Chapter 17 - Worked Solutions

to selected extension questions from the OUP text New Century Senior Physics
by Walding, Rapkins and Rossiter

Q 38.

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

Q 39.

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.

Q 40.
(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.
Q 41.
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
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