1
CHAPTER TWENTY-THREE SOLUTIONS CHAPTER TWENTY-THREE SOLUTIONS Chapter Twenty-Three Readings
Ansbacher, T.H., "Left-Right Semantics?" The Physics Teacher , 30, 70, 1992. Fraser, A.B, and Hirsch, M.W., Mirages M.W., Mirages,, Springer-Verlag, 1975. Galili, I., Goldberg, F. and Bendall, S., "Some Reflections on Plane Mirrors and Images," The Physics Teacher , 29, 471, 1991. Galili, I., Bendall, S. and Goldberg, F., "Author's Response to 'Left-Right Semantics?'", The Physics Teacher , 30, 70, 1992. Galili, I. and Goldberg, F., "Left-Right Conversions in a Plane Mirror," The Physics Teacher Teacher , 31, 463, 1993. Greenslade, T.B., "Multiple Image in Plane Mirrors," The Physics Teacher , January 1982, p. 29. Jones, R.C., "How Images are Detected," Scientific American, American, September 1968, p. 111. Smith, F.D., "How Images are Formed," Scientific American, American, September 1968, p. 97. Thomas, D.E., "Mirror Images," Scientific American, American , December 1980, p. 206. Veldkamp, W.B, and McHugh, T.J., "Binary Optics," Scientific American, American, May 1992, p. 92. Winston, R., "Nonimaging Optics," Scientific American, American, March 1991, p. 76. 23.1
(1) (2)
(3)
23.2
The first first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from from the position of the person. The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is 25.0 ft from the left mirror. mirror. Thus, the second image in the left mirror is 25.0 ft behind the mirror, or 30.0 ft from the person. The first image in the left mirror mirror serves as an object for the right mirror. It is located located 20 ft ft in front of the right mirror and forms an image 20 ft ft behind that mirror. This image then serves as an object for, and forms an image in, the left mirror. The distance from this image to the left mirror is 35.0 ft. The third image in the left mirror is, thus, 35.0 ft behind the mirror, or 40.0 ft from the person.
(See sketch.) (1) angle PRQ = angle P'RQ (90° - = = 90° - ) (2) angle PQR = angle P'QR (both 90°) (3) side QR = side QR Therefore, triangle PQR is congruent to triangle P'QR (angle-side-angle) (angle-side-a ngle) Hence, the distance PQ = QP' or, the image is as far behind the mirror as the object is in front of the mirror.
Q
P
P'
Obj e c t
I m a ge
R Mirror
2
CHAPTER TWENTY-THREE SOLUTIONS 23.3
(a) (b) (c)
23.4
23.5
For a plane mirror, R is infinite. 1 1 2 - p q + = = 0, or q = - p. Additionally, M = - = - = +1. p q R p p Since, q = - p, the image is virtual and as far behind the mirror as the object is in front of the mirror. Also, since M = +1, the image is erect and the same size as the object. These agree with the earlier discussion of plane mirrors.
From the sign conventions for convex mirrors, we see that R = -3.00 cm and f = -1.50 cm. Also, p = 10.0 cm. Thus, 1 1 1 1 1 23.0 = = =, and q = -1.30 cm. -1.50 cm 10.0 cm 30.0 cm q f p q -1.30 cm M = - = - p 10.0 cm = +0.130. R For a concave mirror, both R and f are positive. Also, f = = 10.0 cm. 2 1 1 1 The mirror equation gives: = , q f p pf p(10.0 cm) or q = = p - f p - (10.0 cm) (40.0 cm)(10.0 cm) (a) If p = 40.0 cm, q = = 13.3 cm, (40.0 cm) - (10 cm) q 13.3 cm M = - = = -0.333. 40.0 cm p The image is 13.3 cm in front of the mirror, is real, and inverted. (20.0 cm)(10.0 cm) (b) If p = 20.0 cm, q = = 20.0 cm, (20.0 cm) - (10 cm) q 20.0 cm M = - = = -1.00. 20.0 cm p The image is 20 cm in front of the mirror, is real, and inverted. (10.0 cm)(10.0 cm) (c) If p = 10.0 cm, q = = , (10.0 cm) - (10.0 cm) No image is formed. The rays are reflected parallel to each other.
23.6
1 1 1 1 1 = + = , or f = 1.11 cm 10.0 cm f p q 1.00 cm q -10.0 cm (b) M = - = - p 1.00 cm = +10.0. (a)
and R = 2 f = 2.22 cm.
1 1 1 1 1 = = gives q = - 0.268 m. 0.275 m 10.0 m q f p q -0.268 Thus, the image is virtual. M = = - p 10.0 m = 0.0268 Thus, the image is erect ( M > 0).
23.7
f = R/2 = -0.275 cm.
