# Chapter 19 - Worked Solutions

### to selected extension questions from the OUP text *Senior Physics -
Concepts in Context*

by Walding,
Rapkins and Rossiter

**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

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