Physics 41 Chapter 38 HW Key


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1 Physics 41 Chapter 38 HW Key 1. Helium neon laser light (63..8 nm) is sent through a mmwide single slit. What is the width of the central imum on a screen 1.00 m from the slit? sin θ a y 4. mm 3 y tan θ sin θ θ (for small θ) 1.00 m. A beam of monochromatic green light is diffracted by a slit of width mm. The diffraction pattern forms on a wall.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 4.m. Calculate the wavelength of the light. y P38. The positions of the firstorder minima are sin θ ±. Thus, the spacing between these two minima is L a y L and the wavelength is a 3 3 y a L.06 m 547 nm Coherent light of wavelength nm is sent through two parallel slits in a large, flat wall. Each slit is µ m wide. Their centers are.80 µ m apart. The light then falls on a semicylindrical screen, with its axis at the midline between the slits. (a) Predict the direction of each interference imum on the screen as an angle away from the bisector of the line joining the slits. (b) Describe the pattern of light on the screen, specifying the number of bright fringes and the location of each. (c) Find the intensity of light on the screen at the center of each bright fringe, expressed as a fraction of the light intensity at the center of the pattern. Solution: First some review: The total intensity of the double slit interference pattern is a combination of single slit diffraction and the double slit interference. The cosine squared term gives the intensity of the fringes inside the single slit ima. The sine squared terms is the envelope that shapes the fringes. ( πa θ / ) πd sin θ sin sin cos πa sin θ / Without the diffraction envelope all of the fringes would have equal intensity. This is what we were shown in chapter 37, although this pattern doesn t really exist.
2 So for this problem, you use the full expression to find the intensity of the bright fringes. The single slit minima DOMNATE the double slit interference so if a double ima falls on a single slit minima, the spot will be dark. To find the location of the fringes: Doubleslit interference ima are at angles given by Singleslit diffraction minima, at asin θ m. dsin θ m. (a) The semicylindrical screen is irrelevant. The point is, they include that because it would be difficult to SEE fringes at large angles unless the screen was curved but it doesn t matter when you are solving for the angles. Doubleslit interference ima are at angles given by dsin θ m. For m 0, θ m, (.80 µ m ) sin θ 1( µ m ) : ( ) For 1 θ 1 sin Similarly, for m, 3, 4, 5 and 6, θ 1.0, θ 3 3.5, θ θ, and θ 6 sin ( 1.07) nonexistent Thus, counting the central fringe, there are directions for interference ima. (b) We check for missing orders by looking for singleslit diffraction minima, at asin θ m because the single slit diffraction DOMNATES the double slit interference: For 1 m, ( µ m ) sin θ 1( µ m ) and θ Thus, there is no bright fringe at this angle. There are only nine bright fringes, at θ 0, ± 10.3, ± 1.0, ± 3.5, and ± c) To find the ratio of intensities, you have to use this: ( πa θ / ) πd sin θ sin sin cos πa sin θ /
3 The teacher s solution is using this equation too but they don t explain HOW they use it. That is why reverse engineering the teacher s solution manual can be a waste of time. Let me explain their solution. n part (a) we found angles for which dsin θ m. f you put those angles into this equation, the cosine squared term will always be equal to 1. We solved for the ima. Where the cosine squared term is equal to 1. SO that term drops out and we just end up with using: ( π a sinθ ) sin πasinθ Now what they are doing for the rest of the solution is just solving for the argument and making the units work out because you can t have use both radian and degrees at the same time in your calculator. Check it out. f had just one big equation, see what happens. m going to substitute in the givens and simplify, leaving only the angle: ( π x m θ x m ) ( θ ) sin (.7 10 ) sin ( ) sin sin (.7 10 )sin ( ) sinθ 6 6 π x m θ x m Check it out if you put in an angle, the top argument in the sine is then in radians and you can t have both radians and angles at the same time in your calculator. So the easiest way THNK to do this is to make a substitution: ( β ) sin β where β πa sinθ Then for each angle, solve for β in degree measure and then switch your calculator to radian and put β in and solve for the final answer. Atθ 0, lim sin θ 1and θ Atθ 10.3, use degrees: β ( ) 1.00 π a sinθ π µ m sin which is RADANS µ m Now switch your calculator to radians: ( ) sin OR, if you don t want to switch to radians, then you have to convert β into degrees for the top and then you can keep your calculator in degrees: π asin θ π ( µ ) µ m sin rad m sin And that is what the teacher solution manual did. t works either way but for this intensity formula you can t use both radians and degrees at the same time!!!! So, moral of the story, maybe it is better not to have the teacher solution manual!!!!
4 Similarly, atθ 1.0, Atθ 3.5, Atθ 63.6, π asin θ 1.57 rad 90.0 and π asin θ.36 rad 135 and π asin θ 3.93 rad 5 and The pupil of a cat s eye narrows to a vertical slit of width mm in daylight. What is the angular resolution for horizontally separated mice? Assume that the average wavelength of the light is 500 nm sin θ rad 4 a The mpressionist painter Georges Seurat created paintings with an enormous number of dots of pure pigment, each of which was approximately.00 mm in diameter. The idea was to have colors such as red and green next to each other to form a scintillating canvas. Outside what distance would one be unable to discern individual dots on the canvas? (Assume that 500 nm and that the pupil diameter is 4.00 mm.) Sunday Afternoon on the sland of La Grande Jatte, by Georges Seurat. By Rayleigh s criterion, two dots separated centertocenter by.00 mm would overlap d when θ min 1.. L D Thus, 3 3 (.00 ) ( 4.00 ) 9 ( ) dd L m.