23.8
Since the image is projected on a screen, it is a real image. Therefore, q is positive. We have: 1 1 1 1 +21 = + = + = , and 10.0 cm 200 cm 200 cm p q f = +9.52 cm. Thus, because f is positive, we need a concave mirror. q 200 cm M = - = = -20.0. 10.0 cm p
1 f
3
CHAPTER TWENTY-THREE SOLUTIONS 23.9
We know that for a convex mirror R and f are negative, and we also are given that M = 1/2. M is positive because convex mirrors only form erect virtual images of real objects. Therefore, M = q 1 p 1 1 1 1 -2 = , or q = - . Thus, = = - , which gives: p = 20.0 cm. 2 2 -20.0 cm p p p f q The object should be 20.0 cm in front of the mirror.
23.10
Since we have a real object and want a magnified image, the mirror must be concave. (A convex mirror forms diminished images of all real objects.) Since image is erect, it must be virtual. (All real images formed by concave mirrors of real objects are inverted.) Thus, q< 0, f > 0, and p = 10 cm. h' 4.00 cm q M = = = + 2.00 = - , 2.00 cm h p Q = - 20.0 cm. 1 1 1 1 1 Thus, = + = + , or f = 20.0 cm, and R = 2 f = 40.0 cm. 10.0 cm - 20.0 cm f p q
23.11
For a concave mirror, R and f are positive. Also, h' q M = = - . We are given that h' = 5.0 cm (erect...thus h' is positive), h p h' 5.0 cm h = 2.0 cm, and p = 3.0 cm. Therefore, q = - p = -(3.0 cm) 2.0 cm = -7.5 cm. h 1 1 1 1 1 3.0 And, = + = = , and f = 5.0 cm. 3.0 cm 7.5 cm 15 cm f p q
23.12
Realize that the nearer image (q = -10.0 cm) occurs when using the convex side of the hubcap. Applying the mirror equation to both cases gives: |R| (concave side: f = , q = - 30.0 cm): 2 1 1 2 2 1 1 + = , or = + (1 ) |R| p p q |R| -30.0 |R| (convex side: f = - , q = -10.0 cm): 2 1 1 2 2 1 1 + =, or = - - (2) -10.0 p q |R| |R| p 2 1 1 (a) Equating equations (1) and (2) gives: = + , p 10.0 30.0 or p = +15.0 cm. Thus, his face is 15.0 cm in front of the hubcap. (b) Using the above result (p = 15 cm) in equation (1) gives: 2 1 1 2 1 = + , or = , and |R| = 60.0 cm. 15.0 30.0 cm |R| |R| -30.0 The radius of the hubcap is 60.0 cm
23.13
We know that R and f are positive for a concave mirror. Also, since a concave mirror only forms magnified, erect images when the image is virtual, we know that q is negative. q Thus, M = 2, and M = - = 2.0, or q = -2.0(25 cm) = -50 cm. p 1 1 1 1 1 1 Thus, = + = + 25 cm f p q -50 cm = 50 cm , and f = 50 cm. But, R = 2 f = 100 cm = 1.0 m.
23.14
R A convex mirror has R and f negative. Thus, f = = -5.00 cm. For a virtual image, q is negative. 2 q 1 p Therefore, M = = , and q = - . 3 3 p
4
CHAPTER TWENTY-THREE SOLUTIONS 1 1 1 1 1 3 = + becomes = - , or -5.00 cm p p f p q The object must be 10.0 cm in front of the mirror. Therefore,
23.15
1 1 1 1 1 = + = + , or 1.52 m 0.180 m f p q
p = 10.0 cm.
f = 0.161 m = 16.1 cm.
Now, to get an erect image which is twice the size of the object, we need:
q M = - = +2.00, so p
q = -2.00 p. 1 1 1 1 1 1 f Thus, + = becomes = , or p = . 2 p q f p 2 p f 16.1 cm Thus, p = = 8.05 cm is the required object distance. 2 23.16
For a concave mirror, R and f are positive. Also, for an erect image M is positive. Therefore, M q = - = 4, or q = -4 p. p 1 1 1 1 1 1 3 = + becomes: = = , giving p = 30.0 cm. 40.0 cm p 4 p 4 p f p q
23.17
For a convex mirror, R and f are negative. Since we have a real object, the image will be virtual q 1 p and erect. Thus, q is negative and the magnification is positive. M = - = , and q = - p 2 2 = - 5.00 cm. 1 1 1 1 1 1 Thus, = + = + = , and f = -10.0 cm. 10.0 cm -5.00 cm 10.0 cm f p q But, R = 2 f . Therefore, the radius of curvature is R = -20.0 cm.
