A convex lens and concave lens are placed as shown in figure. For conv

1.	A convex lens and concave lens are placed as shown in figure. For convex lens f = 10 cm for concave it is 5 cm1. Is it converging or diverging why?2. If f1 = 5 cm and f2 =10 cm What change will occur in the optical nature of system?
Answer» 1. \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\) \(\frac{1}{f} = \frac{+1}{10}+\frac{-1}{5}\) = \(\frac{5-10}{50}\) = \(\frac{-1}{10}\) f = -10 cm Effective focal length is negative. Hence this lens is diverging. 2. Effective focal length becomes positive Hence the lens will act as converging.

A convex lens and concave lens are placed as shown in figure. For convex lens f = 10 cm for concave it is 5 cm1. Is it converging or diverging why?2. If f1 = 5 cm and f2 =10 cm What change will occur in the optical nature of system?

Answer»

1. \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\)

\(\frac{1}{f} = \frac{+1}{10}+\frac{-1}{5}\)

= \(\frac{5-10}{50}\) = \(\frac{-1}{10}\)

f = -10 cm

Effective focal length is negative. Hence this lens is diverging.

2. Effective focal length becomes positive Hence the lens will act as converging.