Ch. 27 HW Solutions
1.) Because the lens takes the parallel rays from the sun and focuses them on a point, the
lens is converging. The point at which parallel rays converge is the focus of the lens, 12 cm.
3.) (a) q = 80 cm: real, inverted
(b) q = 200 cm: real, inverted
(c) q = -40 cm: virtual, upright
4.) (a) q = 100 cm: real, inverted
(b) q = 133 cm: real, inverted
(c) q = -33.3 cm: virtual, upright
8.) p = 12 cm, q = 4.0 m. f = (1/p + 1/q)−1 = (1/12 + 1/400)−1 = 11.65 cm. M =
−q/p = −400/12 = −33.3. The image size is magnified 33.3 times, so it is 83.3 cm × 116.7
cm (inverted).
14.) q = 80 cm, p = 12 cm. f = (1/p + 1/q)−1 = (1/12 + 1/80)−1 = 10.4 cm.
15.) f = 80 cm, hi = 2 m = 200 cm, ho = 50 cm. M = −hi /ho = −200/50 = −4
(real images from converging lenses are inverted). Since M = −q/p = −4, q = 4p. Then
1/f = 1/80 = 1/p + 1/4p = 5/4p. So (a) p = (5)(80)/4 = 100 cm. (b) q = 4p = 400 cm.
20.) The object is 150 cm from the screen (the image location) so p + q = 150 cm. With f
= 25 cm, we use the lens equation
1
1
1
150 − p
p
150 − p + p
150
= +
=
+
=
=
.
25
p 150 − p
p(150 − p) p(150 − p)
p(150 − p)
p(150 − p)
25 =
p(150 − p)
−p2 + 150p
=
150
150
p2 − 150p + 3750 = 0
Using the quadratic equation,
150 ±
p
1502 − (4)(3750)
2
and solving yields p = 118.3 cm, 31.7 cm. The lens must be placed this far from the object.
Or instead solving for the image distance, q = 31.7 cm, 118.3 cm, the lens must be placed
this far from the screen.
22.) f = 8 cm, q = -20 cm (images are virtual for magnifying glasses). 1/p = 1/f − 1/q =
1/8 − 1/(−20) so p = 5.71 cm. M = −q/p = 20/5.71 = 3.5.
23.) f1 = 20 cm, f2 = 20 cm, d = 40 cm apart. Object of height ho = 5 cm is placed
p1 = 80 cm in front of the lens 1. To find the location and nature of the final image requires
two steps, finding the image from the first lens and using that image as the object for the
second lens.
To find the image from the first lens, q1 = (1/f1 − 1/p1 )−1 = 26.67 cm behind the first lens.
This image is then d − q1 = 40 − 26.67 = 13.33 cm in front of the second lens, so p2 = 13.33
cm. The final image location, q2 = (1/f2 − 1/p2 )−1 = −40 cm, this is 40 cm in front of (on
the object side) of the second lens or 20 cm in front of the first lens. It must be virtual (q
is negative).
M = M1 M2 = q1 q2 /p1 p2 = (26.67)(−40)/(80)(13.33) = −1.00. The total magnification is
1.00 but the image is inverted. Final image size is 5 cm.
24.) p = 50 m = 5000 cm, f = 15 cm, hi = 3.0 cm. q = (1/f − 1/p)−1 = 15.05 cm.
M = −15.05/5000 = 0.0030. So the actual tower is ho = hi /M = 3.0/0.0030 = 997 cm or
9.97 m tall.
27.) f1 = 20 cm, f2 = −30 cm (diverging), d = 50 cm, ho = 4 cm, p1 = 60 cm.
q1 = (1/f1 − 1/p1 )−1 = 30 cm. p2 = d − q1 = 20 cm.
q2 = (1/f2 − 1/p2 )−1 = −12 cm. The final image is virtual.
M = M1 M2 = q1 q2 /p1 p2 = −0.3. The final image is inverted and hi = ho M = −1.2 cm.
37.) M = 200 = −qo qe /po pe , fo = 0.75 in., fe = 0.25 in. Solve for d, where d = qo + pe . To
solve we first note that the image from the eyepiece will be virtual and roughly at the near
point of the eye, so qe = −10 in, and then we solve backwards for pe and qo .
pe = (1/fe − 1/qe )−1 = 0.2439 in.
From the magnification equation, qo /po = −M pe /qe = 4.878, or qo = po (4.878).
1/fo = 1/po + 1/qo = 1/po + 1/(4.878po ) so po = 0.90375 in.
qo = po (4.878) = 4.4085 in. d = qo + pe = 4.6524 in.
2