Unformatted text preview: Recitation ClassWork 10 solutions Physics 204, morning summer 2008 Chapter 27 Conceptual Questions #11,8,10 Concepts & Calculations #61 Chapter 28 Problems #4,7 11. REASONING AND SOLUTION A transparent coating is deposited on a glass plate and has a refractive index that is larger than that of the glass. For a certain wavelength within the coating, the thickness of the coating is a quarter wavelength. The coating enhances the reflection of the light corresponding to this wavelength. As discussed in Example 3, only the light reflected from the top of the coating is traveling in through a material (air) with lower refractive index toward a material coating with a larger refractive index. Therefore, only the light reflected from the top of the coating undergoes a phase change upon reflection. This change in phase is equivalent to half of a wavelength. Therefore, there is a net phase difference equivalent to half of a wavelength between the light reflected from the top and light reflected from the bottom of the coating. Since the thickness of the coating is a quarter wavelength, the light reflected from the bottom of the coating traverses a distance equal to two times a quarter wavelength, or half a wavelength, as it travels through the coating. The overall relative phase difference between the light reflected from the top and the light reflected from the bottom of the coating is equivalent to half of a wavelength due to the reflection from the top of the coating and another half of a wavelength due to the path difference between the light reflected from the top and the light reflected from the bottom. Thus, the overall phase difference is equivalent to a whole wavelength, and the light reflected from the top is in phase with the light reflected from the bottom. Constructive interference occurs, and the coating enhances light corresponding to that wavelength. 8. REASONING AND SOLUTION Two pieces of the same glass are covered with thin films of different materials. The thickness t of each film is the same. In reflected sunlight, however, the films have different colors. This could occur because, in general, different materials have different refractive indices. Using reasoning similar to (but not identical to) that in Example 3, the wavelength air that is removed from the reflected light corresponds to a value of film in the film that satisfies either of the following conditions: and 2t = mλ film , d i m = 0, 1, 2, 3, . . . 1
2 t = m + 2 λ film , m = 0, 1, 2, 3, . . . n film > n glass , (1) n film < n glass . (2) However, the value of the wavelength film of the light in the film depends on the index of refraction of the film. According to Equation 27.3, λ film = λ vacuum / n , where n is the index of refraction of the film. If the index of refraction of the two materials are different, then conditions (1) and (2) will be satisfied for two different values of film. These values of film correspond to different values of air, and, therefore, different wavelengths will be removed from the reflected light. 10. SSM REASONING AND SOLUTION A thin film of material is floating on water (n = 1.33). When the material has a refractive index of n = 1.20, the film looks bright in reflected light as its thickness approaches zero. But when the material has a refractive index of n = 1.45, the film looks black in reflected light as its thickness approaches zero. When the material has a refractive index of n = 1.20, both the light reflected from the top of the film and the light reflected from the bottom occur under conditions where light is traveling through a material with a smaller refractive index toward a material with a larger refractive index. Therefore, both the light reflected from the top and the bottom undergoes a phase change upon reflection. Both phase changes are equivalent to half of a wavelength. Therefore, the reflections introduce no net phase difference between the light reflected from the top and bottom. As the thickness of the film approaches zero, the path difference between the light reflected from the top and the bottom of the film approaches zero. Since the path difference is close to zero, and there is no relative phase difference due to reflection, the light reflected from the top will be in phase with the light reflected from the bottom of the film; constructive interference will occur, and the film looks bright in reflected light. When the material has a refractive index of n = 1.45, only the light reflected from the top of the film is traveling