23.18 We know that R and f are positive for a concave mirror, and we are given that p = 30.0 cm, and M = 4.00. ( M < 0 since the only magnified real images formed by a concave mirror of real objects are q inverted images.) Hence, M = - = -4.00, or q = 4 p. p Since p = 30.0 cm, q = 120 cm. 1 1 1 1 1 5 Thus, = + = + = , giving f = 24.0 cm, 30.0 cm 120 cm 120 cm f p q and R = 2 f = 48.0 cm. 23.19
(a) (b)
In this case, both the object and the image are real. To form a real image of a real object, a concave mirror must be used. q We are given q = + 5.0 m, and M = - = - 5.0 (a concave mirror forms inverted real images, p hence the negative sign on the value of the magnification.) Therefore, q = 5 p, and since we know that q = 5.0 m, we find that p = 1.0 m as the distance the object must be from the mirror.
23.20
We are given that q = -35.0 cm. The negative sign arises because we have a virtual image on the same side of the boundary as is the object. n1 1.50 p = q = (-35.0 cm) = 52.5 cm. The benzene is 52.5 cm deep. 1.00 n2
23.21
q = -
n2
1 p = (50.0 cm) = -38.2 cm. Thus, the virtual image of the dust speck is 38.2 cm 1.309 n1
below the top surface of the ice.
5
CHAPTER TWENTY-THREE SOLUTIONS 23.22
(a)
For a plane refracting surface, we have: q = -
n2 3 p = - p. 4 n1
Thus, for p = 2.00 m,
(b)
23.23
We have R = -4.0 cm (center of curvature is on side light is coming from), and p = 4.0 cm. Thus, n1 n2 n2 - n1 + = becomes: p q R 1.5 1 1.0 - 1.5 + = . This gives q = -4.0 cm. Note that the location of the 4.0 cm -4.0 cm q virtual image is at the same position as the object. h' h' (1.5)(-4 cm) n1 q M = = - gives: == 1.5, or h' = 3.8 mm. 2.5 mm (1.0)(4 cm) h n2 p
23.24 We use:
(a) (b) (c) (d)
23.25
23.26
q = -1.50 m. The pool appears to be 1.50 m deep. 3 If the pool is half filled, p = 1.00 m, and q = - (1.00 m) = -0.75 m. 4 The bottom surface appears to be 0.75 m below the water surface, or 1.75 m below ground level (or the top of the pool).
n1 p
+
n2
1 20.0 cm 1 8.00 cm 1 4.00 cm 1 2.00 cm
q
=
n2 - n1
1.50 q 1.50 + q 1.50 + q 1.50 + q +
R
.
0.50 8.0 cm 0.50 = 8.00 cm 0.50 = 8.00 cm 0.50 = 8.00 cm =
gives
q = 120 cm.
gives
q = -24.0 cm.
gives
q = -8.00 cm.
gives
q = -3.43 cm.
n1 n2 n2 - n1 + = . But the object is at infinity, and q = 2 R, so p q R n2 n2 - 1.00 = , which gives: n2 = 2.00 2 R R n1 n2 n2 - n1 + = to find the apparent depth of the top and the bottom of the glass plate p q R first, then subtract to find the apparent thickness. Note that for a plane surface, R = , so we have: n2 q = - p. n1 Use
First consider the bottom of the plate as the object and locate the image formed by refraction at the glass-water boundary. Here, p = 8.00 cm, 1.333 n1 = 1.66, and n2 = 1.333, so q1b = - 1.66 (8.00 cm) = - 6.42 cm. Now use this image as the object for refraction at the water-air boundary. In this case, p = 12.0 cm 1.00 + 6.42 cm = 18.42 cm, n1 = 1.333, and n2 = 1.0. Thus, q2b = - 1.333 (18.42 cm) = -13.8 cm. The bottom of the plate appears to be 13.8 cm below the water surface. Now, locate the image formed of the top of the plate by refraction at the water-air boundary. Here p = 12.0 cm, n1 = 1.333, and n2 = 1.00, so 1.00 q2t = - 1.333 (12.0 cm) = -9.00 cm below the water surface and the apparent thickness of the plate is: t = q2b - q2t = 13.8 cm - 9.00 cm = 4.80 cm.
6
CHAPTER TWENTY-THREE SOLUTIONS
23.27
1 1 1 = (n - 1) - becomes: f R1 R2 which gives
23.28
(a)
(c) (d) (e)
1 1 1 = 0.58 , 25.0 cm 1.80 cm R1
R 1= 1.60 cm.
We find the focal length from: 1 1 1 1 1 = (n - 1) - = 0.50 f 15.0 cm -10.0 cm , or f = 12.0 cm. R1 R2
(b) When p =
23.30
n = 1.47.
