Chapter 36, example problems:

Transcription

1 Chapter 6, example problems: (6.0) Light wave with electric field E y (x, t) = E max sin [( m ) x ω t] passes through a slit. First dark band at ±2.6 from the center of the diffraction pattern. (a) Frequency of light f =? We use m = k = 2 π / λ, and obtain λ = 2.6 nm. Thus f = 0 m/s / 2.6 nm = Hz. (b) We use a sin θ = m λ, and put in m = ±, to get a = 2.6 nm / sin 2.6 =. 09 μm. (c) Other dark bands also obey a sin θ = m λ, but with other interger values for m. For m = ±2, we get θ = sin (±2 2.6 nm /.09 μm) = ±7.2. These are the second-order dark bands. For m = ±, we find no solution, because the magnitude of the argument of arcsine cannot be bigger than unity. Thus third order-dark bands and beyond can not be observed in this diffraction pattern. Thus the total number of dark bands observable in this diffraction pattern is four (two on each side of the center). (6.) Monochromatic radiation, wavelength λ. From distant source. Passes through a slit. Screen 2.0 m in front of slit. Width of center maximum = 6.00 mm. (a) λ = 00 nm. a = m λ / sin θ m λ / θ (in rad.) = 00 nm / (.00 mm / 2.0 m) =.7 0 m or 7 μm. (b) λ = 0.0 μm. a = m λ / sin θ m λ / θ (in rad.) = 0.0 μm / (.00 mm / 2.0 m) = m or.7 cm. (c) λ = 0.00 nm. a = m λ / sin θ m λ / θ (in rad.) = 0.00 nm / (.00 mm / 2.0 m) = m or 7 nm. (6.) Single-slit diffraction. Amplitude at O in Fig. 6.a is E 0. Draw phasor diagram and determine amplitude graphically for: (a) sin θ = λ / 2a (or a sin θ = λ / 2): β = (2π / λ) a sin θ = π. That is, the last little vector is rotated by π from the first vector. Hence the phasor diagram: The ratio of cord length to arc length is 2r / πr = 2 / π. Hence the amplitude for this θ is (2 / π) E 0. (Note that in this case E 0 = πr.) Thus for this θ, I = (2 / π) 2 I 0, because the intensity is proportional to the amplitude squared.

2 (b) sin θ = λ / a (or a sin θ = λ ) : β = (2π / λ) a sin θ = 2π. That is, the last little vector is rotated by 2π from the first vector, which means in the same direction. Since the cord length is now zero, the amplitude for this θ is 0. (Note that in this case E 0 = 2πr.) Thus for this θ, I = 0. Note that a sin θ = λ is precisely the condition for the first minimum. (c) sin θ = λ / 2a (or a sin θ = λ / 2): β = (2π / λ) a sin θ = π. That is, the last little vector is rotated by π from the first little vector, which means in the opposite direction to the first little vector. Thus the cord is just like the case (a), yet the arc has gone around a circle one and a half times. Their length ratio is therefore 2r / πr = 2 / π. Hence the amplitude for this θ is (2 / π) E 0. (Note that in this case E 0 = πr.) Thus for this θ, I = (2 / π) 2 I 0 = 0.00 I 0. One aspect of the above drawings is misleading. The total number of little arrows should be the same, and the length of each little arrow should be the same, only their turning angles are different in the three cases. Thus the three circles do not have the same radius, and should become smaller as sin θ increases. It is like wrapping a string of a fixed length (=E 0 ) around a cylinder of smaller and smaller radius. Depending on the radius of the cylinder, the string can cover half of the cylinder (the first case), the full cylinder (the second case), or one-and-a half times the cylinder (the third case). The length of the string is always E 0. The radius of the cylinder in each case is decided by the value of β, which decides how many times the cylinder should be covered by the string.

