A diverging lens has a focal length of magnitude 15.8 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 15.8 cm. (a) Locate the images for each of the following object distances. 31.6 cm distance cmm location G-Select- 15.8 cm distance locationSelect-+ cmm 7.9 cm distance location l-Select- cmm (b) Is the image for the object at distance 31.6 real or virtual? virtual Is the image for the object at distance 15.8 real or virtual? real virtual Is the image for the object at distance 7.9 real or virtual? real virtual (c) Is the image for the object at distance 31.6 upright or inverted upright Is the image for the object at distance 15.8 upright or inverted? upright Is the image for the object at distance 7.9 upright or inverted? upright (d) Find the magnification for the object at distance 31.6 cm. Find the magnification for the object at distance 15.8 cm. Find the magnification for the object at distance 7.9 cm

Explanation / Answer

f = -15.8 cm, do = 31.6 cm

1/f = 1/do + 1/di

1/di = 1/-15.8 - 1/31.6

a) di = -10.53 cm

b)c) image is virtual and upright

m = -di/d0 = 10.53/31.6

d) magnification = 0.33

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f = -15.8 cm, do = 15.8 cm

1/f = 1/do + 1/di

1/di = 1/-15.8 - 1/15.8

a) di = -7.9 cm

b) c) image is virtual and upright

m = -di/d0 = 7.9/15.8

d) magnification = 0.5

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f = -15.8 cm, do = 7.9 cm

1/f = 1/do + 1/di

1/di = 1/-15.8 - 1/7.9

a) di = -5.27 cm

b) c) image is virtual and upright

m = -di/d0 = 5.27/7.9

d) magnification = 0.67

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