A diverging lens has a focal length of -26.0 cm. Locate the images for each of t

Question

A diverging lens has a focal length of -26.0 cm. Locate the images for each of the following object distances. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.

(a) 52.0 cm
cm  --Location of image-- in front of the lens, behind the lens, or no image formed

real erect, real inverted, virtual upright, or virtual inverted

real erect, real inverted, virtual upright, or virtual inverted

real erect, real inverted, virtual upright, or virtual inverted

Explanation / Answer

Part A)

Apply 1/f = 1/p +1/q

1/-26 = 1/52 + 1/q

q = -17.3

Thus the image is virtual and upright and formed in front of the lens

M = -q/p = -(-17.3)/52

M = .333

Part B)

1/-26 = 1/26 + 1/q

q = -13 cm

Thus the image is virtual and upright and formed in front of the lens

M = -q/p = -(-13)/26

M = .5

Part C)

1/-26 = 1/13 + 1/q

q = -8.67 cm

Thus the image is virtual and upright and formed in front of the lens

M = -q/p = -(-8.67)/13

M = .667

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