1
CHAPTER TWENTY-FIVE SOLUTIONS CHAPTER TWENTY-FIVE SOLUTIONS Chapter Twenty-Five Readings
Bandana, E., "The Mystery of Myopia," The Sciences, Sciences, Nov./Dec. 1985, p. 46. Marr, D., Vision, Vision , New York, W.H. Freeman, 1982. Michael, C.R., "Retinal Processing of Visual Images," Scientific American, American, May 1969, p. 105. Neisser, U., "The "The Processes Processes of of Vision," Vision," Scientific American, American, September 1968, p. 204. Price, W.H., "Photographic Lens," Scientific American, American, August 1976, p. 72. Ruiz, M., "Camera Optics," The Physics Teacher , September 1982, p. 372. Shankland, R.S., "Michelson and His Interferometer," Physics Today Today,, April 1976, p. 72. Stix, G., "Pictures Worth a Thousand Cameras," Scientific American, American, November 1996, p. 46. Wald, G., "Eye and Camera," Scientific American, American, August 1950, p. 32. Walker, T.D., Light and its its Uses, Uses , New York, W.H. Freeman, 1980. 1 1 1 1 1 1 1 1 + = , we have: = - = , and from this, p q f q f p 0.250 m 1.50 m 30.0 cm. Thus the image is formed formed 30.0 cm beyond the lens. q 30 cm 1 M = = ==- . 150 cm 5 p
25.1
From
25.2
The magnitude of the magnification is:
h'
M = h
q = 0.300 m=
q = . p
p Therefore h = h' ( ) q From the thin lens equation, pf (100 m)(52.0 mm) q = = = 52.0 mm. p p - f f 100 m - 52.0 x 10 -3 m p 100 m The object height is then: h = h' ( ) = (92.0 mm)( ) = 177 m 52.0 mm q 25.3
Consider rays coming from opposite edges of the object and passing undeviated through the center of the lens as shown at the right. The angle between these rays is the angular width of the object. If the object is very distant, the image distance is equal to the focal length of the lens. Then, if f if f = = 55.0 mm and = = 20.0°, the image size may be
angular width of obj obj ect =
f
h (image size)
2
CHAPTER TWENTY-FIVE SOLUTIONS found as follows: h 2 20° = tan , or h = (2 f ) tan = [2(55.0 mm)] tan = 19.4 mm. 2 2 2 f Therefore, the image easily fits within a 23.5 mm by 35.0 mm space. 25.4
The magnitude of the magnification is M =
h' q = . h p
q Therefore h' = h ( ) p The object distance is: p = 3.84 x 10 8 m and with the object so far away, The object size is the diameter of the moon. Thus, h = 2(1.74 x 10 6 m) = 3.48 x 10 6 m and the image size is: (3.48 x 106 m)(120 x 10-3 m) h' = = 1.09 x 10-3 m = 1.09 mm. (3.84 x 108 m)
25.5
q = f = 120 mm.
1 256 s 1 The exposure time is being reduced by a factor of: = 1 8 32 s Thus, to maintain correct exposure, you need to get 8 times more light on the film. To do this the aperture opening will have to be 8 times the original area, or: (new diameter) = 8 (original diameter) , or D2 =
8 D1.
From the definition of the f -number, f # =
f #lens f #lens f #new = = = D2 8 D1
f #lens D1 8
focal length so Diameter
f #old 4 = = = 1.41. 2.83 8
We must use an f /1.4 setting. 25.6
To properly focus the image of a distant object, the lens must be a distance equal to the focal length away from the film ( q1 = 65.0 mm). For the closer object: 1 1 1 1 1 1 + = becomes + = , and q2 = 67.2 mm. 2000 mm q2 65.0 mm p2 q2 f The lens must be moved away a distance D = q2 - q1 = 2.20 mm.
25.7
To double the energy delivered to the film while using the same exposure time, you must double D2new D2old the area of the aperature. Then, we have A2 = 2 A1 or = 2( ), or Dnew = 2 Dold 4 4 . f lens From the definition of the f -number, f # = . In this case, the focal length of the lens is D constant. f lens f lens f #old 11 Thus, f #new = = = = = 7.78. Dnew 2 Dold 2 2 There is no f /7.78 setting on a camera, so one should use the f /8 setting.
