Question

Two lenses of focal lengths $f_1$ and $f_2$ are placed side by side. What is the focal length of the composite lens?

• A. $f_1+f_2$
• B. $\frac{f_1-f_2}{f_1+f_2}$
• C. $\frac{f_1{f}_2}{f_1+f_2}$
• D. $\frac{f_1{f}_2}{f_1-f_2}$

Answer 1

The correct option is C $\frac{f_1{f}_2}{f_1+f_2}$

The net power of lenses of individual powers $P_1,P_2,P_3,$ and so on, placed in contact is given by,

Net power, $P=P_1+P_2+P_3$ ...

The power $P_1$ of the lens of focal length $f_1$ = $\frac{1}{f_1}$.

The power $P_2$ of the lens of focal length $f_2$ = $\frac{1}{f_2}$.

$P=P_1+P_2$ = $\frac{1}{f_1}+\frac{1}{f_2}$
Let $f$ be the focal length of the combination.

$\Rightarrow$ $\frac{1}{f}$ = $\frac{1}{f_1}$+ $\frac{1}{f_2}$

$\Rightarrow f=\frac{f_1{f}_2}{f_1+f_2}$