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THE NATURE AND PROPAGATION OF LIGHT 33.1 33 IDENTIFY: For reflection, θ r = θ a SET UP: The desired path of the ray is sketched in Figure 33.1 14.0 cm EXECUTE: tan φ = , so φ = 50.6° θ r = 90° − φ = 39.4° and θ r = θ a = 39.4° 11.5 cm EVALUATE: The angle of incidence is measured from the normal to the surface Figure 33.1 33.2 IDENTIFY: The speed and the wavelength of the light will be affected by the vitreous humor, but not the frequency c λ SET UP: n = v = f λ λ = v n λ0, v 400 nm λ0,r 700 nm EXECUTE: (a) λv = = = 299 nm λr = = = 522 nm The range is 299 nm to 1.34 1.34 n n 522 nm c 3.00 × 108 m/s = 4.29 × 1014 Hz (b) Calculate the frequency in air, where v = c = 3.00 × 108 m/s f r = = λr 700 × 10−9 m fv = c λv = 3.00 × 108 m/s 400 × 10−9 m = 7.50 × 1014 Hz The range is 4.29 × 1014 Hz to 7.50 × 1014 Hz c 3.00 × 108 m/s = = 2.24 × 108 m/s n 1.34 EVALUATE: The frequency range in air is the same as in the vitreous humor (c) v = © Copyright 2012 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 33.3 Chapter 33 IDENTIFY and SET UP: Use Eqs (33.1) and (33.5) to calculate v and λ c c 2.998 × 108 m/s so v = = = 2.04 × 108 m/s v n 1.47 λ 650 nm (b) λ = = = 442 nm n 1.47 EVALUATE: Light is slower in the liquid than in vacuum By v = f λ , when v is smaller, λ is smaller EXECUTE: (a) n = 33.4 IDENTIFY: In air, c = f λ0 In glass, λ = λ0 n SET UP: c = 3.00 × 108 m/s EXECUTE: (a) λ0 = 517 nm = 340 nm n 1.52 EVALUATE: In glass the light travels slower than in vacuum and the wavelength is smaller c λ IDENTIFY: n = λ = , where λ0 is the wavelength in vacuum v n (b) λ = 33.5 λ0 c 3.00 × 108 m/s = = 517 nm f 5.80 × 1014 Hz = SET UP: c = 3.00 × 108 m/s n for air is only slightly larger than unity EXECUTE: (a) n = c 3.00 × 108 m/s = = 1.55 v 1.94 × 108 m/s (b) λ0 = nλ = (1.55)(3.55 × 10−7 m) = 5.50 × 10−7 m EVALUATE: In quartz the speed is lower and the wavelength is smaller than in air λ0 33.6 IDENTIFY: λ = 33.7 ⎛ n ⎞ ⎛ 1.333 ⎞ EXECUTE: λwater nwater = λbenzene nbenzene = λ0 λbenzene = λwater ⎜ water ⎟ = (438 nm) ⎜ ⎟ = 389 nm n ⎝ 1.501 ⎠ ⎝ benzene ⎠ EVALUATE: λ is smallest in benzene, since n is largest for benzene IDENTIFY: Apply Eqs (33.2) and (33.4) to calculate θ r and θb The angles in these equations are n SET UP: From Table 33.1, nwater = 1.333 and nbenzene = 1.501 measured with respect to the normal, not the surface (a) SET UP: The incident, reflected and refracted rays are shown in Figure 33.7 EVALUATE: θ r = θ a = 42.5° The reflected ray makes an angle of 90.0° − θ r = 47.5° with the surface of the glass Figure 33.7 (b) na sin θ a = nb sin θb , where the angles are measured from the normal to the interface sin θb = na sin θ a (1.00)(sin 42.5°) = = 0.4070 nb 1.66 θb = 24.0° The refracted ray makes an angle of 90.0° − θb = 66.0° with the surface of the glass © Copyright 2012 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 33.8 33-3 EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index IDENTIFY: The time delay occurs because the beam going through the transparent material travels slower than the beam in air c SET UP: v = in the material, but v = c in air n EXECUTE: The time for the beam traveling in air to reach the detector is d 2.50 m t= = = 8.33 × 10−9 s The light traveling in the block takes time c 3.00 × 108 m/s t = 8.33 × 10−9 s + 6.25 × 10−9 s = 1.46 × 10−8 s The speed of light in the block is c 3.00 × 108 m/s 2.50 m d = = = 1.75 n = = 71 × 10 m/s The refractive index of the block is v 1.71 × 108 m/s t 1.46 × 10−8 s EVALUATE: n > 1, as it must be, and 1.75 is a reasonable index of refraction for a transparent material such as plastic IDENTIFY and SET UP: Use Snell’s law to find the index of refraction of the plastic and then use Eq (33.1) to calculate the speed v of light in the plastic EXECUTE: na sin θ a = nb sin θb v= 33.9 33.10 ⎛ sin θ a ⎞ ⎛ sin 62.7° ⎞ nb = na ⎜ ⎟ = 1.00 ⎜ ⎟ = 1.194 ⎝ sin 48.1° ⎠ ⎝ sin θb ⎠ c c n = so v = = (3.00 × 108 m/s)/1.194 = 2.51 × 108 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 IDENTIFY: Apply Snell’s law at both interfaces SET UP: The path of the ray is sketched in Figure 33.10 Table 33.1 gives n = 1.329 for the methanol EXECUTE: (a) At the air-glass interface (1.00)sin 41.3° = nglass sin α At the glass-methanol interface nglass sin α = (1.329)sin θ Combining these two equations gives sin 41.3° = 1.329sin θ and θ = 29.8° (b) The same figure applies apply as for part (a), except θ = 20.2° (1.00)sin 41.3° = n sin 20.2° and n = 1.91 EVALUATE: The angle α is 25.2° The index of refraction of methanol is less than that of the glass and the ray is bent away from the normal at the glass → methanol interface The unknown liquid has an index of refraction greater than that of the glass, so the ray is bent toward the normal at the glass → liquid interface Figure 33.10 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 θ a = nb sin θb Apply Snell’s law to the refraction from material X into the water and then from the water into the air EXECUTE: (a) Material X to water: na = n X , nb = nw = 1.333 θ a = 25° and θb = 48° ⎛ sin θb ⎞ ⎛ sin 48° ⎞ na = nb ⎜ ⎟ = (1.333) ⎜ ⎟ = 2.34 ⎝ sin 25° ⎠ ⎝ sin θ a ⎠ © Copyright 2012 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 (b) Water to air: As Figure 33.11 shows, θ a = 48° na = 1.333 and nb = 1.00 ⎛n ⎞ sin θb = ⎜ a ⎟ sin θ a = (1.333)sin 48° = 82° ⎝ nb ⎠ Figure 33.11 33.12 EVALUATE: n > for material X, as it must be IDENTIFY: Apply Snell’s law to the refraction at each interface SET UP: nair = 1.00 nwater = 1.333 ⎛ n ⎞ ⎛ 1.00 ⎞ EXECUTE: (a) θ water = arcsin ⎜ air sinθair ⎟ = arcsin ⎜ sin35.0D ⎟ = 25.5D ⎝ 1.333 ⎠ ⎝ nwater ⎠ EVALUATE: (b) This calculation has no dependence on the glass because we can omit that step in the chain: nair sin θ air = nglass sin θglass = nwater sin θ water 33.13 IDENTIFY: When a wave passes from one material into another, the number of waves per second that cross the boundary is the same