Question

a small object is placed 30cm from a diverging (concave) lens of focal length 10cm. determine by scale drawing or by calculation (stating any sign convention used) the position and nature of the image

Answer 1

Let's determine by the calculation the position and nature of the image:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=-\dfrac{1}{f},$$ $$d_i=\dfrac{1}{-\dfrac{1}{f}-\dfrac{1}{d_o}},$$ $$d_i=\dfrac{1}{-\dfrac{1}{10\ cm}-\dfrac{1}{30\ cm}}=-7.5\ cm.$$

The sign minus means that the image is virtual.

Let's find the magnification of the lens:

$$M=\dfrac{-d_i}{d_o}=\dfrac{-(-7.5\ cm)}{30\ cm}=0.25.$$

The sign plus means that the image is upright and diminished in size.

Therefore, the image is virtual, upright and diminished in size.