3 (6.22) Interference pattern due to eight parallel, equally spaced, narrow slits. Phase difference φ between light from adjacent slits is π/. Interference minimum. Phasor diagram given by Fig. 6.(b): From the phasor diagram, it is clear that 6 destructive interference occurs between light from the first slit and the fifth slit, between the second slit and the sixth slit, 7 between the third slit and the seventh slit, and between the fourth slit and the eighth 2 slit. Thus for this phase difference φ between light from adjacent slits, light from the eight slits are pair-wise cancelled, giving I = 0 for this φ. (6.2) Visible range nm. White light falls on a diffraction grating at normal incidence. 0 slits/mm. Find angular width of visible spectrum in (a) The first order: First order bright line decided by d sin θ = λ, or, for very small θ, dθ = λ. Then it is clear that d Δθ = Δλ, where Δλ denoted the visible range of λ, and Δθ denotes to the corresponding range of θ, which is just the angular width of the visible spectrum. Thus Δθ = Δλ / d = 00 nm / (mm / 0) = 0.0 rad. (b) The third order. The equation is now dθ = λ Δθ = 0.0 rad = 0. rad. Thus higher order does give a larger angular width of the visible spectrum. But if it is too large, it could overlap with the neighboring orders. (6.6) Hydrogen, 66. nm. Deuterium, nm. Δλ = 0. nm. Second order, dθ = 2λ the angular separation of the two second-oder lines from the two isotopes is Δθ = 2 Δλ / d = 2 0. nm / d. On the other hand, the half-width of each bright line is the same as the half-width of the center bright line, which is given by the θ in Nd sin θ = λ, or, for very small θ, Ndθ = λ. It gives a halfwidth of 66. nm / Nd. (We have used the average λ since the tiny difference is not important here.) Thus to resolve the two lines, we need to require: 66. nm / Nd < 2 0. nm / d. We see that the factor d is cancelled out from the two sides, and we get: N > 66. nm / 0.6 nm = 2.0. (We have used the average wavelength to get this answer. If we used nm, we would get N > 22.97, and if we used 66. nm, we would get N > 2.7. The average of the two answers is N > Actually, the two bright lines are of slightly different widths, so using the average of the two widths makes the best sense. This is equivalent to using the average wavelength. (6.6) Photography. Telephoto lens. f = mm. Maximum aperture f /.00. s =. m. λ = 0 nm. Bear. m away. (f /.00 means that f / D =.00, where D is the diameter of the aperture. Hence D = f /.00. )

4 (a) Width of the smallest resolved feature =. m angular width =. m (.22 λ / D) =. m [.22 0 nm / ( mm/.00)] = m = mm. (b) f-stop changed to f /22.0. Width of the smallest resolved feature =. m [.22 0 nm / ( mm / 22.0)] = m =.26 mm. (6.0) Searching for starspots. Hale telescope. Mirror diameter = 200 in (=.0 m ). Focuses visible light. Large sunspot is about 0,000 mi (= m). Most distant star to see this sunspot? Denote this distance d. Then m / d =.22 0 nm /.0 m, or d =.2 0 m = light-years. Any star this close? No. The closest star is red dwarf Proxima Centauri, and is.22 light-years away. (6.6) Diffraction grating design. First-order visible spectrum dispersed to an angular range of.0. (a) Find number of slits per centimeter. We can not use d Δθ = Δλ here [See Prob. (6.2) ], since the angle involved is not small. Instead, we must use d sinθ = λ for λ = 00 nm and λ 2 = 700 nm, and get sin θ = λ / d and sin θ 2 = λ 2 / d. But θ 2 = θ +.0, and sin (θ +.0 ) = sin θ cos.0 + cos θ sin.0. Hence (λ / d ) cos.0 + [ (λ / d ) 2 ] sin.0 = λ 2 / d, or, [ (λ / d ) 2 ] sin 2.0 = [(λ 2 / d ) (λ / d ) cos.0 ] 2, or, after multiplying both sides by d 2, and then solving for d 2, we obtain d 2 = [λ 2 λ cos.0 ] 2 / sin λ 2 = 629 nm 2, implying d = 276 nm = cm. Thus the number per centimeter is: / cm = 76 slits/cm. (b) beginning and ending angles of this range: θ = sin (λ / d ) =.27, and θ 2 =.27. You can also get θ 2 from θ 2 = sin (λ 2 / d ) as a double check. (6.6) Phasor diagram for eight equally-spaced, narrow (identical) slits. Consider φ = π/, π/, π/2, 7π/. (a) If φ = π/ = π/2 + π/ = 90 +, then we should draw eight vectors of the same length, each rotated from the previous one by 90 + (counterclockwise), 2 6 7

5 If φ = π/ = π + π/ = 0 +, then we should draw eight vectors of the same length, each rotated from the previous one by 0 + (counterclockwise), If φ = π/2 = π + π/2 = , then we should draw eight vectors of the same length, each rotated from the previous one by (counterclockwise), It is clear that destructive interference occurs between the first and the third slits, between the second and the fourth slits, between the fifth and the seventh slits, and between the sixth and the eighth slits. Therefore the net amplitude is zero, If φ = 7π/ = π + π/2 + π/ = , then we should draw eight vectors of the same length, each rotated from the previous one by (counterclockwise), and then connect them end to end. These eight vectors are: Each case is different, so it is impossible for you to attempt to remember the answers to all cases. However, once you learned the general concepts, it should not be difficult for you to do any such case, for any number of slits.