through a material with lower refractive index toward a material with a larger refractive index. Therefore, only the light reflected from the top undergoes a phase change upon reflection. This phase change is equivalent to half of a wavelength. Therefore, there is a net phase difference equivalent to half of a wavelength between the light reflected from the top and bottom of the film. As the thickness of the film approaches zero, the path difference between the light reflected from the top and the light reflected from the bottom of the film approaches zero. Since the path difference is close to zero, and there is a relative phase difference due to reflection that is equal to half of a wavelength, the light reflected from the top will be exactly out of phase with the light reflected from the bottom of the film, and destructive interference will occur. Therefore, the film looks black in reflected light. 61. CONCEPT QUESTIONS a. In air the index of refraction is nearly n = 1, while in the film it is n = 1.33. A phase change occurs whenever light travels through a material with a smaller refractive index toward a material with a 1
larger refractive index and reflects from the boundary between the two. The phase change is equivalent to 2 λ film , where λfilm is the wavelength in the film. b. In the film the index of refraction is n = 1.33, while in the glass it is n = 1.52. This situation is like that in 1
Question (a) and, once again, the phase change is equivalent to 2 λ film , where λfilm is the wavelength in the film. c. The wavelength of the light in the film (refractive index = n) is given by Equation 27.3 as λfilm = λvacuum/n. Since the refractive index of the film is n = 1.33, the wavelength in the film is less than the wavelength in a vacuum. SOLUTION Both the light reflected from the air‐film interface and from the film‐glass interface experience phase 1
changes, each of which is equivalent to 2 λ film . In other words, there is no net phase change between the waves reflected from the two interfaces, and only the extra travel distance of the light within the film leads to the destructive interference. The extra distance is 2t, where t is the film thickness. The condition for destructive interference in this case is c h 1
3
5
1
2t
= 2 λ film , 2 λ , 2 λ film , ... = m + 2 λ film
123 1444 film4444
4
2
3 Extra travel
distance in film where m = 0 , 1, 2 , 3, ... Condition for
destructive interference Solving for the wavelength gives λ film = 2t
1
m+ 2 m = 0, 1, 2, 3, ... According to Equation 27.3, we have λfilm = λvacuum/n. Using this substitution in our result for λfilm, we obtain λ vacuum = 2 nt
1
m+ 2 m = 0, 1, 2, 3, ... For the first four values of m and the given values for n and t, we find b gb g 2 1.33 465 nm
2 nt
=
= 2470 nm
1
1
0+ 2
m+ 2 m=0 λ vacuum = m=1 λ vacuum = 825 nm m=2 λ vacuum = 495 nm m=3 λ vacuum = 353 nm The range of visible wavelengths (in vacuum) extends from 380 to 750 nm. Therefore, the only visible wavelength in which the film appears dark due to destructive interference is 495 nm . 4. REASONING The total time for the trip is one year. This time is the proper time interval t0, because it is measured by an observer (the astronaut) who is at rest relative to the beginning and ending events (the times when the trip started and ended) and who sees them at the same location in spacecraft. On the other hand, the astronaut measures the clocks on earth to run at the dilated time interval t, which is the time interval of one hundred years. The relation between the two time intervals is given by Equation 28.1, which can be used to find the speed of the spacecraft. ( ) SOLUTION The dilated time interval t is related to the proper time interval t0 by Δt = Δt0 / 1 − v 2 / c 2 . Solving this equation for the speed v of the spacecraft yields ⎛ Δt ⎞
v = c 1− ⎜ 0 ⎟
⎝ Δt ⎠ 2 2 ⎛ 1 yr ⎞
= c 1− ⎜
⎟ = 0.999 95c ⎝ 100 yr ⎠ (28.1) 7. REASONING AND SOLUTION The proper time is the time it takes for the bacteria to double its number, i.e., t0 = 24.0 hours. For the earth based sample to grow to 256 bacteria, it would take 8 days (2n = 256 or n = 8). The "doubling time" for the space culture would be Δt0 Δt = ⎛v⎞
1– ⎜ ⎟
⎝c⎠ 2 = 24.0 h
⎛ 0.866 c ⎞
1– ⎜
⎟
⎝ c ⎠ 2 = 48.0 h In eight earth days, the space bacteria would undergo n' = or 2 days ( 1 ) 8 = 4 "doublings". The number of space bacteria 2 is 4
Number of Space Bacteria = 2n' = 2 = 16 ...
View
Full
Document
This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.
 Spring '09
 rollino
 Physics

Click to edit the document details