25.8
She needs a lens that forms a virtual image at the near point
3
CHAPTER TWENTY-FIVE SOLUTIONS (q = - 60.0 cm) when the paper is held at 24.0 cm.
1 -60 cm
1 = , so f
1 1 1 + = : becomes: p q f
1 + 24.0 cm
f = + 40.0 cm.
25.9
For the right eye, a virtual image of the most distant object should be 8.44 cm in front of the lens (i.e., q = -8.44 cm when p = ). Thus, 1 1 1 1 1 + = gives: 0 + = , or f = -8.44 cm = -0.0844 m, and -8.44 cm f p q f 1 P = = -11.8 diopters. -0.0844 m For the left eye, as in the above, the needed lens has a focal length of: 1 f = -(far point), so f = -0.122 m, and P = = -8.20 diopters. -0.122 m
25.10
(a)
25.11
(a)
When an object is at 25.0 cm in front of the lens ( p = 25.0 cm), the image must be virtual and 100 cm in front of the lens so that the eye can focus on it. (q = -100 cm). 1 1 1 1 1 1 Thus, + = becomes: + = , 25.0 cm p q f -100 cm f from which, f = 33.3 cm. 1 1 (b) P = = = +3.00 diopters. 0.333 m f
(b)
If the far point is at 50.0 cm, we need an image distance of 1 1 1 q = -50.0 cm when p = infinity. Thus, + = becomes: p q f 1 1 0 + = , and f = -50.0 cm = -0.500 m, so the power is: -50.0 cm f 1 1 P = = = - 2.00 diopters. f -0.500 m To be seen by the eye, the virtual image cannot be any closer than 13.0 cm to the lens. Thus, 1 1 1 1 let us find the smallest value the object distance can have: + = becomes: + p q f p 1 = 1 , which gives p = 0.176 m = 17.6 cm. Thus, the near point when -0.130 m -0.500 m glasses are worn is 17.6 cm.
25.12
Considering the image formed by the eye as a virtual object for the implanted lens, we have: p = -(2.53 cm + 2.80 cm) =- 5.33 cm, and q = 2.80 cm 1 1 1 1 1 1 Thus, + = becomes: + = p q f -5.33 cm 2.80 cm f , which gives 1 f = +5.90 cm and P = = +17.0 diopters. 0.059 m
25.13
(a)
To correct nearsightedness, the image of distant objects ( p = ) should be virtual and located at the far point (q = -1.5 m). Thus, 1 1 1 1 1 1 + = becomes: 0 + = , or f = -1.5 m. The power is then P = = p q f f -1.5 m f 1 = -0.67 diopters. -1.5 m (b) To correct farsightedness, objects at p = 25 cm should form a virtual image at the near point 1 1 1 (q = -0.30 m). + = becomes: p q f 1 1 1 + 0.25 m -0.30 m = f , from which f = +1.5 m, and
4
CHAPTER TWENTY-FIVE SOLUTIONS P = 25.14
(a)
1 1 = = +0.67 diopters. 1.5 m f
Objects at the far point ( p = 125 cm) are focused on the retina ( q = 2.00 cm). The focal 1 1 1 length of the lens-cornea combination in this case is: + = , or f far 125 cm 2.00 cm f f ar = 1.97 cm, and
P far =
1 1 = = + 50.8 diopters. 0.0197 m f far
Objects at the near point ( p = 10.0 cm) are also focused on the retina (q = 2.00 cm). Thus, 1 1 1 + = , or 10.0 cm 2.00 cm f near f near = 1.67 cm, and P near = +60.0 diopters. (b) A diverging corrective lens must be used to form virtual images (located at the eye's far point) of very distant objects, ie., 1 1 1 (q = - 125 cm when p = ). Thus, + = gives: p q f 1 1 1 = , or f = - 125 cm, and the power of the needed corrective lens is: 125 cm f P =
25.15
25.16
1 1 = = - 0.800 diopters. f -1.25 m
1 1 1 1 1 1 + = becomes: + = p q f p -25 cm 7.5 cm ,giving p = 5.8 cm. (b) With the virtual image at the normal near point, the angular magnification of a simple 25 cm 25.0 cm magnifier is M = 1 + . In this case, we have: M = 1 + = 5.3. 5.8 cm f (a)
(a)
With the image at the normal near point, the angular magnification is: M = 1 +
25 cm =1+ f
25 cm = 2.0. 25 cm
25.17
When the eye is relaxed:
(a)
We are given that: p = 3.50 cm, and q = -25.0 cm. 1 1 1 1 1 1 Thus, + = becomes: + = , so f = 4.07 cm. p q f 3.50 cm -25.0 cm f With the image at the normal near point, the angular 25.0 cm 25.0 cm magnification is: M = 1 + =1+ = 7.14 f 4.07 cm
