| ⇦ | 
| ⇨ |
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: - When light of wavelength 300 nm falls on a photoelectric emitter,
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- A galvanometer is converted into an ammeter by connecting
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- An electron of a stationary hydrogen atom passes from the fifth energy level to
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When light of wavelength 300 nm falls on a photoelectric emitter,
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- A galvanometer is converted into an ammeter by connecting
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- An electron of a stationary hydrogen atom passes from the fifth energy level to
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