Question

a. A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the image height, if the image is real or virtual (explain in detail) and if the image is erect or inverted (explain in detail).

b. The real object is now placed 13.0 cm in front of the converging lens. Again find the paramaters listed above. Graphically show the optic configuration for this second object placement.

Answer #1

a)

h_{o} = object height = 4 cm

d_{o} = object distance = 30 cm

f = focal length = 23 cm

d_{i} = image distance = ?

h_{i} = image height = ?

Using the lens equation

1/d_{i} + 1/d_{o} = 1/f

1/d_{i} + 1/30 = 1/23

d_{i} = 98.6 cm

Using the equation

h_{i}/h_{o} = - d_{i}/d_{o}

h_{i}/4 = - 98.6/30

h_{i} = - 13.15 cm

the image is real and inverted

b)

d_{o} = object distance = 13 cm

h_{o} = object height = 13.15 cm

f = 23 cm

using the lens equation

1/d_{i} + 1/d_{o} = 1/f

1/d_{i} + 1/13 = 1/23

d_{i} = - 29.9 cm

image height is given as

h_{i}/h_{o} = - d_{i}/d_{o}

h_{i}/13.15 = - (- 29.9)/13

h_{i} = 30.25 cm

the image is virtual and erect

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