(b)
25.18
(a)
M =
25 cm 25 cm = = 1.0. f 25 cm
(b)
First, locate the image being formed by the lens. 1 1 1 1 1 1 + = becomes: + = , so q = 86.5 cm 71.0 cm q 39.0 cm p q f The magnitude of the lateral magnification is then: h' q 86.5 cm = h = p 71.0 cm = 1.22. Thus, h' = 1.22 h. (b) The angular size of the leaf if viewed directly from a distance of h h d = 71.0 cm + 126 cm = 197 would be o= = . When the image is viewed from a d 197 cm h' h' distance of d ' = 126 cm - 86.5 cm = 39.5 cm, its angular size is: = = . The d' 39.5 cm
5
CHAPTER TWENTY-FIVE SOLUTIONS
angular magnification achieved by use of the lens is:
o
=
h'/39.5 cm h' 197 cm = ( ) = h 39.5 cm h/197 cm
1.22(4.99) = 6.08. 25.19
(a)
L M 1 = - = -50.0. Thus, f o
(b) M e = (c) 25.20
f o =
25.0 cm = 20.0, and f e
L 20.0 cm = = 0.400 cm. 50.0 50.0
f e =
25.0 cm = 1.25 cm. 20.0
M = M 1 M e = - (50.0)(20.0) = - 1000.
First, find the size of the final image: angular size =
he
qe
=
he = 1.43 x 10-3 rad, so he = 29.0 cm
4.15 x 10-2 cm. Now apply thin lens equation to each lens to find the lateral magnification produced by each lens and hence the overall magnification: 1 1 1 + = becomes: p q f 1 1 1 + = , so pe = 0.950 cm pe -29.0 cm 0.920 cm, and qe 29.0 cm M e = == 31.5 0.920 cm pe For the objective lens, we have:
1 1 1 + = , or po qo f o
1 1 1 + = , so po = 1.72 cm, 28.1 cm 1.622 cm po qo 28.1 cm The magnitude of the lateral magnification produced by the objective is: M o = = = po 1.72 cm 16.3, and the overall magnification is: M = M e M o = (31.5)(16.3) = 513. he 4.15 x 10-2 cm Therefore, ho = = = 0.810 m. 513 M 25.21
If the eye is relaxed, the final image is at infinity. Thus the object position for the eyepiece is: pe = 25.0 cm f e =2.50 cm, and the angular magnification of the eyepiece is: me = = 10.0 f e The image position for the objective lens is: qo = L - f e = 15.0 cm - 2.50 cm = 12.5 cm. [ Note: The approximation that the the image distance for the objective is nearly equal to the length L of the microscope is a poor one in this case. Hence, we should not use M 1 = .] f o We now find the object position for the objective: 1 1 1 1 1 1 + = , becomes + = , so po = 1.09 cm. 12.5 cm 1.00 cm po qo f o po Thus, the lateral magnification produced by the objective lens is: qo 12.5 cm M 1 = == -11.5, and the overall magnification of the microscope is: 1.09 cm po M overall = M 1me = (-11.5)(10) = -115
6
CHAPTER TWENTY-FIVE SOLUTIONS
25.22
f o 75 cm M = == -19 4.0 cm f e
25.23
(a)
(b)
25.24
The diameter of the lens is 5.00 in = 127 mm, so f o 1250 mm f number = = =9.84 . 127 mm D f o 1250 mm The angular magnification is: M = == - 50 25 mm f e
(See sketch.) The length, x, of an object on the moon, is found as x 1.00 x 10-2 m = ,and x = 15.0 m 3.80 x 108 m
f = 15 m o X
2.53 x 105 m = 157 miles. 25.25
3.8 x 108 m
The length of the telescope is: L = f o + f e = 92 cm. (1) f o The angular magnification is: M = = 45, so f o = 45 f e f e
}
1 cm
(2)
Substitute (2) into (1) to obtain: 45 f e + f e = 92 cm, or f e = 2.0 cm Then (2) gives: f o = 45(2.0 cm) or f o = 90 cm. 25.26
Use the larger focal length (lowest power) lens as the objective element and the shorter focal length 1 (largest power) lens for the eye piece. The focal lengths are: f o = = 0.833 m = 83.3 1.20 diopter cm, and 1 f e = = 0.111 m = 11.1 cm. 9.00 diopters f o (a) The angular magnification (or magnifying power) of the telescope is then: M = = f e
(b)
25.27
83.3 cm = 7.50. 11.1 cm The length of the telescope is: L = f o + f e = 83.3 cm + 11.1 cm = 94.4 cm.
