Question

A thin, diverging lens having a focal length of magnitude 45.0 cm has the same principal axis as a concave mirror with a radius of 60.0 cm. The center of the mirror is 20.0 cm from the lens, with the lens in front of the mirror. An object is placed 22.0 cm in front of the lens. ANSWERS IN CM

1) Where is the final image due to 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) Suppose now that the concave mirror is replaced by a convex mirror of the same radius. Where is the final image due to the lens–mirror combination? Enter the image distance with respect to the mirror. Follow the sign convention. (Express your answer to three significant figures.)

Answer 1

for lens

object distance s1 = 20 cm

focal length f1 = -45 cm

from lens equation

1/s1 + 1/s1' = 1/f1

1/22 + 1/s1' = -1/45

s1' = -14.8 cm

this image serves as object for mirror

for mirror

object distance s2 = 20 + 14.8 = 34.8 cm

radius of curvature R = 60 cm

from mirror equation

1/s2 + 1/s2' = 2/R

1/34.8 + 1/s2' = 2/60

s2' = 217.5 cm = 218 cm   <<<--------ANSWER

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for lens

object distance s1 = 20 cm

focal length f1 = -45 cm

from lens equation

1/s1 + 1/s1' = 1/f1

1/22 + 1/s1' = -1/45

s1' = -14.8 cm

this image serves as object for mirror

for mirror

object distance s2 = 20 + 14.8 = 34.8 cm

radius of curvature R = -60 cm

from mirror equation

1/s2 + 1/s2' = 2/R

1/34.8 + 1/s2' = -2/60

s2' = -16.1 cm <<<--------ANSWER

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