1 1 1 = (n - 1) - becomes: f R1 R2 which gives
23.29
1 1 1 = (n - 1) 60.0 cm 52.5 cm -61.9 cm ,
,
1 1 1 = q f p 1 1 1 = q f p 1 1 1 = q f p
1 1 = , or q f 1 = 12.0 cm 1 = 12.0 cm 1 = 12.0 cm
q = f = 12.0 cm. 1 2 = , or q = 18.0 cm. 36.0 cm 36.0 cm 1 = 0, or q = infinity (no image formed) 12.0 cm 1 1 =, gives q = -12.0 cm. 6.00 cm 12.0 cm -
For a converging lens, f is positive. We use:
1 1 1 + = . p q f
1 1 1 1 1 1 = = = , or q = 40.0 cm. 20.0 cm 40.0 cm 40.0 cm q f p q 40.0 M = - = = -1.00. (real, inverted, and 40.0 cm past the lens) 40.0 p 1 1 1 1 1 (b) = = = 0, giving q = infinity. 20.0 cm 20.0 cm q f p No image is formed. (parallel rays emerge from the lens) 1 1 1 1 1 1 (c) = = =, or q = -20.0 cm. 20.0 cm 10.0 cm 20.0 cm q f p q -20.0 M = - = - p 10.0 = 2.00. (erect, virtual, and 20.0 cm in front of lens) (a)
23.31
(a) (b)
1 1 1 1 1 = (n - 1) = 0.50 , or f = -12.0 cm. 15.0 cm 10.0 cm f R R 1 2
When p = infinity, 1 (c) p = 36.0 cm, = q 1 (d) p = 12.0 cm, = q 1 (e) p = 6.00 cm, = q 23.32
q = f = -12.0 cm. 1 1 1 1 1 == , or q = - 9.00 cm. 12.0 cm 36.0 cm 9.00 cm f p 1 1 1 1 1 == , or q = - 6.00 cm. 12.0 cm 12.0 cm 6.00 cm f p 1 1 1 1 1 == , or q = - 4.00 cm. 12.0 cm 6.00 cm 4.00 cm f p
For a diverging lens, f is negative. 1 1 1 1 1 3 (a) = ==, or q = -13.3 cm. 20.0 cm 40.0 cm 40.0 cm q f p q -13.3 cm 1 M = - = = . 40 cm 3 p The image is virtual, erect, and 13.3 cm in front of the lens. 1 1 1 1 1 1 (b) = ==, or q = -10.0 cm. 20.0 cm 20.0 cm 10.0 cm q f p
7
CHAPTER TWENTY-THREE SOLUTIONS
(c)
q -10.0 cm 1 M = - = = 20.0 cm 2 p The image is virtual, erect, and 10 cm in front of the lens. 1 1 1 1 1 3 = ==, or q = -6.67 cm. 20.0 cm 10.0 cm 20.0 cm q f p q -6.67 cm M = - = = 0.667. 10.0 cm p The image is virtual, erect, and 6.67 cm in front of the lens.
q 1 1 1 23.33 We want |M | = 1. Using M = - , and + = , we have: p q f p 1 1 1 1 = , or p = f 1 - . p Mp f M 1 (a) converging lens: p = (12.00 cm) 1 - . M If M = + 1, (object against lens). p = 0 If M = -1, p = 24.0 cm 1 (b) diverging lens: p = (-12.00 cm) 1 - . M If M = + 1, p = 0 (object against lens). If M = -1, p = -24.0 cm. Note that in both cases, M approaches +1.00 only in the limit as p approaches zero. Note also that in part (b), the object must be virtual to obtain M = -1.00. 23.34
We want M = + 2.00, so 1 1 1 + = becomes: p q f
23.35
q M = - = 2.00, or q = -2.00 p. p 1 1 1 + = , giving p =+ 7.50 cm. p -2.00 p + 15.0 cm
We must first realize that we are looking at an erect, magnified, virtual image. Thus, we have a real object located inside the focal point of a converging lens. Thus, p > 0, q < 0, and f > 0. q M = 2.00 = - , or q = -2.00p = -2(2.84 cm) = - 5.68 cm. p 1 1 1 1 1 1 Thus, + = , or = , giving f = 5.68 cm. 2.84 cm 5.68 cm p q f f
23.36 Start with the thin lens equation
1 1 1 + = and solve for the image p q f
pf . In this case, f = 10.0 cm and the object distance p - f can vary from a minimum of p = 10.2 cm to a maximum of p = 11.0 cm. (11.0 cm)(10.0 cm) When p = 11.0 cm, q = = 110 cm. When p = 10.2 cm, 11.0 cm - 10.0 cm (10.2 cm)(10.0 cm) q = = 510 cm. Thus, the screen should be between 10.2 cm - 10.0 cm 1.10 m and 5.10 m from the lens. distance to get q =
23.37
Given f = - 40.0 cm, and
q
= 30.0 cm. But for a real object, a diverging lens can only form
virtual images (q< 0). Therefore, q = - 30.0 cm. 1 1 1 1 1 = = p f q -40.0 cm - -30.0 cm , and The object should be 120 cm in front of lens. q -30.0 cm 1 M = - = - = +4. 120 cm p
p = 120 cm.