5 5. A circular radar antenna on a Coast Guard ship has a diameter of. and radiates at a frequency of 15.0 GHz. Two small boats are located 9.00 km away from the ship. How close together could the boats be and still be detected as two objects? d P c m D L f D. L m ( m ) ( m ) d m. 6. The hydrogen spectrum includes a red line at 656 n and a blueviolet line at 434 nm. What are the angular separations between these two spectral lines obtained with a diffraction grating that has grooves/cm? P38.1The grating spacing is m d n the 1storder spectrum, diffraction angles are given by sin θ : sin θ d. so that for red θ and for blue so that θ sin θ The angular separation is in firstorder, θ n the secondorder spectrum, θ sin sin d d 13. Again, in the third order, θ sin sin d d 6.5 Since the red does not appear in the fourthorder spectrum, the answer is complete. 7. A helium neon laser ( 63.8 nm) is used to calibrate a diffraction grating. f the firstorder imum occurs at 0.5, what is the spacing between adjacent grooves in the grating? P38. sin θ : Line spacing 1.81 µ m 63.8 nm nm d sin θ A diffraction grating has 4 00 rulings/cm. On a screen.00 m from the grating, it is found that for a particular order m, the ima corresponding to two closely spaced wavelengths of sodium (589.0 nm and nm) are separated by 1.59 mm. Determine the value of m.
6 P d nm 4 00 cm dsin θ m or 1 m y Ltan θ Ltan sin d 1 m θ sin d and Thus, 1 m 1 m1 y L tan sin tan sin d d For m 1, y (.00 m ) tan sin tan sin mm For m, ( ) ( ) y (.00 m ) tan sin tan sin 1.54 mm For m 3, ( ) ( ) y (.00 m ) tan sin tan sin 5.04 mm Thus, the observed order must be m. 9. Unpolarized light passes through two ideal Polaroid sheets.the axis of the first is vertical and that of the second is at 30.0 to the vertical. What fraction of the incident light is transmitted? P38.34 The average value of the cosinesquared function is onehalf, so the first polarizer transmits 1 the light. The second 3 transmits cos f i i 10. The critical angle for total internal reflection for sapphire surrounded by air is Calculate the polarizing angle for sapphire. P sin θ c or n 1 1 n sin θ sin 34.4 Also, tan θ p n. Thus, 1 θ tan ( ) tan 1 p n ( 1.77) c
7 11. The Very Large Array (VLA) is a set of 7 radio telescope dishes in Caton and Socorro Counties, New Mexico (Fig. P38.54). The antennas can be moved apart on railroad tracks, and their combined signals give the resolving power of a synthetic aperture 36.0 km in diameter. (a) f the detectors are tuned to a frequency of 1.40 GHz, what is the angular resolution of the VLA? (b) Clouds of hydrogen radiate at this frequency. What must be the separation distance of two clouds at the center of the galaxy, lightyears away, if they are to be resolved? (c) What f? As the telescope looks up, a circling hawk looks down. Find the angular resolution of the hawk s eye. Assume that that the hawk is most sensitive to green light having a wavelength of 500 nm and that it has a pupil of diameter 1.0 mm. (d) A mouse is on the ground 30.0 m below. By what distance must the mouse s whiskers be separated if the hawk can resolve them? P38.50 (a) v : f s 0.14 m s (b) θ min : min 4 min 1. D 0.14 m θ µ rad s θ min 7.6 µ rad 1.50 arc seconds π d d θminl rad ly ly L θ : ( ) ( ) (c) min 1. D m 1.0 θ θ µ rad ( 10.5 seconds of arc) min 3 (d) d θ L ( )( ) 6 3 min rad 30.0 m mm 1. An American standard television picture is composed of about 485 horizontal lines of varying light intensity. Assume that your ability to resolve the lines is limited only by the Rayleigh criterion and that the pupils of your eyes are 5.00 mm in diameter. Calculate the ratio of minimum viewing distance to the vertical dimension of the picture such that you will not be able to resolve the lines. Assume that the average wavelength of the light coming from the screen is 550 nm. P38.57 The limiting resolution between lines θ 9 ( 550 ) ( 5.00 ) rad. D min 3 Assuming a picture screen with vertical dimension, the minimum viewing distance for no visible lines is found from 485 θ min. The desired ratio is then L L θ rad min ( ) 13. f the interplanar spacing of NaCl is , what is the predicted angle at which Xrays of wavelength nm will be diffracted in a firstorder imum? Answer: 14.4
8 14. Suppose a 5meter diameter telescope were constructed on the dark side of the moon. The viewing there (except for brief periods of sunlight) would be excellent. As an example, what would be the separation between two objects that could just be resolved on the planet Mars in 500 nm light? [The distance to Mars at closest approach is 50 million miles]. Answer: 9.8 km 15. Suppose a 5meter diameter telescope were constructed on the dark side of the moon. The viewing there (except for brief periods of sunlight) would be excellent. As an example, what would be the separation between two objects that could just be resolved on the planet Mars in 500 nm light? [The distance to Mars at closest approach is 50 million miles]. 16. An unpolarized beam of light is incident on a stack ofideal polarizing filters. The axis of the first filter is perpendicular to the axis of the last filter in the stack. Find the fraction by which the transmitted beam s intensity is reduced in the following three cases. (a) Three filters are in the stack, each with its transmission axis at 45.0 relative to the preceding filter. (b) Four filters are in the stack, each with its transmission axis at 30.0 relative to the preceding filter. (c) Seven filters are in the stack, each with its axis at 15.0 relative to the preceding filter.(d) Comment on comparing the answers to parts (a), (b), and (c). 17. Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.38 mm, causing the interferometer pattern to reproduce itself times. Determine the wavelength of the light. What color is it?
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