A diverging lens has a focal length of magnitude 22.0 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 22.0 cm.

(a) Locate the images for each of the following object distances.

44.0 cm (in front )

. 22.0 cm (in front)

11.0 cm (in front)

(b) Is the image for the object at distance 44.0 real or virtual? virtual

Is the image for the object at distance 22.0 real or virtual? virtual

Is the image for the object at distance 11.0 real or virtual? virtual

(c) Is the image for the object at distance 44.0 upright or inverted? upright

Is the image for the object at distance 22.0 upright or inverted? upright

Is the image for the object at distance 11.0 upright or inverted? upright

(d) Find the magnification for the object at distance 44.0 cm.

Find the magnification for the object at distance 22.0 cm.

Find the magnification for the object at distance 11.0 cm.

Explanation / Answer

f= - 22 cm

1/v- 1/u = 1/f

1/v - 1/-44 = 1/-22

v= -14.67

m = v/ u = -14.67 /-44 = 0.33

virtual upright

case 2

u= -22 we got v = -11 virtual upright

m = -v/u = -11/ -22 = 0.5

case 2

u = -11 so we got v= -7.33 cm

virtual upright

m = v/u = 7.33/11 = 0.33

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