on both sides of the boundary, so the frequency does not change The wavelength and speed of the wave, however, change SET UP: In a material having index of refraction n, the wavelength is λ = wavelength in vacuum, and the speed is λ0 n , where λ0 is the c n EXECUTE: (a) The frequency is the same, so it is still f The wavelength becomes λ = λ0 n , so λ0 = nλ c The speed is v = , so c = nv n (b) The frequency is still f The wavelength becomes λ ′ = 33.14 λ0 n′ = nλ ⎛ n ⎞ = ⎜ ⎟ λ and the speed becomes n′ ⎝ n′ ⎠ c nv ⎛ n ⎞ v′ = = = ⎜ ⎟v n′ n′ ⎝ n′ ⎠ EVALUATE: These results give the speed and wavelength in a new medium in terms of the original medium without referring them to the values in vacuum (or air) IDENTIFY: The wavelength of the light depends on the index of refraction of the material through which it is traveling, and Snell’s law applies at the water-glass interface SET UP: λ0 = λ n so λw nw = λgl ngl Snell’s law gives nglsinθgl = nw sinθ w © Copyright 2012 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 ⎛λ ⎞ ⎛ 726 nm ⎞ EXECUTE: ngl = nw ⎜ w ⎟ = (1.333) ⎜ ⎟ = 1.779 Now apply ngl sinθgl = nw sinθ w ⎜ λgl ⎟ ⎝ 544 nm ⎠ ⎝ ⎠ ⎛n ⎞ ⎛ 1.333 ⎞ sinθgl = ⎜ w ⎟ sinθ w = ⎜ ⎟ sin42.0° = 0.5014 θ gl = 30.1° ⎜ ngl ⎟ ⎝ 1.779 ⎠ ⎝ ⎠ EVALUATE: θ gl < θ air because ngl > nair 33.15 IDENTIFY: Apply na sin θ a = nb sin θb SET UP: na = 1.70, θ a = 62.0° nb = 1.58 33.16 ⎛n ⎞ ⎛ 1.70 ⎞ EXECUTE: sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 62.0° = 0.950 and θb = 71.8° n ⎝ 1.58 ⎠ ⎝ b⎠ EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal IDENTIFY: No light will enter the water if total internal reflection occurs at the glass-water boundary Snell’s law applies at the boundary SET UP: Find ng , the refractive index of the glass Then apply Snell’s law at the boundary na sin θ a = nb sin θb ⎛ sin49.8° ⎞ EXECUTE: ng sin36.2° = nw sin49.8° ng = (1.333) ⎜ ⎟ = 1.724 Now find θ crit for the glass to ⎝ sin36.2° ⎠ 1.333 and θ crit = 50.6° water refraction ng sinθ crit = nw sin90.0° sin θ crit = 1.724 33.17 EVALUATE: For θ > 50.6o at the glass-water boundary, no light is refracted into the water IDENTIFY: The critical angle for total internal reflection is θ a that gives θb = 90° in Snell’s law SET UP: In Figure 33.17 the angle of incidence θ a is related to angle θ by θ a + θ = 90° EXECUTE: (a) Calculate θ a that gives θb = 90° na = 1.60, nb = 1.00 so na sin θ a = nb sin θb gives 1.00 and θ a = 38.7° θ = 90° − θ a = 51.3° 1.60 1.333 (b) na = 1.60, nb = 1.333 (1.60)sin θ a = (1.333)sin 90° sin θ a = and θ a = 56.4° 1.60 θ = 90° − θ a = 33.6° (1.60)sin θ a = (1.00)sin 90° sin θ a = EVALUATE: The critical angle increases when the ratio na decreases nb Figure 33.17 © Copyright 2012 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 33-6 Chapter 33 33.18 IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will occur at the cube surface if the angle of incidence is greater than or equal to the critical angle SET UP: At the critical angle θ crit , Snell’s law gives nglass sin θcrit = nair sin 90° and likewise for water EXECUTE: (a) At the critical angle θ crit , nglass sin θcrit = nair sin 90° 1.53 sinθ crit = (1.00)(1) and θ crit = 40.8° (b) Using the same procedure as in part (a), we have 1.53 sinθ crit = 1.333 sin 90° and θcrit = 60.6° 33.19 EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air IDENTIFY: Use the critical angle to find the index of refraction of the liquid SET UP: Total internal reflection requires that the light be incident on the material with the larger n, in this case the liquid Apply na sin θ a = nb sin θb with a = liquid and b = air, so na = nliq and nb = 1.0 EXECUTE: θ a = θ crit when θb = 90°, so nliq sin θ crit = (1.0)sin 90° nliq = sin θ crit = = 1.48 sin 42.5° (a) na sin θ a = nb sin θb (a = liquid, b = air) sin θb = na sin θ a (1.48)sin 35.0° = = 0.8489 and θb = 58.1° nb 1.0 (b) Now na sin θ a = nb sin θb with a = air, b = liquid sin θb = 33.20 na sin θ a (1.0)sin 35.0° = = 0.3876 and θb = 22.8° nb 1.48 EVALUATE: For light traveling liquid → air the light is bent away from the normal For light traveling air → liquid the light is bent toward the normal IDENTIFY: The largest angle of incidence for which any light refracts into the air is the critical angle for water → air SET UP: Figure 33.20 shows a ray incident at the critical angle and therefore at the edge of the circle of light The radius of this circle is r and d = 10.0 m is the distance from the ring to the surface of the water EXECUTE: From the figure, r = d tan θcrit θ crit is calculated from na sin θ a = nb sin θb with na = 1.333, θ a = θcrit , nb = 1.00 and θb = 90° sin θcrit = r = (10.0 m) tan 48.6° = 11.3 m (1.00)sin 90° and θ crit = 48.6° 1.333 A = π r = π (11.3 m) = 401 m EVALUATE: When the incident angle in the water is larger than the critical angle, no light refracts into the air Figure 33.20 © Copyright 2012 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 33.21 33-7 IDENTIFY and SET UP: For glass → water, θ crit = 48.7° Apply Snell’s law with θ a = θ crit to calculate the index of refraction na of the glass EXECUTE: na sin θcrit = nb sin 90°, so na = nb 1.333 = = 1.77 sin θ crit sin 48.7° 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 33.22 IDENTIFY: If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface is θcrit SET UP: The ray has an angle of incidence of 0° at the first surface of the glass, so enters the glass without being bent, as shown in Figure 33.22 The figure shows that α + θ crit = 90° EXECUTE: (a) For the glass-air interface θ a = θ crit , na = 1.52, nb = 1.00 and θb = 90° (1.00)(sin 90°) and θ crit = 41.1° α = 90° − θcrit = 48.9° 1.52 (b) Now the second interface is glass → water and nb = 1.333 na sin θ a = nb sin θb gives na sin θ a = nb sin θb gives sin θ crit = (1.333)(sin 90°) and θ crit = 61.3° α = 90° − θ crit = 28.7° 1.52 EVALUATE: The critical angle increases when the air is replaced by water sin θcrit = Figure 33.22 33.23 IDENTIFY: Total internal reflection must be occurring at the glass-water boundary Snell’s law applies there SET UP: na sin θ a = nb sin θb λ = λ0 /n EXECUTE: Apply Snell’s law to find ngl : ngl sin 62.0° = nw sin 90.0° and ngl = 1.510 Then ⎛ ngl ⎞ ⎛ 1.510 ⎞ ⎟ = (408 nm) ⎜ ⎟ = 462 nm n 1.333 ⎠ ⎝ ⎝ w⎠ λw nw = λglngl and λw = λgl ⎜ EVALUATE: The wavelength is greater in the water than it is in the glass, as it must be, since nw < ngl 33.24 IDENTIFY: We apply Snell’s law to sound waves, making an appropriate definition of the index of refraction for sound We cannot use the speed of sound in vacuum because sound does not travel through a vacuum v SET UP: n = air When θ a = θ crit , θb = 90° na sin θ a = nb sin θb v v 344 m/s v = 0.261 Air has a larger index of EXECUTE: (a) For air, n = air = 1.00 For water, n = air = v 1320 m/s v refraction for sound waves © Copyright 2012 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 33-8 Chapter 33 (b) Total internal reflection requires that the waves be incident in the material of larger refractive index na = 1.00, nb = 0.261, θ a = θcrit , and θb = 90° Applying na sin θ a = nb sin θb gives 33.25 ⎛ 0.261 ⎞ sin θ crit = ⎜ ⎟ sin 90°, so θ crit = 15.1° ⎝ 1.00 ⎠ (c) The sound wave must be traveling in air (d) Sound waves can be totally reflected from the surface of the water EVALUATE: Light travels faster in vacuum than in any material and n is always greater than 1.00 Sound travels faster in solids and liquids than in air and n for sound is less than 1.00 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 sin θ a = nb sin θb From the graph in Figure 33.18 in the textbook, for λ = 400 nm (the violet end of the visible spectrum), n = 1.67 and for λ = 700 nm (the red end of the visible spectrum), n = 1.62 The path of a ray with a single wavelength is sketched in Figure 33.25 Figure 33.25 EXECUTE: For λ = 400 nm, sin θb = na 1.00 sin θ a = sin 35.0°, so θb = 20.1° For λ = 700 nm, nb 1.67 1.00 sin 35.0°, so θb = 20.7° Δθ is about 0.6° 1.62 EVALUATE: This angle is small, but the separation of the beams could be fairly large if the light travels through a fairly large slab c IDENTIFY: Snell's law is na sin θ a = nb sin θb v = n SET UP: a = air, b = glass sin θb = 33.26 EXECUTE: (a) red: nb = na sin θ a (1.00)sin 57.0° (1.00)sin 57.0° = = 1.36 violet: nb = = 1.40 sin 36.7° sin θb sin 38.1° c 3.00 × 108 m/s c 3.00 × 108 m/s = = 2.21× 108 m/s; violet: v = = = 2.14 × 108 m/s n 1.36 n 1.40 EVALUATE: n is larger for the violet light and therefore this light is bent more toward the normal, and the violet light has a smaller speed in the glass than the red light IDENTIFY: The first polarizer filters out half the incident light The fraction filtered out by the second polarizer depends on the angle between the axes of the two filters SET UP: I = I cos φ (b) red: v = 33.27 EXECUTE: After the first filter, I = ⎛1 ⎞ I After the second filter, I = ⎜ I ⎟ cos φ , which gives ⎝2 ⎠ ⎛1 ⎞ I = ⎜ I ⎟ cos 30.0° = 0.375I ⎝2 ⎠ EVALUATE: The only variable that affects the answer is the angle between the axes of the two polarizers © Copyright 2012 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 33.28 33-9 IDENTIFY: The sunlight must be striking the lake surface at the Brewster’s angle (the polarizing angle) since the reflected light is completely polarized n SET UP: The reflected beam is completely polarized when θ a = θ p , with tan θ p = b na =1.00, na nb = 1.333 θ p is measured relative to the normal to the surface EXECUTE: (a) tan θ p = 33.29 1.333 , so θ p = 53.1° The sunlight is incident at an angle of 90° − 53.1° = 36.9° 1.00 above the horizontal (b) Figure 33.27 in the text shows that the plane of the electric field vector in the reflected light is horizontal EVALUATE: To reduce the glare (intensity of reflected light), sunglasses with polarizing filters should have the filter axis vertical IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12 and the transmitted light is polarized along the axis of the polarizer When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter SET UP: For the second polarizer φ = 60° For the third polarizer, φ = 90° − 60° = 30° EXECUTE: (a) At point A the intensity is I /2 and the light is polarized along the vertical direction At point B the intensity is ( I /2)(cos60°) = 0.125I , and the light is polarized along the axis of the second 33.30 polarizer At point C the intensity is (0.125I )(cos30°) = 0.0938I (b) Now for the last filter φ = 90° and I = EVALUATE: Adding the middle filter increases the transmitted intensity IDENTIFY: Apply Snell’s law SET UP: The incident, reflected and refracted rays are shown in Figure 33.30 sin θ a sin 53° EXECUTE: From the figure, θb = 37.0° and nb = na = 1.33 = 1.77 sin θb sin 37° EVALUATE: The refractive index of b is greater than that of a, and the ray is bent toward the normal when it refracts Figure 33.30 33.31 IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals the polarizing angle, so θ p = 54.5° Use Eq (33.8) to calculate the refractive index of the glass Then use Snell’s law to calculate the angle of refraction n EXECUTE: (a) tan θ p = b gives nglass = nair tan θ p = (1.00) tan 54.5° = 1.40 na (b) na sin θ a = nb sin θb sin θb = na sin θ a (1.00)sin 54.5° = = 0.5815 and θb = 35.5° nb 1.40 © Copyright 2012 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 33-10 Chapter 33 EVALUATE: Note: φ = 180.0° − θ r − θb and θ r = θ a Thus φ = 180.0° − 54.5° − 35.5° = 90.0°; the reflected ray and the refracted ray are perpendicular to each other This agrees with Figure 33.28 in the text book Figure 33.31 33.32 IDENTIFY: Set I = I /10, where I is the intensity of light passed by the second polarizer SET UP: When unpolarized light passes through a polarizer the intensity is reduced by a factor of and the transmitted light is polarized along the axis of the polarizer When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter I EXECUTE: (a) After the first filter I = and the light is polarized along the vertical direction After the I0 I0 ⎛ I0 ⎞ second filter we want I = , so = ⎜ ⎟ (cos φ ) cos φ = 2/10 and φ = 63.4° 10 10 ⎝ ⎠ (b) Now the first filter passes the full intensity I of the incident light For the second filter 33.33 I0 = I (cos φ )2 cos φ = 1/10 and φ = 71.6° 10 EVALUATE: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the same overall reduction in intensity than when the incident light is unpolarized IDENTIFY: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters SET UP: If the angle between the two axes is θ , the intensity of the emerging light is I = I max cos θ EXECUTE: At angle θ , I = I max cos 2θ , and at the new angle α , intensities gives I max cos α I max cos θ = 1I I , which gives us cos α = 1I = I max cos 2α Taking the ratio of the cosθ Solving for α yields ⎛ cosθ ⎞ ⎟ ⎝ ⎠ EVALUATE: For θ = 0°, I = I max The expression we derived then gives α = 45° and for this angle α = arccos ⎜ between the axes of the two filters, I = I max /2 So, our expression is seen to be correct for this special case 33.34 IDENTIFY: The reflected light is completely polarized when the angle of incidence equals the polarizing n angle θ p , where tan θ p = b na SET UP: nb = 1.66 EXECUTE: (a) na = 1.00 tan θ p = 1.66 and θ p = 58.9° 1.00 1.66 and θ p = 51.2° 1.333 EVALUATE: The polarizing angle depends on the refractive indicies of both materials at the interface (b) na = 1.333 tan θ p = © Copyright 2012 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 33.35 33-11 IDENTIFY: When unpolarized light of intensity I is incident on a polarizing filter, the transmitted light has intensity 1I and is polarized along the filter axis When polarized light of intensity I is incident on a polarizing filter the transmitted light has intensity I cos φ SET UP: For the second filter, φ = 62.0° − 25.0° = 37.0° EXECUTE: After the first filter the intensity is 1I = 10.0 W/m and the light is polarized along the axis of the first filter The intensity after the second filter is I = I 0cos 2φ , where I = 10.0 W/m and φ = 37.0° This gives I = 6.38 W/m 33.36 EVALUATE: The transmitted intensity depends on the angle between the axes of the two filters IDENTIFY: Use the transmitted intensity when all three polairzers are present to solve for the incident intensity I Then repeat the calculation with only the first and third polarizers SET UP: For unpolarized light incident on a filter, I = 12 I and the light is linearly polarized along the filter axis For polarized light incident on a filter, I = I max (cos φ ) , where I max is the intensity of the incident light, and the emerging light is linearly polarized along the filter axis EXECUTE: With all three polarizers, if the incident intensity is I the transmitted intensity is I = ( 12 I )(cos 23.0°)2 (cos[62.0° − 23.0°]) = 0.256 I I = I 75.0 W/cm = = 293 W/cm With only 0.256 0.256 the first and third polarizers, I = ( 12 I )(cos62.0°) = 0.110 I = (0.110)(293 W/cm ) = 32.2 W/cm 33.37 EVALUATE: The transmitted intensity is greater when all three filters are present IDENTIFY and SET UP: Apply Eq (33.7) to polarizers #2 and #3 The light incident on the first polarizer is unpolarized, so the transmitted light has half the intensity of the incident light, and the transmitted light is polarized (a) EXECUTE: The axes of the three filters are shown in Figure 33.37a I = I max cos φ Figure 33.37a After the first filter the intensity is I1 = 12 I and the light is linearly polarized along the axis of the first polarizer After the second filter the intensity is I = I1 cos φ = ( 12 I )(cos 45.0°)2 = 0.250 I and the light is linearly polarized along the axis of the second polarizer After the third filter the intensity is I = I cos φ = 0.250 I (cos 45.0°) = 0.125 I and the light is linearly polarized along the axis of the third polarizer (b) The axes of the remaining two filters are shown in Figure 33.37b After the first filter the intensity is I1 = 12 I and the light is linearly polarized along the axis of the first polarizer Figure 33.37b © Copyright 2012 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 33-12 Chapter 33 After the next filter the intensity is I = I1 cos φ = 33.38 ( 12 I0 ) (cos 90.0°)2 = No light is passed EVALUATE: Light is transmitted through all three filters, but no light is transmitted if the middle polarizer is removed IDENTIFY: The shorter the wavelength of light, the more it is scattered The intensity is inversely proportional to the fourth power of the wavelength SET UP: The intensity of the scattered light is proportional to 1/λ ; we can write it as I = (constant)/λ EXECUTE: (a) Since I is proportional to 1/λ , we have I = (constant)/λ Taking the ratio of the intensity of the red light to that of the green light gives 4 I R (constant)/λR4 ⎛ λG ⎞ ⎛ 520 nm ⎞ = =⎜ ⎟ =⎜ ⎟ = 0.374, so I R = 0.374I I (constant)/λG4 ⎝ λR ⎠ ⎝ 665 nm ⎠ 4 I V ⎛ λG ⎞ ⎛ 520 nm ⎞ =⎜ ⎟ =⎜ ⎟ = 2.35, so I V = 2.35I I ⎝ λV ⎠ ⎝ 420 nm ⎠ EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about times as great as that of the red light, so this scattered light will have a blue-violet color IDENTIFY: Reflection reverses the sign of the component of light velocity perpendicular to the reflecting surface but leaves the other components unchanged SET UP: Consider three mirrors, M1 in the (x,y)-plane, M in the (y,z)-plane and M in the (x,z)-plane (b) Following the same procedure as in part (a) gives 33.39 EXECUTE: A light ray reflecting from M1 changes the sign of the z-component of the velocity, reflecting from M changes the x-component and from M changes the y-component Thus the velocity, and hence 33.40 also the path, of the light beam flips by 180° EVALUATE: Example 33.3 discusses some uses of corner reflectors IDENTIFY: The light travels slower in the jelly than in the air and hence will take longer to travel the length of the tube when it is filled with jelly than when it contains just air SET UP: The