| ⇦ |
| ⇨ |
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 moment of inertia of a semicircular ring about the centre is
- When you make ice cubes, the entropy of water
- The breaking stress of a wire depends upon
- The conductivity in the intrinsic semiconductor does not depend on
- The force of repulsion between two electrons at a certain distance is F.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a semicircular ring about the centre is
- When you make ice cubes, the entropy of water
- The breaking stress of a wire depends upon
- The conductivity in the intrinsic semiconductor does not depend on
- The force of repulsion between two electrons at a certain distance is F.
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