Consider first the lens used for the left eye. The near point is at 50.0 cm (q = -50.0 cm when p = 25.0 cm). Thus, 1 1 1 1 1 1 + = becomes: + 25.0 cm p q f -50.0 cm = f , and f = 50.0 cm. For the right eye lens, we have a near point of 100 cm ( q = -100 cm when p = 25 cm). 1 1 1 1 1 1 + = becomes: + 25.0 cm p q f -100 cm = f , and f = 33.3 cm. (a) The angular magnification of the telescope (using the longest focal length lens for the f o 50.0 cm objective) is: M = = = 1.50. 33.3 cm f e (b)
We shall use the 50.0 cm lens as the objective, and we shall require that a virtual final image be formed at qe = -25.0 cm for maximum magnification. The object distance for the eyepiece is: 1 1 1 1 1 1 + = , or + = pe qe f e pe -25.0 cm 33.3 cm , yielding pe = 14.3 cm.
7
CHAPTER TWENTY-FIVE SOLUTIONS When the image of the eyepiece (acting as a simple magnifier) is at the normal near point ( qe 25.0 cm = -25.0 cm), the angular magnification produced by the eyepiece is: me = 1 + =1 f e 25.0 cm = 1.75. 33.3 cm The image position for the objective is: +
qo = L - pe = 10.0 cm - 14.3 cm = - 4.30 cm, and the object position is found as:
1 + po
1 1 1 1 1 = , or + qo f o po -4.30 cm = 50.0 cm , yielding po = 3.96 cm. The lateral qo -4.30 cm magnification produced by the objective lens is: M 1 = = - po 3.96 cm = 1.09. The overall magnification is then found to be: M overall = M 1me = (1.09)(1.75) = 1.91. These steps can be repeated with the 33.3 cm lens as the objective and the 50 cm as the eyepiece of the microscope. For that case, the overall magnification will be 1.80. Thus, the arrangement used above gives the greater magnification. 25.28
The angular resolution needed is: s 300 m = = 7.9 x 10-7 rad. m = r 3.8 x 108 m For a circular aperture: m =
1.22
D
, so D = 1.22
500 x 10-9 m 7.9 x 10-7 rad
= 1.22
m
= 0.77 m (about 30 inches.) 25.29
If just resolved:
= m =
Thus the altitude is: h =
25.30
d
1.22 =
noil
= 1.22
D 1.00 m
500 x 10-9 m 0.300 m
2.03 x 10-6 rad
The limit of resolution in air is: air
air =
1.22
D
= 2.03 x 10-6 rad.
= 4.93 x 105 m = 493 km.
= 0.60 rad. In oil, the wavelength becomes oil =
, so the limiting angle in oil is: air
oil =
25.31
s = r
=
1.22
oil
D
=
1.22( ) noil
2.0 m 10.0 x 103 m
D
=
air
noil
=
0.60 rad = 0.40 rad. 1.5
= 2.0 x 10-4 rad.
If the two lights are to be just barely resolved, 1.22 , so D = , or = m = 1.22 D
8.85 x 10-7 m = 5.4 x 10-3 m = 5.4 mm. D = 1.22 2.0 x 10-4 rad 25.32
(a)
medium =
vac
nmedium
=
500 nm = 376 nm. The limiting angle is then: 1.33
8
CHAPTER TWENTY-FIVE SOLUTIONS m =
(b) r = 25.33
s m
=
3.76 x 10-7 m = 2.29 x 10-4 rad. 1.22 = 1.22 D 2.00 x 10-3 m
1.00 x 10-2 m = 43.7 m 2.29 x 10-4rad
The diameter of the aperature is: D = 20 in = 51 cm = 0.51 m. Since the two stars are barely resolved, = m
500 x 10-9 m = 1.2 x 10-6 rad. = 1.22 = 1.22 D 0.51 m
Thus, s = r = (8.0 ly)
25.34
If just resolved:
9.461 x 1012 km (1.2 x 10-6 rad.) 1 ly
= m =
= 9.1 x 107 km
550 x 10-9 m = 1.9 x 10-6 rad. 1.22 = 1.22 D 0.35 m
and s = r = (1.9 x 10-6 rad.)(2.00 x 105 m) = 0.38 m = 38 cm. 25.35
Under the conditions of this problem, the limit of resolution is: 500 x 10-9 m = 1.22 x 10-7 rad. Thus, the minimum observable = 1.22 = 1.22 D 5.00 m separation is: s = r m = (1.22 x 10 -7 rad)(8.0 x 107 km) = 9.8 km m
27.36
A fringe shift occurs when the mirror is moved a distance of
4
. Thus, if the mirror is moved a
distance L = 0.180 mm = 1.80 x 10 -4 m and the wavelength is = 550 nm, the number of fringe shifts observed is: 4( L) 4(1.80 x 10-4 m) L fringe shifts = = = = 1.31 x 103 m ( /4) 550 x 10-9 m 25.37
A fringe shift occurs when the mirror is moved a distance of . Thus if 310 fringe shifts are 4 counted, the interferometer mirror has moved a distance of L =d = 310
650 x 10-9 m 4
= 50.4 m
(length of the amoeba). 25.38
If the number of wavelengths that will fit in the length of the tube changes by N , there will be 4( N ) fringe shifts observed. When the tube is evacuated, the number of wavelengths in its length L is N = When the tube is filled with gas, the number of wavelengths is: vac
N ' =
L gas
=
ngas L L = ( vac/ngas) vac
Thus, the number of fringe shifts observed as the tube fills
will be: ngas L fringe shifts = 4( N ) = 4( vac
L vac
)=
4 L
(ngas - 1) .
vac
Solving for the index of refraction gives: ngas = 1 +
vac
(#shifts) 4L If 600 nm light is used, L = 5.00 cm and 160 fringe shifts are observed, we have: +
600 x 10-9 m (160) = 1.0005 4(5.00 x 10-2 m)
ngas = 1
9
CHAPTER TWENTY-FIVE SOLUTIONS 25.39
Note that changing the number of wavelengths that will fit within the length of the cell by N will produce 4( N ) fringe shifts since this is equivalent to increasing the length of the interferometer arm by ( N ) . As the air is evacuated from the cell, the wavelength of the light within the cell changes from air =
vac
nair
to vac.
Thus, the number of wavelengths that will fit within the length of the cell changes from N = L
nair
vac
to N ' =
L vac
L air
and the number of fringe shifts observed will be:
#fringe shifts = 4( N - N ') nair L 4 L = 4( L )= (nair - 1) , or vac
# fringe shifts =
25.40
vac
vac
4(5.00 x 10-2 m) (1.00029 - 1) = 98.3, or 98 complete shifts. 590 x 10-9 m
In a vacuum, the number of wavelengths in a distance t is N =
t vac
When this space is filled by a medium of refractive index nm , the number of wavelengths in the nmt t same distance t is N ' = = Thus, the number of fringe shifts that will occur is: medium
# fringe shifts = 4( N ) = 4(
vac
nmt vac
-
t
4t
) = (nm - 1) . vac vac
If t = 15 x 10-6 m, = 600 nm, and nm = 1.4, we have: # fringe shifts =
4(15 x 10-6 m) (1.40 - 1) = 40 600 x 10-9 m
25.41 We first find the grating spacing as: d =
n (2)(502.nm) = = 2008 nm. sin30.0° sin
Then for the 668 nm line, we have: n (1)(668 nm) sin = = = 0.333, from which: 2008 nm d 25.42
=
19.4°.
1 = 2.732 x 10-4 cm = 2.732 x 10 -6 m = 2732 nm. 3660 lines/cm d sin The wavelength found at angle is: = . n At = 10.1°, = 479 nm; at = 13.7°, = 647 nm; and at = 14.8°, = 698 nm.
(a)
d =
(b)
The grating spacing is: d =
sin
, and
2 for the second order: (2) = d sin Combining these equations gives: sin = = 2 d sin = 2 sin .