definition of the index of refraction is n = c /v, where v is the speed of light in the jelly EXECUTE: First get the length L of the tube using air In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m The speed in the jelly is L c = (2.616 m)/(8.72 ns + 2.04 ns) = 2.431 × 108 m/s n = = (3.00 × 108 m/s)/(2.431 × 108 m/s) = 1.23 t v EVALUATE: A high-speed timer would be needed to measure times as short as a few nanoseconds IDENTIFY: Snell’s law applies to the sound waves in the heart (See Exercise 33.24.) SET UP: na sin θ a = nb sin θb If θ a is the critical angle then θb = 90° For air, nair = 1.00 For heart v= 33.41 muscle, nmus = 344 m/s = 0.2324 1480 m/s EXECUTE: (a) na sin θ a = nb sin θb gives (1.00)sin (9.73°) = (0.2324)sin θb sin θb = θb = 46.7° sin (9.73°) so 0.2324 (b) (1.00)sin θ crit = (0.2324)sin 90° gives θ crit = 13.4° 33.42 EVALUATE: To interpret a sonogram, it should be important to know the true direction of travel of the sound waves within muscle This would require knowledge of the refractive index of the muscle c IDENTIFY: Use the change in transit time to find the speed v of light in the slab, and then apply n = v and λ = λ0 n SET UP: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam © Copyright 2012 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 EXECUTE: 33-13 0.840 m 0.840 m 0.840 m − = (n − 1) = 4.2 ns We can now solve for the index of refraction: c /n c c (4.2 × 10−9 s)(3.00 × 108 m/s) 490 nm + = 2.50 The wavelength inside of the glass is λ = = 196 nm 0.840 m 2.50 EVALUATE: Light travels slower in the slab than in air and the wavelength is shorter IDENTIFY: The angle of incidence at A is to be the critical angle Apply Snell’s law at the air to glass refraction at the top of the block SET UP: The ray is sketched in Figure 33.43 EXECUTE: For glass → air at point A, Snell’s law gives (1.38)sin θcrit = (1.00)sin 90° and θ crit = 46.4° n= 33.43 θb = 90° − θcrit = 43.6° Snell’s law applied to the refraction from air to glass at the top of the block gives (1.00)sin θ a = (1.38)sin(43.6°) and θ a = 72.1° EVALUATE: If θ a is larger than 72.1° then the angle of incidence at point A is less than the initial critical angle and total internal reflection doesn’t occur Figure 33.43 33.44 IDENTIFY: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell’s law SET UP: Snell’s law is na sin θa = nb sinθ b and n = 1.000 for air At point B the angle of the prism is 30.0° EXECUTE: Apply Snell’s law at AB The prism angle at A is 60.0°, so for the upper ray, the angle of incidence at AB is 60.0° + 12.0° = 72.0° Using this value gives n1 sin 60.0° = sin 72.0° and n1 = 1.10 For the lower ray, the angle of incidence at AB is 60.0° + 12.0° + 8.50° = 80.5°, giving n2 sin 60.0° = sin 80.5° and n2 = 1.14 EVALUATE: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater index of refraction than the upper ray © Copyright 2012 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 33-14 33.45 Chapter 33 IDENTIFY: For total internal reflection, the angle of incidence must be at least as large as the critical angle SET UP: The angle of incidence for the glass-oil interface must be the critical angle, so θb = 90° na sin θ a = nb sin θb EXECUTE: na sin θ a = nb sin θb gives (1.52)sin 57.2° = noil sin 90° noil = (1.52)sin 57.2° = 1.28 EVALUATE: noil > 1, which it must be, and 1.28 is a reasonable value for an oil 33.46 IDENTIFY: Apply λ = λ0 The number of wavelengths in a distance d of a material is n wavelength in the material SET UP: The distance in glass is d glass = 0.00250 m The distance in air is d λ where λ is the d air = 0.0180 m − 0.00250 m = 0.0155 m 33.47 EXECUTE: number of wavelengths = number in air + number in glass d glass d 0.0155 m 0.00250 m n= number of wavelengths = air + (1.40) = 3.52 × 104 + λ λ 5.40 × 10−7 m 5.40 × 10−7 m EVALUATE: Without the glass plate the number of wavelengths between the source and screen is 0.0180 m = 3.33 × 104 The wavelength is shorter in the glass so there are more wavelengths in a 5.40 × 10−3 m distance in glass than there are in the same distance in air IDENTIFY: Find the critical angle for glass → air Light incident at this critical angle is reflected back to the edge of the halo SET UP: The ray incident at the critical angle is sketched in Figure 33.47 Figure 33.47 2.67 mm = 0.8613; θ crit = 40.7° 3.10 mm Apply Snell’s law to the total internal reflection to find the refractive index of the glass: na sin θ a = nb sin θb nglass sin θcrit = 1.00sin 90° EXECUTE: From the distances given in the sketch, tan θ crit = nglass = 1 = = 1.53 sin θ crit sin 40.7° EVALUATE: Light incident on the back surface is also totally reflected if it is incident at angles greater than θcrit If it is incident at less than θ crit it refracts into the air and does not reflect back to the emulsion 33.48 IDENTIFY: Apply Snell's law to the refraction of the light as it passes from water into air ⎛ 1.5 m ⎞ SET UP: θ a = arctan ⎜ ⎟ = 51° na = 1.00 nb = 1.333 ⎝ 1.2 m ⎠ © Copyright 2012 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 33-15 ⎛n ⎞ ⎛ 1.00 ⎞ EXECUTE: θb = arcsin ⎜ a sin θ a ⎟ = arcsin ⎜ sin 51° ⎟ = 36° Therefore, the distance along the bottom n 333 ⎝ ⎠ ⎝ b ⎠ of the pool from directly below where the light enters to where it hits the bottom is x = (4.0 m) tan θb = (4.0 m) tan 36° = 2.9 m xtotal = 1.5 m + x = 1.5 m + 2.9 m = 4.4 m 33.49 EVALUATE: The light ray from the flashlight is bent toward the normal when it refracts into the water 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 33.49a 8.0 cm = 0.500 16.0 cm θ a = 26.57° tan θ a = Figure 33.49a After the liquid is poured in, θ a is the same and the refracted ray passes through the center of the bottom of the glass, as shown in Figure 33.49b 4.0 cm = 0.250 16.0 cm θb = 14.04° tan θb = Figure 33.49b EXECUTE: Use Snell’s law to find nb , the refractive index of the liquid: na sin θ a = nb sin θb nb = na sin θ a (1.00)(sin 26.57°) = = 1.84 sin θb