| ⇦ | 
| ⇨ |
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: - The ratio of minimum wavelengths of Lyman and Balmer series will be
- If the binding energy of the nucleon in ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06 MeV
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- The mean free path of molecules of a gas,(radius ‘r’) is inversely proportional to
- Monochromatic light of wavelength 667 nm is produced by a helium neon laser
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ratio of minimum wavelengths of Lyman and Balmer series will be
- If the binding energy of the nucleon in ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06 MeV
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- The mean free path of molecules of a gas,(radius ‘r’) is inversely proportional to
- Monochromatic light of wavelength 667 nm is produced by a helium neon laser
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