Question

The power of a combination of two lenses A and B is 5D.The focal length of A is 15cm. What is the focal length of B?
$
  {\text{A}}{\text{. }} + 15cm \\
  {\text{B}}{\text{. }} - 20cm \\
  {\text{C}}{\text{. }} - 60cm \\
  {\text{D}}{\text{. }} + 3cm \\
 $

Answer 1

Hint: Here we go through by the formula of finding the power of lenses that is equal to $\dfrac{1}{f}$ and we know that the formula $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$ where ${f_1}$ and ${f_2}$ are the focal length of the two lenses.
Formula used: $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$

Complete Step-by-Step solution:
Here in the question it is given that the power of a combination of two lenses A and B is 5D.
As we know Power of a lens is defined as the inverse of focal length (in meters) of the lens. It is a measure of the amount of deviation of light ray produced by a lens, more the power, more is the deviation. Unit of power is the Diopter (D).
As we know when two lens are joined the formula for finding the power is $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Where the total power $\dfrac{1}{f} = 5D$ given in the question.
And by the definition of the power of lens we know that $P = \dfrac{1}{f}$ i.e. 1D where f=1m. The units of dioptre is 1/m.
Here in the question the focal length of A is given as 15cm let it as ${f_1}$ .
$\therefore {f_1} = 15cm$
As we know 1D=1/m, so we can say 5D=5/m.
And now we convert the units meter into centimeter for solving the question
$\therefore P = \dfrac{1}{f} = 5D = \dfrac{5}{{100cm}}$
And now put these value in the formula $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
$
   \Rightarrow \dfrac{5}{{100cm}} = \dfrac{1}{{15cm}} + \dfrac{1}{{{f_2}}} \\
   \Rightarrow \dfrac{1}{{{f_2}}} = \dfrac{5}{{100cm}} - \dfrac{1}{{15cm}} \\
   \Rightarrow \dfrac{1}{{{f_2}}} = \dfrac{{15 - 20}}{{300cm}} = \dfrac{{ - 5}}{{300cm}} \\
  \therefore {f_2} = - 60cm \\
 $
Therefore, the focal length of B is -60 cm.
Hence option C is the correct answer.

Note: - Whenever we face such a type of question the key concept for solving the question is to put the value of focal lengths in the formula to find out the unknown focal length and also keep in mind that convex lens has positive power and concave lens has negative power. Power of a plane glass plate is 0. Power of combinations of lenses kept close to each other is equal to the sum of individual powers of each lens.