sin14.04° EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal © Copyright 2012 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 33-16 33.50 Chapter 33 IDENTIFY: The incident angle at the prism → water interface is to be the critical angle SET UP: The path of the ray is sketched in Figure 33.50 The ray enters the prism at normal incidence so is not bent For water, nwater = 1.333 EXECUTE: From the figure, θ crit = 45° na sin θ a = nb sin θb gives nglass sin 45° = (1.333)sin 90° 1.333 = 1.89 sin 45° EVALUATE: For total internal reflection the ray must be incident in the material of greater refractive index nglass > nwater , so that is the case here nglass = Figure 33.50 33.51 IDENTIFY: Apply Snell’s law to the water → ice and ice → air interfaces (a) SET UP: Consider the ray shown in Figure 33.51 We want to find the incident angle θ a at the water-ice interface that causes the incident angle at the ice-air interface to be the critical angle Figure 33.51 EXECUTE: ice-air interface: nice sin θ crit = 1.0 sin 90° nice sin θ crit = 1.0 so sin θ crit = nice But from the diagram we see that θb = θ crit , so sin θb = nice water-ice interface: nw sin θ a = nice sin θb But sin θb = 33.52 1 so nw sin θ a = 1.0 sin θ a = = = 0.7502 and θ a = 48.6° nice nw 1.333 (b) EVALUATE: The angle calculated in part (a) is the critical angle for a water-air interface; the answer would be the same if the ice layer wasn’t there! IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle SET UP: θb = 90° The incident angle for the ray in the figure is 60° © Copyright 2012 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 33.53 33-17 ⎛ n sin θ a ⎞ ⎛ 1.62 sin 60° ⎞ EXECUTE: na sin θ a = nb sin θb gives nb = ⎜ a ⎟=⎜ ⎟ = 1.40 ⎝ sin θb ⎠ ⎝ sin 90° ⎠ EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater refractive index IDENTIFY: Apply Snell’s law to the refraction of each ray as it emerges from the glass The angle of incidence equals the angle A = 25.0° SET UP: The paths of the two rays are sketched in Figure 33.53 Figure 33.53 EXECUTE: na sin θ a = nb sin θb nglass sin 25.0° = 1.00sin θb sin θb = nglass sin 25.0° sin θb = 1.66sin 25.0° = 0.7015 θb = 44.55° β = 90.0° − θb = 45.45° Then δ = 90.0° − A − β = 90.0° − 25.0° − 45.45° = 19.55° The angle between the two rays is 2δ = 39.1° 33.54 EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism IDENTIFY: No light enters the gas because total internal reflection must have occurred at the water-gas interface SET UP: At the minimum value of S, the light strikes the water-gas interface at the critical angle We apply Snell’s law, na sin θ a = nb sin θ b , at that surface S = (1.09 m)/(1.10 m) = 0.991 rad = 56.77° This is the critical angle R So, using the refractive index for water from Table 33.1, we get n = (1.333)sin 56.77° = 1.12 EXECUTE: (a) In the water, θ = (b) (i) The laser beam stays in the water all the time, so ⎛ c ⎞ Dnwater t = 2R/v = 2R/ ⎜ = (2.20 m)(1.333)/(3.00 × 108 m/s) = 9.78 ns ⎟= n c ⎝ water ⎠ (ii) The beam is in the water half the time and in the gas the other half of the time Rngas = (1.10 m)(1.12)/(3.00 × 108 m/s) = 4.09 ns c The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns tgas = EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12 © Copyright 2012 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 33-18 33.55 Chapter 33 (a) IDENTIFY: Apply Snell’s law to the refraction of the light as it enters the atmosphere SET UP: The path of a ray from the sun is sketched in Figure 33.55 δ = θ a − θb From the diagram sin θb = R R+h ⎛ R ⎞ ⎟ ⎝R+h⎠ θb = arcsin ⎜ Figure 33.55 EXECUTE: Apply Snell’s law to the refraction that occurs at the top of the atmosphere: na sinθa = nb sinθb (a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n) ⎛ R ⎞ ⎛ nR ⎞ sin θ a = n sin θb = n ⎜ ⎟ soθ a = arcsin ⎜ ⎟ ⎝R+h⎠ ⎝ R+h⎠ ⎛ nR ⎞ ⎛ R ⎞ ⎟ − arcsin ⎜ ⎟ R + h ⎝ ⎠ ⎝R+h⎠ δ = θ a − θb = arcsin ⎜ R 6.38 × 106 m = = 0.99688 R + h 6.38 × 106 m + 20 × 103 m nR = 1.0003(0.99688) = 0.99718 R+h ⎛ R ⎞ θb = arcsin ⎜ ⎟ = 85.47° ⎝R+h⎠ (b) ⎛ nR ⎞ ⎟ = 85.70° ⎝ R+h⎠ δ = θ a − θb = 85.70° − 85.47° = 0.23° EVALUATE: The calculated δ is about the same as the angular radius of the sun IDENTIFY and SET UP: Follow the steps specified in the problem EXECUTE: (a) The distance traveled by the light ray is the sum of the two diagonal segments: d = ( x + y12 )1/ + ((l − x ) + y22 )1/2 Then the time taken to travel that distance is θ a = arcsin ⎜ 33.56 d ( x + y12 )1/2 + ((l − x) + y22 )1/2 = c c (b) Taking the derivative with respect to x of the time and setting it to zero yields dt d ⎡ = ( x + y12 )1/2 + ((l − x) + y22 )1/2 ⎤ = ⎡ x( x + y12 ) −1/ − (l − x)((l − x ) + y22 ) −1/2 ⎤ = This gives ⎦ c⎣ ⎦ dx c dx ⎣ t= x x + 33.57 y12 = (l − x ) (l − x ) + y22 , sin θ1 = sin θ and θ1 = θ EVALUATE: For any other path between points and 2, that includes a point on the reflective surface, the distance traveled and therefore the travel time is greater than for this path IDENTIFY and SET UP: Find the distance that the ray travels in each medium The travel time in each medium is the distance divided by the speed in that medium (a) EXECUTE: The light travels a distance h12 + x in traveling from point A to the interface Along this path the speed of the light is v1, so the time it takes to travel this distance is t1 = h12 + x The light v1 © Copyright 2012 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 h22 + (l − x) in traveling from the interface to point B Along this path the speed of travels a distance the light is v2 , so the time it takes to travel this distance is t2 = h22 + (l − x)2 The total time to go from v2 h12 + x h + (l + x ) + v1 v2 A to B is t = t1 + t2 = (b) 33-19 dt ⎛ ⎞ ⎛ 1⎞ = ⎜ ⎟ ( h1 + x ) −1/2 (2 x ) + ⎜ ⎟ (h22 + (l − x ) )−1/2 2(l − x)(−1) = ⎝ ⎠ dx v1 v2 ⎝ ⎠ x v1 h12 +x l−x = v2 h22 + (l − x) Multiplying both sides by c gives c x c = 2 v1 h + x v2 l−x h22 + (l − x ) c c = n1 and = n2 (Eq 33.1) v1 v2 From Figure P33.57 in the textbook, sin θ1 = x