Question

A converging and a diverging lens of equal focal lengths are placed coaxially in contact. Find the power and focal length of the combination.

Answer 1

To answer this question, we first need to know the general formula of focal length when two or more than two lenses are placed coaxially $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$.

Complete answer:
Converging lens - The converging lens is a type of lens that converges rays of light that are parallel to its principal axis.
Diverging lens - The diverging lens is a type of lens that diverges rays of light that are parallel to its principal axis.
When two lenses are placed in contact with each other then the net focal length is given by,
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$.
Taking LCM ,
$\dfrac{1}{f} = \dfrac{{{f_1} + {f_2}}}{{{f_1}{f_2}}}$
Therefore $f = \dfrac{{{f_1}{f_2}}}{{{f_1} + {f_2}}}$
As given ${f_1} = - {f_2}$(negative sign is due to the converging and diverging lens)
Therefore ${f_1} + {f_2} = 0$.
So, $f = \dfrac{{ - {f_1}{f_2}}}{0}$.
$f = \infty$ (As (constant/0) is infinity).
And also $P = \dfrac{1}{f}$.
Hence, $P = 0$.
So, the final Power is 0 and the focal length is $\infty$.

Note: When two or more lenses are held in contact, the combined lens's resultant power is equal to the algebraic number of the individual powers. The focal length of a convex lens is positive, while the focal length of a concave lens is negative.