A convex lens of focal length ${f_1}$ is kept in contact with a concave lens of focal length ${f_2}$. Find the focal length of the combination.

Hint:
Focal length of combination can be found by power formula of the combination as given –
${P_{eq}} = {P_1} + {P_2} + {P_3} + ......$

Complete step by step solution: Let the focal length of convex lens is ${f_1}$ and focal length of concave lens is ${f_2}$ using sign convention ${f_1}$ will be positive and ${f_2}$will be negative.
Now we can write the power of lens P as $P = \dfrac{1}{f}$, $f$ will be with sign.
Now for the combination
${P_{eq}} = {P_1} + {P_2}$
\[ \Rightarrow \dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{( - {f_2})}}\]
\[ \Rightarrow \dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{{f_1}}} - \dfrac{1}{{{f_2}}}\]
\[ \Rightarrow {f_{eq}} = \dfrac{{{f_1}{f_2}}}{{{f_2} - {f_1}}}\]

Note:
A. The formula ${P_{eq}} = {P_1} + {P_2} + {P_3} + ......$ can be used for both lens and mirror combination
B. ( for mirror $P = - \dfrac{1}{f}$ and for lens $P = \dfrac{1}{f}$
for both mirror and lens $f$ should be placed with sign
C. The number of terms ${P_1},{P_2},{P_3},.......$ are considered according as the number of reflection and refraction through the combination.
D. Formula ${P_{eq}} = {P_1} + {P_2} + {P_3} + ......$ can be used for both combination of lenses and Mirrors