Careers360 Logo
JCECE 2025 - Date, Registration, Seat Allotment, Choice Filling, Merit List, Counselling, Cutoff

Power Of Lens And Mirror - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Power of lens and mirror is considered one of the most asked concept.

  • 11 Questions around this concept.

Solve by difficulty

A convex lens and a concave lens, each having the same focal length of \mathrm{25\ cm} are put in contact to form a combination of lenses. The power in diopters of the combination is:

Two lenses are placed in contact with each other and the focal length of the combination is \mathrm{80\ cm}. If the focal length of one is \mathrm{20\ cm}, then the power of the other will be:

An achromatic convergent doublet of two lenses in contact has a power of \mathrm{+ 2\ D}. The convex lens has power \mathrm{+ 5\ D}. What is the ratio of the dispersive powers of the convergent and divergent lenses?

The nearer point of the hypermetropic eye is \mathrm{40\ cm}. The lens to be used for its correction should have the power:

Concepts Covered - 1

Power of lens and mirror

Power and lens and mirror-

Power of a lens is defined as the reciprocal of focal length and Power of a mirror is defined as the negative of reciprocal of focal length.

                                                        So, 

                                                                P_{Lens}=\frac{1}{f} \ \ \ and \ \ \ P_{Mirror} = \frac{-1}{f}  

                                                                                  Where, f = focal length

The unit of power is Diopter, 1D = 1 m-1

The sign of the focal length will be taken according to the sign convention which we have discussed in the previous concept. 

The power of a combination of lenses in contact is the algebraic sum of the powers of individual lenses, so it can be written as -

                                                             P_{Total } = P_{I}+P_{2}+P_{3}+\dots+P_{n}

 

Study it with Videos

Power of lens and mirror

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top