Therefore, if = 10.1° then sin = 2sin (10.1°) gives Similarly, for = 13.7°, = 28.3°; and
=
20.5°.
for = 14.8°, = 30.7°.
=
10
CHAPTER TWENTY-FIVE SOLUTIONS 25.43
(a) The longest wavelength in the visible spectrum is 700 nm. The number of complete spectra that can be seen is therefore the same as the number of visible orders of the 700 nm light. From the grating equation, we have: d sin max nmax = d sin max or nmax = . The slit spacing is:
d =
1 = 1.67 x 10-3 mm. 600 line/mm
(1.67 x 10-6 m)sin90° = 2.39, 700 x 10-9 m and we see that only two complete orders of the visible spectrum can be seen. (b) For the violet edge of the first order: n (1)(400 x 10-9 m) sin V1 = = = 0.240, and V1 = 13.9°. d 1.67 x 10-6 m For the red edge of the first order: n (1)(700 x 10-9 m) sin R1 = = = 0.419, and V1 = 24.8°. d 1.67 x 10-6 m Therefore, the angular width of the first order visible spectrum is: = R1 - V1 = 24.8° - 13.9° = 10.9° Thus, nmax =
25.44
(a)
(b)
25.45
1 = 6.667 x 10-4 cm = 6.667 x 10 -6 m = 6667 nm, and 1500 lines/cm d sin (6667 nm)sin90° n = = = 13.3. 500 nm Thus, 13 complete orders will be observed. For 15,000 lines/cm: d = 666.7 nm, and dsin (666.7 nm)sin90° n = = = 1.33. 500 nm Only one complete order can be seen. d =
The wavelengths for the sodium doublet are: B = 589.592
A =
588.995 nm, and
nm. For the given grating,
1 cm = 4.0 x 10-4 cm = 4000 nm. Using the grating equation, 2500 m m . d sin = m , or sin = , and = arcsin d d Thus, for m = 1: (1)(588.995 nm) A = arcsin 4000 nm = arcsin(0.14725) = 8.4675°, and (1)(589.592 nm) B = arcsin 4000 nm = arcsin(0.14740) = 8.4762°. Thus, = B - A = 0.0087°. d =
For m = 2:
A
= 17.1274°,
B =
17.1453°; and = 0.0179°.
For m = 3:
A
= 26.2154°,
B =
26.2440°, and
= 0.0286°.
n (1)(546.1 nm) = = 1.524 x 103 nm = 1.524 x 10 -3 mm, and the number of lines per sin(21.0)° sin 1 1 millimeter is: = = 656 line/mm. d 1.524 x 10-3 mm
25.46
d =
25.47
We use d sin = m , with m = 1, and d =
1 cm = 2.5 x 10-6 m = 2500 nm. 4000
11
CHAPTER TWENTY-FIVE SOLUTIONS (a) (b)
25.48
m (1)400 nm = = 0.160 and blue = 9.21°. 2500 nm d m (1)650 nm For the red light sin = = = 0.260 and red = 15.1°. 2500 nm d
For the blue light, sin =
1 = 2.00 x 10-4 cm = 2000 nm. 5000 slits/cm 2 1 2(480 nm) In the second order, 2 = d sin so sin 1 = = = 0.480, and 2000 nm d 2(610 nm) sin 2 = = 0.610, giving 2 = 37.6°. 1 = 28.7°. Also, 2000 nm y1 = L tan 1 = (2.00 m) tan28.7° = 1.095 m, d =
y2 = L tan 2 = (2.00 m) tan37.6° = 1.540 m, and
y = y2 - y1 = 0.445 m = 44.5 cm 25.49
(a)
The resolving power is given by:
R =
= mN . Thus,
N =
. m
656.20 nm = 3646 slits. (1) (0.18 nm) 656.20 nm For the second order, N = = 1823 slits. (2) (0.18 nm) For the first order: N =
(b) 25.50
R = Nm = Error! The highest order of 600 nm light visible is: d sin max (1670 nm)sin90° mmax = = 2.78 or 2 orders. 600 nm The resolving power of this grating in the second order is R = [(6000 slits/cm)(15.0 cm)] (2) = 1.80 x 10 5 , and the resolving power needed to separate these two wavelengths is: R =
=
600.000 nm = 0.003 nm
2.00 x 105. These two spectral lines cannot be resolved using this grating. 25.51
(a)
Corrective lenses must form a virtual image at q = -75.0 cm for objects which are 25.0 cm in front of the lens ( p = 25.0 cm). 1 1 1 1 1 1 + = becomes: + = , 25.0 cm -75.0 cm f p q f and f = 37.5 cm = 0.375 m. 1 1 Thus, P = = = 2.67 diopters. f 0.375 m (b) If q = -75.0 cm when p = 26.0 cm rather than 25.0 cm as assumed in part(a), we have: 1 1 1 + = , 26.0 cm -75.0 cm f ' and f ' = 39.8 cm = 0.398 m. 1 1 Thus, P = = = 2.51 diopters, (0.16 diopters too low). f ' 0.398 m
