Can someone help me please? Use the worked example above to help you solve this

Question

Can someone help me please?

Use the worked example above to help you solve this problem. A diverging lens of focal length f = minus9.9 cm forms imaqes of an obiect situated at various distances. If the object is placed p_1 = 29.7 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. q = cm M = Repeat the problem when the object is at p_2 = 9.9 cm. q = cm M = Repeat the problem again when the object is 4.95 cm from the lens. q = cm M = Use the values from PRACTICE IT to help you work this exercise. Repeat the calculation, finding the position of the image and the magnification if the object is 19.7 cm from the lens. q = cm M =

Explanation / Answer

For thin lens:

1/do + 1/di = 1/f

=> image distance, di = fdo/(do - f)

magnification, m = -di/do

a) di = -9.9 * 29.7 / [29.7 - (-9.9)] = -7.4 cm

Since the image distance is negative, image is virtual.

m = -(-7.4) / 29.7 = 0.25

Since magnification is positive, image is upright.

b) di = -9.9 * 9.9 / [9.9 - (-9.9)] = -4.95 cm

Since the image distance is negative, image is virtual.

m = -(-4.95) / 9.9 = 0.5

Since magnification is positive, image is upright.

c) di = -9.9 * 4.95 / [4.95 - (-9.9)] = -3.3 cm

Since the image distance is negative, image is virtual.

m = -(-3.3) / 4.95 = 0.67

Since magnification is positive, image is upright.

d) di = -9.9 * 19.7 / [19.7 - (-9.9)] = -6.6 cm

Since the image distance is negative, image is virtual.

m = -(-6.6) / 19.7 = 0.34

Since magnification is positive, image is upright.

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