(a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from:
(i) a diverging lens of focal length 40 cm.
(ii) a converging lens of focal length 40 cm.
(b) Draw labelled ray diagrams to show the formation of images in cases (i) and (ii) above (The diagrams may not be according to scale).
(a) \(h_1\) = 2 cm
u = -20 cm
(i) f = -40 cm (Diverging lens)
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-20}\) = \(\frac{1}{-40}\)
\(\frac{1}{v}\) = -\(\frac{1}{40}\) - \(\frac{1}{20}\)
\(\frac{1}{v}\) = \(\frac{-3}{40}\)
v = -13.33 cm
m = \(\frac{v}{u}\) = \(\frac{h_2}{h_1}\)
\(\frac{-13.33}{-20}\) = \(\frac{h_2}{2}\)
\(h_2\) = 1.33 cm
(ii) f = 40 cm (Diverging lens)
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) - \(\frac{1}{-20}\) = \(\frac{1}{40}\)
\(\frac{1}{v}\) = \(\frac{1}{40}\) - \(\frac{1}{20}\)
\(\frac{1}{v}\) = \(\frac{-1}{40}\)
v = -40 cm
m = \(\frac{v}{u}\) = \(\frac{h_2}{h_1}\)
\(\frac{-40}{-20}\) = \(\frac{h_2}{2}\)
\(h_2\) = 4 cm
(b) Formation of image in case (i):