New Century Senior Physics

New Century Senior Physics

Second Edition

Worked solutions

Chapter 17 - Mirrors

from Richard Walding

Luminous: sun, stars, a gas flame and an electric light filament. Non-luminous: Moon, Earth, yourself.

(a) CBD, DBA, CB, BA, DB

(b) XWY, YWZ, XW, WZ, YW

(a) 50, (b) 70, (c) 65

She approaches door at 2 m s-1 and the image approaches door from opposite direct at 2 m s-1. The speed of the image relative to the person is this 4 m s-1.

The image is laterally inverted twice so it appear non-reversed.

Q10

Q11

Q12

Q13

*14

Laws of reflection hold but the light has diffuse reflection by the uneven surface

*15

It is laterally reversed (left to right), is upright, not magnified, and is as far behind the mirror as the object is in front.

*16

Four metres to the image. Autofocus may have to be told to focus on the image. If the mirror has specks of dust on it the camera may focus on the dust.

*17

It should be half the size of your body length and placed so that the bottom of the mirror is placed on the wall at a height that is half the distance between the bottom of your feet and your eyes.

*18

A real image is one through which the rays actually pass. A virtual image is one through which the ray only appear to pass.

Q19

Q20

Q21

Q22

Q23

Q24

*25

In a rough surface you get diffuse reflection so the rays do not reflect in any uniform way. In a smooth surface you get specular reflection which produces an image as the reflection is regular.

*26

They could put the chart 3.0 m from the mirror and have the observer 3,0 m from the mirror. The chart would have to be printed backwards as there would be lateral inversion.

*27

So that when they come up behind a driver, the sign is laterally reversed in the driver’s rear view mirror and reads correctly.

**28

Q29

Q30

Q31

Q32

Q33

Q34

Q35

Q36

(a) parabolic shape; (b) place at focus

Q37

(a)

Image distance (cm) / 60 / 37.5 / 26.3 / 24 / 21.4 / 20 / 18.5
Image height (cm) / 12 / 6 / 3 / 2.4 / 1.7 / 1.3 / 1
Magnification / 3.00 / 1.50 / 0.75 / 0.60 / 0.43 / 0.33 / 0.25

(b)

Can check using equation: y = 0.0666x - 1.0

M = 0.0666 × 45 - 1.0 = 2.0

(d) M = 0.5, v = 22.5 cm from graph. or 22.5 using equation

(e) When M = 1, v = u = 30 cm

1/f = 1/v + 1/u

1/f = 1/30 + 1/30 = 2/30

f = 15 cm

Q38

If the object is 20 cm in front of the plane mirror then the image in the plane mirror is 20 cm behind the mirror.
The image of the object in the convex mirror is 10 cm behind the concave mirror. This is because it then produced an image in the plane mirror and this image is 60 cm behind the plane mirror. (40 cm between the two images.)
Therefore the image in the convex mirror is 10 cm behind the convex mirror.
Therefore the image distance = -10 cm
1/v +1/u = 1/f
1/-10 + 1/30 = 1/f
1/f = - 3/30 + 1/30 = -2/30
f = -15 cm

Q39

The image in the convex mirror ( f = - 30 cm):
1/v + 1/u = 1/f
1/v + 1/20 = 1/-30
1/v = -1/30 - 1/20
v = -12 cm : that is the image is 12 cm behind the mirror
total distance from the object to its image = 20 + 12 = 32 cm
therefore the plane mirror is half way = 16 cm
that is 4 cm in front of the convex mirror.

Q40
(a)
f = 20 cm, Virtual image and M = 2
ie. v/u = 2 therefore v = 2u and v is negative.
1/v + 1/u = 1/f
1/-2u + 1/u = 1/20
-1/2u + 2/2u = 1/20
1/2u = 1/20
2u = 20
u = 10
therefore v = -20 cm, Object distance = 10 cm , Image distance = 20 cm behind the mirror.
(b)
Real image and M = 2
ie. v/u = 2 , v = 2u and positive.
1/v + 1/u = 1/f
1/2u + 1/u = 1/20
1/2u + 2/2u = 1/20
3/2u = 1/20
u = 30 cm
v = 60 cm
Object distance = 30 cm, Image distance = 60 cm in front of the mirror.

Q41

The plot of the data is as follows

When the distance between the object and mirror (u) equals the distance between the image and mirror (v) the object/image must be on the centre of curvature. This occurs when u = v = 16 cm, hence, the centre of curvature C = 16 cm. The focal length is half this value, so f = 8 cm