25.52
1 1 1 1 1 1 (a) + = gives: + = , and f = 1.96 cm 100 cm 2.00 cm f p q f f 1.96 cm (b) and (c) f max = = = 9.80, 0.200 cm Dmin
12
CHAPTER TWENTY-FIVE SOLUTIONS f 1.96 cm and f min = = = 3.27 0.600 cm Dmax 25.53
(a)
25.54
d =
25.55
(a)
n (3)(546.1 nm) = = 1.66 x 103 nm= 1.66 x 10-3 mm. sin81.0° sin 1 1 Thus, lines/mm = = = 602. d 1.66 x 10-3 mm The resolving power is given by: Thus, (b) N =
25.56
f o 100 cm (b) M = = = 67. f e 1.50 cm
L = f o + f e = 101.5 cm.
N =
m
=
R =
= mN .
589.00 nm = 980. (1)(0.60 nm)
m 589.00 nm = = 330. (3)(0.60 nm)
In a distance t in a vacuum, the number of wavelengths that can be fitted in is N =
t
. If this
vac
space is then filled with material having a refractive index n, the number of wavelengths which can t nt now be fitted in is N ' = = . Since four fringe shifts occur when the path length increases
vac
by one wavelength, the number of fringe shifts observed as the material is introduced will be: # fringe shifts = 4( N ) = 4( Therefore,
n = 1 +
nt vac
-
t
) . Therefore, n = (1 +
vac
vac
4t
).
vac
4t
(#fringe shifts) .
Thus, if t = 2.5 m, = 580 x 10-9 m, and 12 fringe shifts are observed n = 1 +
(580 x 10-9 m)12 4(2.5 x 10-6 m)
= 1.70 is the index of refraction. 1 cm = 3.64 x 10-4 cm = 3640 nm. 2750 lines The violet end of the second order spectrum occurs at: m 2(400 nm) sin = = = 0.220, or = 12.7° = 0.222 radians. 3640 nm d The red end of the second order spectrum is located at: m 2(700 nm) sin = = = 0.385, or = 22.6° = 0.395 radians. 3640 nm d The angular width of the second order spectrum is thus: = (0.395 - 0.222) radians = 0.173 radians. At distance, L, the arc length included is: s = L( ), so 1.75 cm s L = . If s = 1.75 cm, then: L = = 10.1 cm. 0.173 radians
25.57
The grating spacing is: d =
25.58
(a)
First find the location and size of the image formed by the objective lens: becomes:
1 1 1 + = po qo f o
13
CHAPTER TWENTY-FIVE SOLUTIONS 1 1 1 + = , and qo = 8.02 x 10 -2 m is the image distance. The 40.0 m qo 8.00 x 10-2 m image size is: hqo (30.0 cm)(0.0802 m) h' = == - 6.02 x 10-2 cm. (40.0 m) po (b)
Now, treat the eyepiece lens. With the image at q = -, we have: 1 1 1 + = , so pe = -2.00 cm. ( p < 0 means a virtual object) pe -2.00 cm
(c)
Distance between lenses is qo + pe, or L = 8.02 cm - 2.00 cm = 6.02 cm. f o 8.00 cm M = f = -2.00 cm = 4.00. e
(d)
25.59
The image of a very distant object is formed at q = 5.0 cm. Thus, the focal point is:
0+
1 5.0 cm
1 = , or f = 5.0 cm. f (a) The maximum magnification occurs when the image is at the near point of the eye. Thus, we see that q = -15 cm. We find: 1 1 1 15 + = , yields p = cm. The magnification is: 4 p -15 cm 5.0 cm q -15 cm M = - = = 4.0. 15 p cm 4 h h (b) We have: o = , and = . Therefore, 15 cm f 15 cm 15 cm M = = = = 3.0. f 5.0 cm o
A B
Image plan e d B'
A'
30 cm
f
500 x 10-9 m = 6.10 x 10-6 rad 1.22 = 1.22 D 0.100 m d d = 1 f , and 2 = , so 30.0 cm 1 f 140 = ( 6.10 x 10-6 rad) = 2.85X 10-5 rad = 28.5 rad 2 = 30.0 cm 30.0
25.60
1 = m =
25.61
First, find the longer wavelength: y =
n L, d
and y2 - y1 =
L, d
15 cm , where 0.844 cm = 8.333 x 10-4 cm 1 d= = 8.333 x 10-4 cm = 8333 nm. Thus, 1200 slits/cm or
long
4.689 10
5
cm
469 nm.
