| ⇦ | 
| ⇨ |
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: - A body of mass 0.25 kg is projected with muzzle velocity 100 m/s from a tank
- The displacement of particle is given by x=aₒ+(a₁t/2)-(a₂t²/3) what is acceleration?
- Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen
- In Young’s experiment, the third bright band for light of wavelength 6000Å
- Light is emitted when
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass 0.25 kg is projected with muzzle velocity 100 m/s from a tank
- The displacement of particle is given by x=aₒ+(a₁t/2)-(a₂t²/3) what is acceleration?
- Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen
- In Young’s experiment, the third bright band for light of wavelength 6000Å
- Light is emitted when
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