A diverging lens has a focal length of -14.0 cm. Locate the images for each of t

Question

A diverging lens has a focal length of -14.0 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) 28.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erect

real, inverted    

virtual, erect

virtual, inverted

magnification
?

(b) 14.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erect

real, inverted    

virtual, erect

virtual, inverted

magnification
?

(c) 7.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erect

real, inverted    

virtual, erect

virtual, inverted

magnification
?

Explanation / Answer

For diverging lens,

f = - 14 cm

In each case use, 1/v - 1/u = 1/f

v = image distance

u = object distance

and m = magnification = v/u

a) u = -28 cm

v = - 9.33 cm, in front of the lens

m = 0.33 ( Virtual, upright and diminished image)

b) u = - 14 cm

v = - 7 cm, in front of the lens

m = 0.5 ( Virtual, upright and diminished image)

c) u = - 7 cm

v = - 4.67 cm , in front of the lens

m = 0.67 ( Virtual, upright and diminished image)

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