Question

A diverging lens has a focal length of magnitude 21.4 cm. (a) Locate the images for each of the following object distances. 42.8 cm distance cm location 21.4 cm distance cm location 10.7 cm distance cm location (b) Is the image for the object at distance 42.8 real or virtual? real virtual Is the image for the object at distance 21.4 real or virtual? real virtual Is the image for the object at distance 10.7 real or virtual? real virtual (c) Is the image for the object at distance 42.8 upright or inverted? upright inverted Is the image for the object at distance 21.4 upright or inverted? upright inverted Is the image for the object at distance 10.7 upright or inverted? upright inverted (d) Find the magnification for the object at distance 42.8 cm. Find the magnification for the object at distance 21.4 cm. Find the magnification for the object at distance 10.7 cm.

Answer 1

focal length f = -21.4 cm

(a)

object distance p = 42.8 cm

1/p + 1/q = 1/f

1/42.8 + 1/q = -1/21.4

q = -14.3 cm

object distance p = 21.4 cm

1/p + 1/q = 1/f

1/21.4 + 1/q = -1/21.4

q = -10.7cm

object distance p = 10.7 cm

1/p + 1/q = 1/f

1/10.7 + 1/q = -1/21.4

q = -7.13 cm

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(b)

Is the image for the object at distance 42.8

virtual

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Is the image for the object at distance 21.4

virtual

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Is the image for the object at distance 10.7

virtual

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(c)

Is the image for the object at distance 42.8

upright

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Is the image for the object at distance 21.4

upright

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Is the image for the object at distance 10.7

upright

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(d) the magnification for the object at distance 42.8 cm

M = 14.3/42.8 = 0.334

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the magnification for the object at distance 21.4 cm

M = 10.7/21.4 = 0.5

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the magnification for the object at distance 10.7 cm

M = 7.13/10.7 = 0.666