Answer to Question #148231 in Physics for Tevon.t

Question

A 10-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, and the image height of the image.

Answer 1

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

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_{0}}}=\dfrac{1}{\dfrac{1}{15\ cm}-\dfrac{1}{5\ cm}}=-7.5\ cm.$$

The sign minus means that the image is virtual and formed on the same side of the lens as the object. 

b) We can find the magnification of the image from the magnification equation:

$$M=\dfrac{h_i}{h_o}=-\dfrac{d_i}{d_o},$$ $$M=-\dfrac{-7.5\ cm}{5\ cm}=1.5$$

The sign plus means that the image is upright.

c) Finally, we can find the height of the image from the same magnification equation:

$$h_i=Mh_o=1.5\cdot 10\ cm=15\ cm.$$

The sign plus means that the image is upright.

Answer:

a) $d_i=-7.5\ cm,$ virtual image.

b) $M=1.5$

c) $h_i=15\ cm,$ upright.