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

Question

33. 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 location 22.0 cm distance location Tm 11.0 cm location (b) Is the image for the object at distance 44.0 real or virtual? Is the image for the object at distance 22.0 real or virtual? real virtual Is the image for the object at distance 11.0 real or virtual? real lvirtual (c) Is the image for the object at distance 44.0 upright or inverted? !upright Elinverted Is the image for the object at distance 22.0 upright or inverted? upright ]inverted Is the image for the object at distance 11.0 upright or inverted? upright inverted 31. (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

focal length f = -22 cm

(a)

object distance p1 = 44 cm

image distance q1 = ?

1/p1 + 1/q1 = 1/f

1/44 + 1/q1 = -1/22

q1 = -14.7 cm

object distance p1 = 22 cm

image distance q2 = ?

1/p2 + 1/q2 = 1/f

1/22 + 1/q2 = -1/22

q2 = -11 cm

object distance p3 = 11 cm

image distance q3 = ?

1/p3 + 1/q3 = 1/f

1/11 + 1/q3 = -1/22

q3 = -7.3 cm

(b)

q1 is negative

image is virtual

q2 is negative image is virtual

q3 is negative image is virtual

(c)

magnification m = -q/p

magnification is positive , image is upright

magnification is positive , image is upright

magnification is positive , image is upright

(d)

m1 = 14.7/44 = 0.334

m2 = 11/22 = 0.5

m3 = 7.3/11 = 0.66

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