GOAL Calculate geometric quantities associated with a diverging lens. PROBLEM A

Question

GOAL Calculate geometric quantities associated with a diverging lens. PROBLEM A diverging lens of 750 focal length 10.0 cm forms images of an object situated at various 5.00 cm distances. (a) If the object is placed 30.0 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (b) Repeat the problem when the object is at 10.0 cm and (c) again when the object is 5.00 cm from the lens. STRATEGY Once again, substitution into the thin-lens equation and the associated magnification equation, together with the thin lens conventions, solve the various parts. The only difference is the negative focal length. SOLUTION (A) Locate the image and its magnification if the object is at 30.0 cm. The ray diagram is given in figure a. Apply the thin-lens equation with p f 30.0 cm to locate the image. 30.0 cm 10.0 cm 7.50 cm Solve for a, which is negative and hence virtual. substitute to get the magnification. 7.50 cm +0.250 Because Mis positive and has 30.0 cm absolute value less than 1, the image is upright and smaller than the object. (B) Locate the image and find its magnification if the object is 10.0 cm from the lens Apply the thin-lens equation, taking p 1 1- l 10.0 cm 10.0 cm 10.0 cm

Explanation / Answer

1/v -1/u = 1/f

1/v + 1/16.3 = -1/8.4

v = - 5.543 cm Image is formed on left side of lens that is its virtual Image.

m = v/u = -5.543/-16.3 = 0.340

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