| ⇦ | 
| ⇨ |
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 200 W sodium street lamp emits yellow light of wavelength 0.6 µm
- Imagine a light planet revolving around a very massive star in a circular
- A uniform electric field and uniform magnetic field are acting along the same
- Vessel A is filled with hydrogen while vessel B, whose volume is
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A 200 W sodium street lamp emits yellow light of wavelength 0.6 µm
- Imagine a light planet revolving around a very massive star in a circular
- A uniform electric field and uniform magnetic field are acting along the same
- Vessel A is filled with hydrogen while vessel B, whose volume is
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
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