Question

Some body of the height 2cm is at the distance of 15cm from the diverging lens

with the focal length (distance) 0.1m. Where will be the image of the body? What size will

be the image of the body?

Answer 1

We can find the image distance from the lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}}=\dfrac{1}{-\dfrac{1}{10\ cm}-\dfrac{1}{15\ cm}}=-6\ cm.$$

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:

$$m=-\dfrac{d_i}{d_o}=\dfrac{h_i}{h_o},$$ $$h_i = -\dfrac{h_od_i}{d_o}=-\dfrac{2\ cm\times(-6\ cm)}{10\ cm}=1.2\ cm.$$

The image is upright and diminished in size.