A diverging lens has a focal length of magnitude 19.8 cm (a) Locate the images f

Question

A diverging lens has a focal length of magnitude 19.8 cm (a) Locate the images for each of the following object distances 39.6 cm distance location | Select cm 19.8 cm distance location | Select cm 9.9 cm distance location -Select- cm (b) Is the image for the object at distance 39.6 real or virtual? real virtual Is the image for the object at distance 19.8 real or virtual? real virtual Is the image for the object at distance 9.9 real or virtual? real O virtual (c) Is the image for the object at distance 39.6 upright or inverted? O upright inverted Is the image for the object at distance 19.8 upright or inverted? O upright O inverted Is the image for the object at distance 9.9 upright or inverted? O upright inverted (d) Find the magnification for the object at distance 39.6 cm

Explanation / Answer

1/f = 1/u + 1/v

1/-19.8 = 1/39.6 + 1/ v

1/v = 1/-19.8 - 1/ 39.6

v = -13.2

m = -v/ u = - ( -13.2/ 39.6) = 0.33

image is virtual ,and upright

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for u = 19.8

1/f = 1/u + 1/v

1/-19.8 = 1/19.8 + 1/ v

1/v = 1/-19.8 - 1/ 19.8

v = -9.94

m = -v/ u = - ( -9.94/ 19.8) = 0.50

image is virtual ,and upright

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for u = 9.9

1/f = 1/u + 1/v

1/-19.8 = 1/9.9 + 1/ v

1/v = 1/-19.8 - 1/ 9.9

v = -6.6

m = -v/ u = - ( -6.6/ 9.9) = 0.66

image is virtual ,and upright

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