8
CHAPTER TWENTY-THREE SOLUTIONS 23.38
Given: f > 0, and p = 10 f . 1 1 1 1 1 Thus, = = , and q f p f 10f focal point.
q=
10 1 f , or, the image is distance of f outside the 9 9
23.39
For a virtual image, the image distance is negative and the magnification is positive (assuming a real object). q 1 p Thus, M = - = , and q = - . 3 p 3 1 1 1 1 3 1 So, + = becomes - = , or p = -2 f . p q f p p f
23.40
We are given that p = 80.0 cm and q = -40.0 cm. Thus, 1 1 1 1 1 1 = + = = , and f = -80.0 cm. 80.0 cm 40.0 cm 80.0 cm f p q
23.41
23.42
1 1 1 1 1 = = - , and p = 8.82 cm. 12.5 cm -30.0 cm p f q q -30.0 cm M = - , = - p 8.82 cm = +3.40 (image is erect since M > 0). Let us consider the first lens. We find the image position and magnification as: 1 1 1 1 1 1 = = = , or q1 = 30.0 cm. 15.0 cm 30.0 cm 30.0 cm q1 f 1 p1 q1 30.0 cm M 1 = == -1.00. 30.0 cm p1 Now consider the second lens. The image produced by the first lens becomes the object for this second lens. Thus, 1 p2 = 40.0 cm - q1 = 40.0 cm - 30.0 cm = 10.0 cm. The thin lens equation then gives: = q2 1 1 1 1 1 = = , or 15.0 cm 10.0 cm 30.0 cm f 2 p2 q2 -30.0 cm M2 = = - p2 10.0 cm = 3.00. The overall magnification M is:
23.43
q2 = -30.0 cm.
M = ( M 1)( M 2) = (-1.00)(3.00) = -3.00.
For the diverging lens, (parallel incident rays p1 = ): 1 1 1 Using the thin lens equation, + = q1 - 10 cm , or q1 = - 10.0 cm. The image formed by the diverging lens is the object for the converging lens. Thus, p2 = 30.0 cm q1 = 30.0 cm - (-10.0 cm) = 40.0 cm. Also, since the emerging rays are parallel, q2 = . 1 1 1 For the converging lens: + = 40.0 cm f C , gives
23.44 Consider the first lens.:
f C = + 40.0 cm.
1 1 1 1 1 1 = = =, so 25 cm 20 cm 100 cm q1 f 1 p1
q1 -100 cm and M 1 = = - p1 20 cm = +5. The object position for the second lens is: p2 = 125 cm. Thus, 1 1 1 1 1 = = , and q2 = -9.3 cm. -10 cm 125 cm q2 f 2 p2
q1 = -100 cm,
9
CHAPTER TWENTY-THREE SOLUTIONS q2 -9.3 cm M 2 = = - p2 125 cm = 0.074,
the overall magnification M is:
M = ( M 1)( M 2) = (5.0)(0.074) = 0.37. 23.45
(a)
Note that: q = 12.9 cm - p, so
1 1 1 + = 12.9 - p 2.44 p
which yields: - p2 + 12.9 p = 31.48. This has solutions: p = 9.63 cm, or p= 3.27 cm, and both solutions are valid. (b)
For a virtual image: |q| = -q = p + 12.9 cm, so 1 1 1 = , or 2.44 p 12.9 + p p2 + 12.9 p = 31.48, from which p = 2.10 cm, or p = -15 cm. We must have a real object, so the -15 cm solution must be rejected.
23.46
(a)
The distance from the object to the lens is p, so the image distance is q = 5.00 m - p. Thus, 1 1 1 + = becomes: p q f 1 1 1 + = . This reduces to a quadratic equation: p 5.00 m - p 0.800 m
p2 - 5.00 p + 4.00 = 0, which yields p = 4.00 m, or p = 1.00 m. Thus, there are two possible object distances, both corresponding to real objects. 1.00 1 (b) For p = 4.00 m: q = 5.00 m - 4.00 m = 1.00 m, and M = =- . 4.00 4 4.00 For p = 1.00 m: q = 5.00 m - 1.00 m = 4.00 m, and M = = -4.00. 1.00 Both images are real and inverted, but the magnifications are different, with one being 4 times the size of the object and the other one-fourth the size of the object. 23.47
Treating the first lens: or
1 1 1 1 1 = = , 8.00 cm 4.00 cm q1 f 1 p1
q1 = -8.0 cm. 1 1 1 1 1 = = , or q2 = -7.47 cm -16.0 cm 14.0 cm q2 f 2 p2 q1 q2 M 1 = = 2.00, and M 2 = - = 0.533, and M = M 1 M 2 = 1.07. p1 p2
Thus, p2 = 14.0 cm, and
Thus, if the object is 1.00 cm in height, the erect ( M > 0), virtual image (q2 < 0) is 1.07 cm tall. 23.48
(a)
Start with the second lens and work backward. f 2 = 20.0 cm, q2 = -(50.0 cm - 31.0 cm) = -19.0 cm 1 1 1 1 1 1 + = becomes, = , from which f p2 q2 p2 19.0 20.0
p2 = 9.74 cm.