h12 +x and sin θ = l−x h22 + (l − x ) So n1 sin θ1 = n2 sin θ , which is Snell’s law 33.58 EVALUATE: Snell’s law is a result of a change in speed when light goes from one material to another IDENTIFY: Apply Snell’s law to each refraction SET UP: Refer to the angles and distances defined in the figure that accompanies the problem EXECUTE: (a) For light in air incident on a parallel-faced plate, Snell’s Law yields: n sin θ a = n′ sin θb′ = n′ sin θb = n sin θ a′ ⇒ sin θ a = sin θ a′ ⇒ θ a = θ a′ (b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out The requirement of parallel faces ensures that the angle θ n′ = θ n and the chain of equations can continue (c) The lateral displacement of the beam can be calculated using geometry: t t sin(θ a − θb′ ) d = L sin(θ a − θb′ ) and L = ⇒d = cosθb′ cosθb′ 33.59 (2.40 cm)sin(66.0° − 30.5°) ⎛ n sin θ a ⎞ ⎛ sin 66.0° ⎞ = 1.62 cm = arcsin ⎜ = 30.5° and d = (d) θb′ = arcsin ⎜ ⎟ ⎟ cos30.5° ⎝ n′ ⎠ ⎝ 1.80 ⎠ EVALUATE: The lateral displacement in part (d) is large, of the same order as the thickness of the plate IDENTIFY: Apply Snell’s law to the two refractions of the ray SET UP: Refer to the figure that accompanies the problem A A EXECUTE: (a) na sin θ a = nb sin θb gives sin θ a = nb sin But θ a = + α , so 2 A + 2α A ⎛A ⎞ sin ⎜ + α ⎟ = sin = n sin At each face of the prism the deviation is α , so 2α = δ and 2 ⎝2 ⎠ sin A+δ A = n sin 2 A⎞ 60.0° ⎞ ⎛ ⎛ (b) From part (a), δ = 2arcsin ⎜ n sin ⎟ − A δ = 2arcsin ⎜ (1.52)sin ⎟ − 60.0° = 38.9° 2⎠ ⎠ ⎝ ⎝ (c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ: ⎛ ⎝ δ red = 2arcsin ⎜ (1.61)sin 60.0° ⎞ ⎟ − 60.0° = 47.2° ⎠ © Copyright 2012 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 33-20 Chapter 33 60.0° ⎞ ⎛ ⎟ − 60.0° = 52.2° ⇒ Δδ = 52.2° − 47.2° = 5.0° ⎠ ⎝ EVALUATE: The violet light has a greater refractive index and therefore the angle of deviation is greater for the violet light IDENTIFY: Apply Snell’s law and the results of Problem 33.58 SET UP: From Figure 33.18 in the textbook, nr = 1.61 for red light and nv = 1.66 for violet In the δ violet = 2arcsin ⎜ (1.66)sin 33.60 notation of Problem 33.58, t is the thickness of the glass plate and the lateral displacement is d We want the difference in d for the two colors of light to be 1.0 mm θ a = 70.0° For red light, na sin θ a = nb sin θ b′ (1.00)sin 70.0° (1.00)sin 70.0° and θb′ = 35.71° For violet light, sin θb′ = and θb′ = 34.48° 1.61 1.66 EXECUTE: (a) n decreases with increasing λ , so n is smaller for red than for blue So beam a is the red one sin(θ a − θb′ ) sin(70° − 35.71° ) = 0.6938t and for violet (b) Problem 33.58 says d = t For red light, d r = t cosθb′ cos35.71° gives sin θb′ = sin(70° − 34.48°) 0.10 cm = 0.7048t d v − d r = 1.0 mm gives t = = 9.1 cm cos34.48° 0.7048 − 0.6938 EVALUATE: Our calculation shown that the violet light has greater lateral displacement and this is ray b IDENTIFY and SET UP: The polarizer passes 12 of the intensity of the unpolarized component, light, d v = t 33.61 independent of φ Out of the intensity I p of the polarized component the polarizer passes intensity I p cos (φ − θ ), where φ − θ is the angle between the plane of polarization and the axis of the polarizer (a) Use the angle where the transmitted intensity is maximum or minimum to find θ See Figure 33.61 Figure 33.61 EXECUTE: The total transmitted intensity is I = 12 I + I p cos (φ − θ ) This is maximum when θ = φ and from the table of data this occurs for φ between 30° and 40°, say at 35° and θ = 35° Alternatively, the total transmitted intensity is minimum when φ − θ = 90° and from the data this occurs for φ = 125° Thus, θ = φ − 90° = 125° − 90° = 35°, in agreement with the above (b) IDENTIFY and SET UP: I = 12 I + I p cos (φ − θ ) Use data at two values of φ to determine the two constants I and I p Use data where the I p term is large (φ = 30°) and where it is small (φ = 130°) to have the greatest sensitivity to both I and I p EXECUTE: φ = 30° gives 24.8 W/m = 12 I + I p cos (30° − 35°) © Copyright 2012 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 33-21 24.8 W/m = 0.500 I + 0.9924 I p φ = 130° gives 5.2 W/m = 12 I + I p cos (130° − 35°) 5.2 W/m = 0.500 I + 0.0076 I p Subtracting the second equation from the first gives 19.6 W/m = 0.9848I p and I p = 19.9 W/m And then I = 2(5.2 W/m − 0.0076(19.9 W/m )) = 10.1 W/m EVALUATE: Now that we have I , I p and θ we can verify that I = 12 I + I p cos (φ − θ ) describes that 33.62 data in the table IDENTIFY: The angle by which the plane of polarization of light is rotated depends on the concentration of the compound SET UP: If we follow the hint in the problem and graph (not shown) the concentration C as a function of the rotation angle ϑ , l-leucine and d-glutamic acid both exhibit linear relationships between C and ϑ , with the y-intercept being zero in both cases Using the slope y-intercept form of the equation of a straight line ( y = mx + b), we can find the equation for C as a function of ϑ EXECUTE: For l-leucine, the slope of the graph is m = −9.09 g ⋅ deg −1, so the equation for C as a 100 mL g ⎛ ⎞ function of ϑ is C = − ⎜ 9.09 ⋅ deg −1 ⎟ϑ For d-glutamic acid, the slope is 100 mL ⎝ ⎠ g g ⎛ ⎞ ⋅ deg −1, so the desired equation is C = ⎜ 8.06 ⋅ deg −1 ⎟ϑ The opposite signs in the 100 mL 100 mL ⎝ ⎠ equations tell us that the two compounds rotate the plane of polarization in opposite directions EVALUATE: Inspection of the data indicates that the slope is constant and that the y-intercept is zero (no concentration, no rotation) We can use data points to find the slope For example, using the second and third data points for l-leucine, the slope is ΔC 5.0 g/(100 mL) − 2.0 g/(100 mL) 3.0 g/(100 mL) g = = = −9.09 ⋅ deg −1, m= Δϑ −0.55° − (−0.22°) −0.33° 100 mL which is the same result we got from the graph, leading to the same equation IDENTIFY: The reflected light is totally polarized when light strikes a surface at Brewster’s angle SET UP: At the plastic wall, Brewster’s