Question

The image obtained with a converging lens is erect and three times the length of the object. The focal length of the lens is 20cm. Calculate the object and image distances.

Answer 1

Let's write the magnification equation:

$$M=\dfrac{h_i}{h_o}=\dfrac{-d_i}{d_o}=3,$$ $$d_i=-3d_o.$$

Let's wtrite the equation of thin lens:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$\dfrac{1}{d_o}+\dfrac{1}{-3d_o}=\dfrac{1}{f},$$ $$d_o=\dfrac{2}{3}f=\dfrac{2}{3}\cdot20\ cm=13.3\ cm.$$

Finally, we can find the image distance:

$$d_i=-3d_o=-3\cdot13.3\ cm=-40\ cm.$$

The sign minus means that the image is virtual.