# Chapter 19 - Worked Solutions

### to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.

Solutions-Ch19-Finch.pdf

The rest are by Greg Rapkins:

Q 26.

For the convex lens
1/v + 1/u = 1/f
1/v = 1/f - 1/u
= 1/5 - 1/12
v = 8.57 cm
Therefore the distance to the concave lens = 25 - 8.57 = 16.43 cm.
For the concave lens :
1/v + 1/u = 1/f
1/v = 1/f - 1/u
= 1/-8 - 1/16.43
v = -5.4 cm
therefore the image is 5.4 cm to the right of the concave lens.
Q 27.

(a) See image of light through the lens and the image of the image of the light in the mirror.
Q28.
u 25 30 45 50 60 80 100
v 100 60 36 33.3 30 27 25

When v = u the object and the image are at the center of curvature that is at 2f.
v = u = 40 cm ,therefore f = 20 cm.
Question 29

Let the object distance be 'x'
then: v = (3.6 - x) m
M = 8
M = v/u = (3.6 - x) / x
(3.6 - x) / x = 8
3.6 - x = 8x
3.6 = 9x
x = 40 cm
1/v + 1/u = 1/f
1/320 + 1/40 = 1/f
f = 35.5 cm
Q 30.
M = v/u = 7 ,then v = 7 u, and v + u = 400 cm
7u + u = 400
8u = 400
u = 50 cm
therefore v = 400 - 50 = 350 cm
1/v + 1/u = 1/f
1/50 + 1/350 = 1/f
f = 43.8 cm.
Q31.
(a) Since the image is upright and this only occurs when there is a virtual image, the image is on the same side of the lens. That is 'v' is negative.
f = 20 , M = 5 , M = v/u = 5, v = 5 u ,and v is negative
1/v + 1/u = 1/f
1/-5u + 1/u = 1/20
-1/5u + 5/5u = 1/20
4/5u = 1/20
5u = 80
u = 16 cm
(b) Since the image is inverted it is real , and therefore on the opposite side of the lens. M = v/u = 5
v = 5u and is positive
1/5u + 1/u = 1/20
1/5u + 5/5u = 1/20
6/5u = 1/20
5u = 120
u = 24 cm
Q 32.

Using a few values: M = v/u
when u = 80 cm M = 0.2 that is 0.2 = v/80
or v = 0.2 x 80 = 16 cm
1/v + 1/u = 1/f
1/-16 + 1/80 = 1/f
-5/80 + 1/80 = 1/f
-4/80 = 1/f
f = -20 cm
For u = 20 cm, M = 0.5
that is 0.5 = v/20
or v = 0.5 x 20 = 10 cm
1/v + 1/u = 1/f
1/-10 + 1/20 = 1/f
-2/20 + 1/20 = 1/f
-1/20 = 1/f
f = -20 cm, the focal length is 20 cm
Graphically: Since 1/v + 1/u = 1/f or 1/v = - 1/u + 1/f
that is y = -x +c where y = 1/v, x = 1/u, and c = 1/f.
Graphing 1/v against 1/u gives an intercept on the y (1/v) axis of c (1/f).

The graph crosses the y axis at approximately 0.055 this is 1/f
1/f = 0.055 , f = 18.2cm
Q 33.
M = 3, that is v/u = 3 or v = 3u
1/v + 1/u = 1/f
1/3u + 1/u = 1/f
1/3u + 3/3u = 1/f
4/3u = 1/f
Now M = 2.5, that is v/u = 2.5 or v = 2.5 u
that is (3u -20) = 2.5 x, where x = new object distance.
(3u-20)/2.5 = x
and 1/v + 1/u = 1/f,
Combining and reducing to 1/f = 3.5/(3u-20)
substitute for 1/f = 4/3u
10.5 u = 12u - 80
1.5u = 80
u = 53.3 cm
therefore 1/f = 4/3u = 4 / 3 x 53.3
f = 40 cm