angle obeys the equation tan θ p = nb /na , and Snell’s law, m = 8.06 33.63 na sin θ a = nb sin θb , applies at the air-water surface EXECUTE: To be totally polarized, the reflected sunlight must have struck the wall at Brewster’s angle tan θ p = nb /na = (1.61)/(1.00) and θ p = 58.15° This is the angle of incidence at the wall A little geometry tells us that the angle of incidence at the water surface is 90.00° – 58.15° = 31.85° Applying Snell’s law at the water surface gives (1.00) sin31.85° = 1.333 sinθ and θ = 23.3° 33.64 EVALUATE: We have two different principles involved here: Reflection at Brewster’s angle at the wall and Snell’s law at the water surface IDENTIFY: The number of wavelengths in a distance D of material is D /λ , where λ is the wavelength of the light in the material D D = + , where we have assumed n1 > n2 so SET UP: The condition for a quarter-wave plate is λ1 λ2 λ2 > λ1 EXECUTE: (a) (b) D = n1D λ0 λ0 4(n1 − n2 ) = = n2 D λ0 + λ0 and D = 4( n1 − n2 ) 5.89 × 10−7 m = 6.14 × 10−7 m 4(1.875 − 1.635) © Copyright 2012 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 33-22 33.65 Chapter 33 EVALUATE: The thickness of the quarter-wave plate in part (b) is 614 nm, which is of the same order as the wavelength in vacuum of the light IDENTIFY: Follow the steps specified in the problem SET UP: cos(α − β ) = sin α sin β + cos α cos β sin(α − β ) = sin α cos β − cos α sin β EXECUTE: (a) Multiplying Eq (1) by sin β and Eq (2) by sin α yields: x y (1): sin β = sin ωt cos α sin β − cos ωt sin α sin β and (2): sinα = sin ωt cos β sin α − cos ωt sin β sin α a a x sin β − y sin α Subtracting yields: = sin ωt (cos α sin β − cos β sin α ) a (b) Multiplying Eq (1) by cos β and Eq (2) by cos α yields: x y (1): cos β = sin ωt cos α cos β − cos ωt sin α cos β and (2): cos α = sin ωt cos β cos α − cos ωt sin β cos α a a x cos β − y cos α Subtracting yields: = − cos ωt (sin α cos β − sin β cos α ) a (c) Squaring and adding the results of parts (a) and (b) yields: ( x sin β − y sin α ) + ( x cos β − y cos α ) = a (sin α cos β − sin β cos α ) (d) Expanding the left-hand side, we have: x (sin β + cos β ) + y (sin α + cos α ) − xy (sin α sin β + cos α cos β ) = x + y − xy (sin α sin β + cos α cos β ) = x + y − xy cos(α − β ) The right-hand side can be rewritten: a (sin α cos β − sin β cos α ) = a sin (α − β ) Therefore, x + y − xy cos(α − β ) = a sin (α − β ) Or, x + y − xy cos δ = a sin δ , where δ = α − β EVALUATE: (e) δ = : x + y − xy = ( x − y ) = ⇒ x = y, which is a straight diagonal line π a2 δ = : x + y − xy = , which is an ellipse π δ = : x + y = a ,which is a circle This pattern repeats for the remaining phase differences 33.66 2 2 IDENTIFY: Apply Snell’s law to each refraction SET UP: Refer to the figure that accompanies the problem EXECUTE: (a) By the symmetry of the triangles, θbA = θ aB , and θ aC = θ rB = θ aB = θbA Therefore, sin θbC = n sin θ aC = n sin θbA = sin θ aA = θbC = θ aA (b) The total angular deflection of the ray is Δ = θ aA − θbA + π − 2θ aB + θbC − θ aC = 2θ aA − 4θbA + π ⎛1 ⎞ (c) From Snell’s Law, sin θ aA = n sin θbA ⇒ θbA = arcsin ⎜ sin θ aA ⎟ ⎝n ⎠ ⎛1 ⎞ Δ = 2θ aA − 4θbA + π = 2θ aA − 4arcsin ⎜ sin θ aA ⎟ + π ⎝n ⎠ (d) dΔ dθ aA =0= 2−4 d ⎛ ⎛1 ⎞⎞ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ = − A⎜ dθ a ⎝ ⎝n ⎠⎠ 2 ⎛ cosθ1 ⎞ ⎛ sin θ1 ⎞ ⎛ 16cos θ1 ⎞ ⋅⎜ ⋅ ⎜1 − = ⎟ ⎜ ⎟⎟ ⎟ ⎜ n ⎟⎠ ⎜⎝ n2 sin θ1 ⎝ n ⎠ ⎝ ⎠ 1− n 4cos θ1 = n − + cos θ1 3cos θ1 = n − cos θ1 = (n − 1) ⎛ ⎞ ⎛ ⎞ (e) For violet: θ1 = arccos ⎜⎜ ( n − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 58.89° ⎝ ⎠ ⎝ ⎠ Δ violet = 139.2° ⇒ θ violet = 40.8° © Copyright 2012 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 33-23 ⎛ ⎞ ⎛ ⎞ For red: θ1 = arccos ⎜⎜ ( n − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 59.58° ⎝ ⎠ ⎝ ⎠ Δ red = 137.5° ⇒ θ red = 42.5° EVALUATE: The angles we have calculated agree with the values given in Figure 33.20d in the textbook θ1 is larger for red than for violet, so red in the rainbow is higher above the horizon 33.67 IDENTIFY: Follow similar steps to Challenge Problem 33.66 SET UP: Refer to Figure 33.20e in the textbook EXECUTE: (a) The total angular deflection of the ray is Δ = θ aA − θbA + π − 2θbA + π − 2θbA + θ aA − θbA = 2θ aA − 6θbA + 2π , where we have used the fact from the previous problem that all the internal angles are equal and the two external equals are equal Also using the Snell’s law relationship, ⎛1 ⎞ ⎛1 ⎞ we have: θbA = arcsin ⎜ sin θ aA ⎟ Δ = 2θ aA − 6θbA + 2π = 2θ aA − 6arcsin ⎜ sin θ aA ⎟ + 2π n n ⎝ ⎠ ⎝ ⎠ (b) dΔ dθ aA =0= 2−6 d ⎛ ⎛1 ⎞⎞ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ = − A⎜ ⎝ ⎠⎠ n dθ a ⎝ ⎛ cos θ ⎞ ⎜ ⎟ sin θ ⎝ n ⎠ 1− 2 n ⎛ sin θ ⎞ = (n − + cos θ ) = 9cos θ cos θ = (n − 1) n ⎜1 − ⎟ ⎜ ⎟ n ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ (c) For violet, θ = arccos ⎜⎜ (n − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 71.55° Δ violet = 233.2° and ⎝ ⎠ ⎝ ⎠ θ violet = 53.2° ⎛ ⎞ ⎛ ⎞ For red, θ = arccos ⎜⎜ (n − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 71.94° Δ red = 230.1° and θ red = 50.1° ⎝ ⎠ ⎝ ⎠ EVALUATE: The angles we calculated agree with those given in Figure 33.20e in the textbook The color that appears higher above the horizon is violet The colors appear in reverse order in a secondary rainbow compared to a primary rainbow © Copyright 2012 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 ... sin θa = nb sinθ b and n = 1.000 for air At point B the angle of the prism is 30.0° EXECUTE: Apply Snell’s law at AB The prism angle at A is 60.0°, so for the upper ray, the angle of incidence... IDENTIFY: The incident angle at the prism → water interface is to be the critical angle SET UP: The path of the ray is sketched in Figure 33.50 The ray enters the prism at normal incidence so is not... 33.54 EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism IDENTIFY: No light enters the gas because total internal reflection
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