Answer to Question #191825 in Physics for Luca

Question

An object is located 10 cm from a lens of focal length 12 cm

a.) Solve for the image distance

b.) Solve for the magnification

c.) Is the image real or virtual?

d.) Is the image reduced or enlarged?

Answer 1

(a) We can find the image distance from the lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}},$$ $$d_i=\dfrac{1}{\dfrac{1}{12\ cm}-\dfrac{1}{10\ cm}}=-60\ cm.$$

The sign minus means that the image is virtual.

(b) We can find the magnification of the lens from the magnification equation:

$$M=-\dfrac{d_i}{d_o},$$ $$M=-\dfrac{-60\ cm}{10\ cm}=6.$$

The positive magnification means that the image is upright and enlarged.

(c) The image is virtual.

(d) The image is upright and enlarged.