Question

A converging lens forms an image of an 8.50 mm -tall real object. The image is 12.5 cm to the left of the lens, 3.50 cm tall, and upright.

A) What is the focal length of the lens?

Express your answer in centimeters to three significant figures.

B) Where is the object located?

Express your answer in centimeters to three significant figures.

Answer 1

Part A.

Using the lens equation for first converging lens

1/f = 1/u + 1/v

Magnification is given by:

M = -v/u = hi/ho

hi = height of image = 3.50 mm

ho = height of object = 8.50 cm

v = image distance = -12.5 cm (Since image is on the left of lens)

M = -v/u = 3.50*10^-3/(8.50*10^-2)

u = -(-12.5)*8.50*10^-3/(3.50*10^-2)

u = 3.04 cm = object distance

Now using lens equation:

1/f = 1/u + 1/v

f = u*v/(u + v) = 3.04*(-12.5)/(3.04 - 12.5)

f = focal length of lens = 4.02 cm

Part B.

from above solved part

u = object distance = 3.04 cm

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