Now, treat the first lens, recognizing that the object of the second lens is the image of the first lens: q1 = (50.0 cm - 9.74 cm) = 40.26 cm, and f 1 = 10.0 cm, 1 1 1 1 1 so = = , 10.0 cm 40.26 cm p1 f 1 q1 or p1 = 13.3 cm.
10
CHAPTER TWENTY-THREE SOLUTIONS q1 40.26 cm (b) M 1 = == - 3.03, and 13.3 cm p1
Thus, M overall = M 1 M 2 = -5.91. The overall magnification is less than 0, so the final image is inverted. Also, the final image is virtual (before final lens).
(c) 23.49
q2 -19.0 cm M 2 = == 1.95. 9.74 cm p2
The final image will be on the film, so let us start with the second lens. If d = 5.00 cm, then the lenses are 7.00 cm apart and q2 = 5.00 cm. 1 1 1 1 1 = = , so p2 = -8.01 cm. 13.3 cm 5.00 cm p2 f 2 q2 Thus,
q1 = 7.00 cm + 8.01 cm = 15.01 cm. 1 1 1 1 1 Then, = = , and p1 = 225 m (max dist). 15.0 cm 15.01 cm p1 f 1 q1 Similarly, if d = 10.0 cm, the lenses are 2.00 cm apart and q2 = 10 cm. 1 1 1 1 1 = = , giving p2 = -40.3 cm. 13.3 cm 10.0 cm p2 f 2 q2 Thus, q1 = 2.00 cm + 40.3 cm = 42.3 cm, so 1 1 1 1 1 = = , and 15.0 cm 42.3 cm p1 f 1 q1
p1 = 0.232 m (min dist).
23.50 The image, I 1, formed by the first lens serves as the object for the second lens. With the two lenses in contact with each other, this is a virtual object. If the thicknesses of the two lenses are ignored, the image distance of the first lens and the object distance for the second lens have the same magnitude but opposite signs (i.e., p2 = - q1). Thus, the thin lens equation applied to the second lens gives: 1 1 1 + = - q1 q2 f 2
Applying the thin lens equation to the first lens, we have 1 1 1 + = Adding these two equations yields: p1 q1 f 1 1 1 1 1 + = + . This equation gives the relation between the p1 q2 f 1 f 2 position of the object, O, and the position of the final image, I 2. Note that this is in the form of a thin lens equation. Thus, we see that a pair of thin lenses in contact behaves like a thin lens having a focal length, 1 1 1 f , given by = + . f f 1 f 2 23.51
We first find the focal length of the mirror. 1 1 1 1 1 9.00 = + = + = , and f = 4.44 cm. 10.0 cm 8.00 cm 40.0 cm f p q Hence, if p = 20.0 cm: 1 1 1 1 1 15.56 = = = , and q = 5.71 cm (real image) 4.44 cm 20.0 cm 88.8 cm q f p
11
CHAPTER TWENTY-THREE SOLUTIONS
23.52
For the first lens,
1 1 1 = + f p q
becomes:
1 1 1 = + , giving q1= - 4.0 cm. -6.0 cm 12 cm q1
When we require that q2 approach infinity, we see that p2 must be the focal length of the second lens. In this case: p2 = d - (-4.0 cm). Therefore, d + 4.0 cm = f 2 = 12 cm, and d = 8.0 cm. 23.53 Since the object is very far from
the lens, we approximate p , and the thin lens equation becomes 1 1 1 + = 50.0 mm q which gives q = 50.0 mm. Then, if the angular field of view is as shown at the right, we have 1 (35.0 mm) 2 tan( ) = = 0.350 . 2 50.0 mm Therefore, 23.54
2
= 19.3°, or the angular field of view is:
=
38.6°.
1 1 1 1 1 = = , 10.0 cm 12.5 cm q1 f 1 p1 giving q1 = 50.0 cm. (to left of mirror) This serves as a virtual object for the lens, so p2 = 25.0 cm - q1 = -25.0 cm. 1 1 1 1 1 = = - , or q2 = -50.3 cm. (to right of lens) Thus, the final -16.7 cm - 25.0 cm q2 f 2 p2 image is located 25.3 cm to right of mirror. q1 q2 50.0 cm -50.3 cm M 1 = == - 4.00, M 2 = = - 12.5 cm p1 p2 -25.0 cm = - 2.01, and M = M 1 M 2 = 8.05. Thus, the final image is virtual, erect, 8.05 times the size of object, and 25.3 cm to right of the mirror.
