A thin, converging lens having a focal length of magnitude 23.00 cm is placed 1.

Question

A thin, converging lens having a focal length of magnitude 23.00 cm is placed 1.000 m from a plane mirror that is oriented perpendicular to the principal axis of the lens. A flower, 8.400 cm tall, is 1.50 m from the mirror on the principal axis of the lens.

1) Where is the final image of the flower produced by the lens–mirror combination? Enter the image distance with respect to the mirror. Follow the sign convention. (Express your answer to three significant figures.)

2)How tall is the image? Follow the sign convention. (Express your answer to three significant figures.)

3)If the converging lens is replaced by a diverging lens having a focal length of the same magnitude as the original lens, where is the final image of the flower produced by the lens-mirror combination? Enter the image distance with respect to the mirror. Follow the sign convention. (Express your answer to three significant figures.)

4)How tall is the image? Follow the sign convention. (Express your answer to three significant figures.)

Explanation / Answer

1] Use the lens formula,

1/f = 1/v + 1/u

u = 1.5m - 1.0m = 50 cm

=> 1/23 = 1/v + 1/50

=> v = 42.5926 cm

this image gets reflected. So, u' = - 42.5926 cm

1/23 = 1/v' - 1/42.5926

=> v' = 14.935 cm to the left.

So, the final distance with respect to the mirror is 100 + 14.935 = 114.935 cm = 115cm. (2sf). This is the final image.

Magnification of the setup M = vv'/uu' = 14.935(42.5926)/(42.5926)(50) = 0.2987

2] so Height of the image of flower H = Mh = 2.51 cm

3] Repeat the same for f = - 23 cm

- 1/23 = 1/v + 1/50

=> v = - 15.753 cm

which will be to the left of the diverging lens. Hence its image shall not reach the mirror and will thus not get reflected.

so, the final distance from the mirror will be: 100 + 15.753 = 115.753 cm = 116 cm (3 sf).

4] Height of the image of the flower H = Mh = (15.753/50)8.4 = 2.65 cm.

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