTE: For 400 nm, sin b na 100 sin a sin350, so b 201. For 700 nm, nb 167 100 sin350, so b 207. is about 0.6. 162 EVALUATE: This angle is small, but the separation of the beams could be fairly large if the light travels through a fairly large slab. sinb 33.29. IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence n equals the polarizing angle, so p 545. Use Brewster’s law, tanp b , to calculate the refractive na index of the glass. Then use Snell’s law to calculate the angle of refraction. See Figure 33.29. n EXECUTE: (a) tanp b gives nglass nair tanp (100) tan545 140. na (b) na sin a nb sin b . sinb na sin a (100)sin545 05815 and b 355. nb 140 EVALUATE: © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 33-4 Chapter 33 Note: 1800 r b and r a . Thus 1800 545 355 900; the reflected ray and the refracted ray are perpendicular to each other. This agrees with Figure 33.28 in the textbook. Figure 33.29 33.42. IDENTIFY: Because the prism is a right-angle prism, the normals at point A and at surface BC are perpendicular to each other (see Figure 33.42). Therefore the angle of incidence at A is 50.0°, and this is the critical angle at that surface. Apply Snell’s law at A and at surface BC. For light incident at the critical angle, the angle of refraction is 90°. Figure 33.42 SET UP: Apply Snell’s law: na sin a nb sin b . Use n = 1.00 for air, and let n be the index of refraction of the glass. EXECUTE: Apply Snell’s law at point A. n sin(50.0°) = (1.00) sin(90°) = 1.00. n = 1.305. Now apply Snell’s law at surface BC. (1.00) sin = (1.305) sin(40.0°). = 57.0°. EVALUATE: The critical angle at A would not be 50.0° if the prism were not a right-angle prism. 33.45. IDENTIFY: Use Snell’s law to determine the effect of the liquid on the direction of travel of the light as it enters the liquid. SET UP: Use geometry to find the angles of incidence and refraction. Before the liquid is poured in, the ray along your line of sight has the path shown in Figure 33.45a (next page). 80 cm 0500. 160 cm a 2657. tana © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The Nature and Propagation of Light 33-5 Figure 33.45a After the liquid is poured in, a is the same and the refracted ray passes through the center of the bottom of the glass, as shown in Figure 33.45b. 40 cm 0250. 160 cm b 1404. tanb Figure 33.45b EXECUTE: Use Snell’s law to find nb , the refractive index of the liquid: na sin a nb sin b . nb na sin a (100)(sin 2657) 184. sin b sin1404 EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal. 33.50. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle. SET UP: b 90. The incident angle for the ray in the figure is 60. n sin a 156 sin 60 EXECUTE: na sina nb sinb gives nb a 135. sin b sin 90 EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater refractive index. © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.