23.55
Start with the first pass through the lens: 1 1 1 1 1 = = , so q1 = 400 cm. 80.0 cm 100 cm q1 f 1 p1
(to right of lens)
1 1 1 1 1 = = . -50.0 cm (- 300 cm) q2 f 2 p2 This yields: q2 = -60.0 cm. For the second pass through the lens: p3 = +160 cm 1 1 1 1 1 = = , giving q3 = 160 cm. (to left of lens) 80.0 cm 160 cm q3 f 1 p3 For the mirror, p2 = -300 cm, so
q1 400 cm M 1 = == - 4.00, 100 cm p1
q2 -60.0 cm 1 M 2 = = - p2 -300 cm = - 5.00
q3 160 cm M 3 = == -1.00, 160 cm p3
and
M = M 1 M 2 M 3 = -0.800.
Since M < 0, the final image is inverted relative to the original object. 23.56
(a)
We find the image position produced by the first lens.
12
CHAPTER TWENTY-THREE SOLUTIONS 1 1 1 1 1 1 = = = , so q1 = 40.0 cm, and 20.0 cm 40.0 cm 40 cm q1 f 1 p1 q1 40.0 cm M 1 = == -1.00. 40.0 cm p1 For the second lens, p2 = 10.0 cm. Thus, 1 1 1 1 1 = = , and q2 = 10.0 cm (to right of 2nd lens) 5.00 cm 10.0 cm q2 f 2 p2 q2 10.0 cm M 2 = = = -1.00. 10.0 cm p2 (b)
(c)
The overall magnification M is: M = ( M 1)( M 2) = (-1.00)(-1.00) = +1.00. Thus, the final image is real (q 2 > 0), erect, and the same size (2.00 cm) as the object. With the two lenses in contact, the effective focal length is calculated using 1 1 1 1 1 = + = + yielding f eff = 4.00 cm 5.00 cm f eff f 1 f 2 20.0 cm The final image is then located by:
1 1 1 1 1 = - = q f eff p 4.00 cm 5.00 cm
which gives q = 20.0 cm. 23.57
We use
n1 p
+
n2 q
=
n2 - n1 R
At the first surface, we have:
applied to each surface in turn. 1 1.50 1.50 - 1 + = ,or q1 = -2.00 cm. 1.00 cm q1 2.00 cm
Thus, the image is formed 2.00 cm to the left of the surface, a distance of 16 cm from the second surface. This image becomes the object for the second surface (hence, p2 = + 16.0 cm). n1 n2 n2 - n1 We again use, + = : p q R 1.50 1 1.00 - 1.50 + = , or q2 = 32.0 cm. 16.0 cm -4.00 cm q2 The final image is a real image 32.0 cm to the right of the second surface. 23.58
First locate the image of the coins formed by the upper mirror: 1 1 1 1 1 1 + = gives: + = , so q1 = , or the reflected rays are 7.5 cm q1 7.5 cm p1 q1 f parallel as they leave this mirror. Now, locate the final image, realizing that parallel rays are reflected toward the lower mirror by the upper mirror (thus. p2 = ). 1 1 1 1 1 1 + = , gives p2 q2 f + q2 = 7.5 cm , and q2 = 7.5 cm. Thus, the final image is located at the center of the upper mirror, is real, and is erect (having undergone two inversions — one due to each mirror). It is also the same size as the object.
23.59
(a)
(b)
1 1 1 R 1.00 m + = we make the following table, with f = = = 0.500 m 2 2 p q f Between p = 3.00 m and p = 0.50 m above the mirror, p = 3 m q = 0.6 m the ball's image is real and moves from + 0.60 m to + p = 0.5 m q = +/- infinity. Between 0.50 m and contact with the mirror, p = 0 q=0 the ball's image is virtual and moves from - infinity to 0. (Note the "jump" of image location as the ball passes through the focal point of the mirror.) For the ball and its image to coincide, the image must be real with
Using
13
CHAPTER TWENTY-THREE SOLUTIONS 1 1 1 1 1 1 2 1 + = becomes: + = ; = ; p = 2 f = 1.00 m. p p f p q f p f The time it takes for the ball to fall 2.00 m to this point is found using 1 1 x = vot + at 2: - 2.00 m = 0 + (-9.80 m/s2) t 2, or t = 0.639 s. 2 2 q = p, so
23.60
(a)
The lensmakers equation, 1 = (n - 1) 5.00 cm
(b)
1 1 1 = (n - 1) - , becomes: f R 1 R2
1 - 1 , 9.00 cm - 11.0 cm
giving
n = 1.99.
