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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
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Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The equations of motion of a projectile are given by x=36 t metre and 2y=96t-9.8t²
- The ionisation energy of an electron in the ground state of helium atom is 24.6 eV.
- The electric field in a certain region is given by E=5i ̂-3ĵ kV/m. The potential
- A short linear object of length L lies on the axis of a spherical mirror of focal
- At what height h above earth, the value of g becomes g/2 (where R=radius of earth)?
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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