Question

Suppose a book is held 5.8 cm from a 12-cm focal length lens.

(a) Find the magnification of the image of the book.
(b) Find the magnification of the lens when the book is held 8.4 cm from the magnifier.

Answer 1

a) Let's first find the image distance from the lens equation:

$$\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{f},$$

here, $p=5.8\ cm$ is the object distance, $q$ is the image distance and $f=12\ cm$ is the focal length of the lens.

Then, from this equation we can find $q$:

$$q=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{p}}=\dfrac{1}{\dfrac{1}{12\ cm}-\dfrac{1}{5.8\ cm}}=-11.2\ cm.$$

The sign minus means that the image is virtual.

Finally, we can calculate the magnification of the image of the book from the magnification equation:

$$M=-\dfrac{q}{p}=-\dfrac{-11.2\ cm}{5.8\ cm}=1.93$$

b) Again, let's first find $q$ from the lens equation:

$$q=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{p}}=\dfrac{1}{\dfrac{1}{12\ cm}-\dfrac{1}{8.4\ cm}}=-28\ cm.$$

The sign minus means that the image is virtual.

Finally, we can calculate the magnification of the image of the book from the magnification equation:

$$M=-\dfrac{q}{p}=-\dfrac{-28\ cm}{8.4\ cm}=3.3$$

Answer:

a) $M=1.93$

b) $M=3.3$