Answer to Question #286056 in Physics for lovely

Question

what is the image distance and the image size if a 4:00 cm tall light bulb is placed a distance of 30.5 cm from a diverging lens having a focal length of -11.2 cm?

Answer 1

We can find the image distance from the lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}}=\dfrac{1}{-\dfrac{1}{11.2\ cm}-\dfrac{1}{30.5\ cm}}=-8.2\ cm.$$

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

We can find the image size from the magnification equation:

$$m=-\dfrac{d_i}{d_o}=\dfrac{h_i}{h_o},$$ $$h_i = -\dfrac{h_od_i}{d_o}=-\dfrac{4\ cm\times(-8.2\ cm)}{30.5\ cm}=1.07\ cm.$$

The image is upright and diminished in size.