Question

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object.
(a) What is the focal length of the lens? Is the lens converging or diverging?
(b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted?
(c) Draw a principal-ray diagram.

Answer 1

(a) -48.0 cm, diverging

We can use the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}$

$f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm$

The lens is diverging, since the focal length is negative.

(b) 6.38 mm, erect

We can use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the size of the image

y = 8.50 mm is the size of the object

Substituting p and q that we used in the previous part of the problem, we find y':

$y'=-y\frac{q}{p}=-(8.50 mm)\frac{-12.0 cm}{16.0 cm}=6.38 mm$

and the image is erect, since the sign is positive.

(c)

See attached picture.