The minima in intensity occur at angles of deviation for which the path difference for light passing through adjacent slits on the grating is
14
CHAPTER TWENTY-FIVE SOLUTIONS
1 d sin m with m 0, 1, 2, 3, . . . 2
minimum for the longer wavelength, sin
Hence, at the first-order
long 1 long 0 2 2d d
. The angle at which
the third-order maximum for the shorter wavelength short occurs is sin
then
m
3 short
d
3 short
d
d
long
2d
. If this maximum coincides with the first-order minimum for long ,
, or short
long
469 nm
6
6
Obje ctive
A
Ey e piece
h'
B
Object D po = 300 cm
qo o
C
pe = f e
f o = 20 c m
25.62 The angular magnification is m =
78.2 nm.
par a lle l ra y s emerge
fe = 2 cm
, where is the angle subtended by the final image, and o is
the angle subtended by the object. These angles are labeled in the sketch. Note that since the telescope is adjusted for minimum eyestrain, the final image is at infinity. Thus, the image formed by the objective lens (serves as object for eyepiece) must be located at the focal point of the h' h' eyepiece. From triangle ABD, tan o = , and using triangle BCD, tan = . qo f e Assuming small angles h' , and qo
h'
f ,
e
so
tan and M =
o
o
qo
f
e
tan o , these equations become
o
.
To find qo, apply the thin lens equation to the objective lens: 1 1 1 1 1 1 + = becomes: + = , giving 300 cm qo 20.0 cm po qo f o qo =21.4 cm. Therefore,
M =
21.4 cm = 10.7. 2.00 cm
n1 n2 n2 - n1 + = , with p = infinity. p q R n2 - n1 = 2 cm 1.34 - 1.00 = 0.507 cm = 5.07 mm. Thus, R = q 1.34 n2
25.63
We use:
25.64
From the grating equation with d = 2500 nm, we find: m m sin = = . 2500 nm d Using the given wavelengths, with the restriction sin 1, we obtain the following table: sin Wavelength, (color) m=1 m=2 m=3 m=4 m=5 400 nm V 0.160 0.320 0.480 0.640 0.800 550 nm G 0.220 0.440 0.660 0.880 700 nm R 0.280 0.560 0.840
m=6 0.960
15
CHAPTER TWENTY-FIVE SOLUTIONS The order in which these lines are observed is: V1 G1 R 1 V2 G2 V3 R 2 V4 G3 V5 G4 R 3 V6 ANSWERS TO CONCEPTUAL QUESTIONS
2. The light from the flashlight consists of many different wavelengths with random time
differences between the individual waves. Thus, there is no coherence between the two sources and no possibility of interference. 4. The shutter of a camera is a close approximation to the iris of the eye. The retina of the
eye corresponds to the film of the camera, and a close approximation to the cornea of the eye is the lens of the camera. 6. You want a real image formed at the location of the paper. To form such an image, the
light source should be farther away from the paper than the focal point of the lens. 8. The telescope forms an inverted image. Thus, the astronauts could find themselves
exploring the northern hemisphere when they were planning to explore the southern hemisphere. It is highly unlikely that a civilization advanced enough to travel to the Moon would ever face such a problem. 10. A magnified image of an object is produced by a converging lens when the object is
placed somewhere between the focal point and the lens. Hence, the distance of the object from the lens should be less than 15 cm. 12. The image formed on the retina by the lens and cornea is already inverted.
