| ⇦ |
| ⇨ |
When an object is placed 40cm from a diverging lens, its virtual image is formed 20 cm from the lens. The focal length and power of lens are
Options
(a) F=-20 cm, P=-5 D
(b) F=-40 cm, P=-5 D
(c) F=-40 cm, P=-2.5 D
(d) F=-20 cm, P=-2.5 D
Correct Answer:
F=-40 cm, P=-2.5 D
Explanation:
We have, 1 / f = (1/v) – (1/u) ⇒ 1 / f = (1/-20) – (-1/40) = (-1 + 1) / 40 = 1 / 40
f = – 40 cm
Power of the lens, P = – (200 / 0.40) = – 2.5 D
Related Questions: - For a particle in a non-uniform accelerated circular motion correct statement is
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
- In a nuclear reactor, the number of U²³⁵ nuclei undergoing fissions per second
- In Young’s double slit experiment, the wavelength of the light used is doubled
- Figure below shows two paths that may be taken by a gas to go from a state A
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For a particle in a non-uniform accelerated circular motion correct statement is
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
- In a nuclear reactor, the number of U²³⁵ nuclei undergoing fissions per second
- In Young’s double slit experiment, the wavelength of the light used is doubled
- Figure below shows two paths that may be taken by a gas to go from a state A
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply