1. A 10 cm tall object is located in front of a diverging lens with a power of -

Question

1. A 10 cm tall object is located in front of a diverging lens with a power of -5 dipoters. Describe the type, location and height of the image created, and draw the ray diagrams if the object is located (a) 5cm from the lens, (b) 10 cm from the lens, (c) 20cm from the lens, and (d) 50 cm from the lens.

2.A 2.00 cm tall object is 18 cm in front of a converging lens with a focal length of 30.0 cm. (a) Use the lens equation and (b) a ray trace diagram to describe the type, location, and height of the image formed.

Explanation / Answer

1) power = -5 so f = -1/5 = -0.2 m = -20 cm, height of object = 10 cm

use formula: 1/f = 1/u + 1/v

a) u = 5cm, f= - 20 cm so we get: v = 4cm (virtual so -ve sign is not shown but is there)

M = - (-4/5) = 0.8 so height of image = 8 cm

b) u =10 cm, f = -20 cm so we get: v = 6.67 cm (virtual so -ve sign is not shown but is there)

M = - (-6.67/10) = 0.667 so height of image = 6.67 cm

c) u = 20cm, f = -20 cm so we get: v = 10 cm (virtual so -ve sign is not shown but is there)

M = - (-10/20) = 0.5 so height of image = 5 cm

d) u = 50 cm, f = -20 cm so we get: v = 14.286 cm (virtual so -ve sign is not shown but is there)

M = - (-14.286/50) = 0.286 so height of image = 2.86 cm

2) converging lens, f = 30 cm, height of object = 2 cm

a) 1/f = 1/u + 1/v

u = 18 cm and f = 30 cm

so we get: v = 45 cm (virtual so -ve sign is not shown but is there)

height of image = - (-45/18) * 2 = 5 cm

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