A diverging lens has a focal length of magnitude 18.0 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 18.0 cm. (a) Locate the images for each of the following object distances. 36.0 cm ocabion 18.0 cm 9.0 cm (b) Is the image for the object at distance 36.0 real or virtual? o real o virtual Is the image for the object at distance 18.0 real or virtual? O real O virtual Is the image for the object at distance 9.0 real or virtual? o real o virtual (c) Is the image for the object at distance 36.0 upright or inverted? o upright O inverted Is the image for the object at distance 18.0 upright or inverted? O upright O inverted Is the image for the object at distance 9.0 upright or inverted? o upright O inverted (d) Find the magnification for the object at distance 36.0 cm. Find the magnification for the object at distance 18.0 cm.

Explanation / Answer

we know, the focal length of a diverging lens is negative.

so, f = -18.0 cm

a) u = 36 cm (object distance)

let v is the image distance.

use, 1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/(-18) - 1/36

v = -12 cm

image distance = 12 cm
location : same side as object

u = 18 cm (object distance)

use, 1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/(-18) - 1/18

v = -9 cm

image distance = 9 cm
location : same side as object

u = 9 cm (object distance)

use, 1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/(-18) - 1/9

v = -6 cm

image distance = 6 cm
location : same side as object

b)

virtual

virtual

virtual

c) upright

upright

upright

d)

when u = 36 cm,

magnification, m = -v/u = -(-12)/36 = 0.33

when u = 18 cm,

magnification, m = -v/u = -(-9)/2 = 0.50

when u = 9 cm,

magnification, m = -v/u = -(-6)/9 = 0.67

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