A diverging lens has a focal length of magnitude 16.8 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 16.8 cm. (a) Locate the images for each of the following object distances. 33.6 cm distance cm location 16.8 cm distance cm location 8.4 cm distance cm location (b) Is the image for the object at distance 33.6 real or virtual? real virtual Is the image for the object at distance 16.8 real or virtual? real virtual Is the image for the object at distance 8.4 real or virtual? real virtual (c) Is the image for the object at distance 33.6 upright or inverted? upright inverted Is the image for the object at distance 16.8 upright or inverted? upright inverted Is the image for the object at distance 8.4 upright or inverted? upright inverted (d) Find the magnification for the object at distance 33.6 cm. Find the magnification for the object at distance 16.8 cm. Find the magnification for the object at distance 8.4 cm.

Explanation / Answer

f = 16.8 cm

1/f = 1/do + 1/di

Magnification M = -di/do

Note that for real images, M is negative and the image is inverted. For virtual images, M is positive and the image is upright.

For do = 33.6 cm

di = 33.6 cm

Magnification M = -di/do = -1

M is negative and the image is real and inverted

For do = 16.8 cm

di = infinty/NA

Magnification M = -di/do = Not available

For do = 8.4 cm

di = -16.8 cm

Magnification M = -di/do = 0.5

M is positive and the image is virtual and upright.

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