As the light passes through the lens for the first time, the thin lens equation, 1 becomes: f
1 1 + = p1 q1
1 1 1 + = , or 8.00 cm q1 5.00 cm
q1 = 13.33 cm, and
M 1 =
-q1 13.33 cm == -1.67. This image becomes the 8.00 cm p1
object for the concave mirror with: pm = 20.0 cm - q1 = 20.0 cm - 13.33 cm = 6.67 cm, R and f = = + 4.00 cm. 2 1 1 1 The mirror equation becomes: + = , 6.67 cm qm 4.00 cm qm = 10.0 cm, and
M 2 =
giving
-qm 10.0 cm == -1.50. 6.67 cm pm
The image formed by the mirror serves as a real object for the lens on the second pass of the light through the lens with: p3 = 20 cm - qm = + 10.0 cm. The thin lens equation yields: 1 1 1 + = , or q3 = 10.0 cm, 10.0 cm q3 5.00 cm and
M 3 =
-q3 p3
=-
10.0 cm = -1.00. The final image is a real image located 10.0 cm to the 10.0 cm
left of the lens. Since the total magnification ( M total = M 1 M 1 M 1 = -2.50) is negative, this final image is inverted. 23.61
(a)
1 1 1 1 1 = = , giving q1 = 30.0 cm. 10.0 cm 15.0 cm q1 f 1 p1 Therefore: p2 = -20.0 cm, and q2 = -25.0 cm. 1 1 1 Thus, = + , and f 2 = -11.1 cm. -20.0 cm -25.0 cm f 2
q1 30.0 cm (b) M 1 = == -2.00, 15.0 cm p1 (c) 23.62
q2 -25.0 cm M 2 = == -1.25, and -20.0 cm p2
Moverall = (M1)(M2) = (-2.00)(-1.25) = 2.50. The final image is virtual, erect, and enlarged.
With the given radii of curvature, the lens maker's equation becomes: 1 n1 1 1 n1 -2 + 1 n1 - 1 , = - 1 = - 1 = - 1 f n2 -3.0 m -6.0 m n2 6.0 m n2 6.0 m or (a)
1 1 n1 = 1. f 6.0 m n2 When n1 = 1.5 and n2 = 1, the focal length of the lens is:
(1)
14
CHAPTER TWENTY-THREE SOLUTIONS
(b) (c) (d)
1 1 = (-0.5 m) , or f = -12.0 m, so the thin lens equation gives f 6.0 m 1 1 1 + = , which yields: q = -5.50 m, or 10.0 m q -12 m the image is 5.50 m to left of the lens. When n1 = 1.5, and n2 = 1.333, the above yields: f = -48.0 m, and q = -8.30 m, or 8.30 m to left of lens. When n1 = 1.5, and n2 = 2.0, the above yields: f = 24 m, and q = -17 m, or 17 m to left of lens. As shown in equation (1), the focal length will be positive if n1 n1 > 0, or < 1, meaning n1 < n2. That is; the surrounding material should 1 - n2 n2 have a higher index of refraction than the lens material.
23.63
1 1 1 fq q = + for p to find: p = and we use M = - . f p q (q - f ) p f (4 f ) 4 f For q = 4 f : p = = . 3 (4 f - f ) f (- 3 f ) 3 f For q = - 3 f : p= = . 4 (-3 f - f ) 4 f - 3 f For case (a), M = = - 3.00, and for case (b), M = = 4.00 4 f 3 f 3 4
We solve (a) (b) (c)
ANSWERS TO CONCEPTUAL QUESTIONS
2. Because of refraction, the fish appears to be at a depth that is less than its actual depth.
Therefore, the person should aim below the image of the fish to hit it. 4. Chromatic aberration is produced when light passes through a material, as it does when
passing through the glass of a lens. A mirror, silvered on its front surface never has light passing through it, so this aberration cannot occur. This is only one of many reasons why large telescopes use mirrors rather than lenses for their optical elements. 6. Make the mirror an efficient reflector (shiny); use a parabolic shaped mirror so that it
reflects all rays to the image point, even those far from the axis; most important, use a large-diameter mirror in order to collect more solar power. 8. A convex mirror is being used. Since a demagnified image is produced, the driver may
think the small truck bearing down on him is far away; thus, the warning. This kind of mirror is used so that a large area behind the car can be seen. 10. All objects beneath the stream appear to be closer to the surface than they really are
because of refraction. Thus, the pebbles on the bottom of the stream appear to be close to the surface of a shallow stream. 12. An effect similar to a mirage is produced except the "mirage" is seen hovering in the
air. Ghost lighthouses in the sky have been seen over bodies of water by this effect.
15
CHAPTER TWENTY-THREE SOLUTIONS 14. Actually no physics is involved here. The design is chosen so your eyelashes will not
brush against the glass as you blink. A reason involving a little physics is that with this design, when you direct your gaze near the outer circumference of the lens you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally distorts the direction to the object you are looking at. 16. Both words are inverted. However OXIDE looks the same right side up and upside
down. LEAD does not.
