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A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperature of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:
Options
(a) f and I/4
(b) 3f/4 and I/2
(c) f and 3I/4
(d) f/2 and I/2
Correct Answer:
f and 3I/4
Explanation:
By covering aperture, focal length does not change. But intensity is reduced by 1/4 times, as aperture diameter d/2 is covered.
I’ = I – I / 4 = 3I / 4
New focal length = f and intensity = 3I / 4
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Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A long straight wire carries a certain current and produces a magnetic field
- The radius of the convex surface of a plano-convex lens is 20 cm and the refractive index
- The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at time
- In common base mode of a transistor, the collector current is 5.488 mA
- One milligram of matter on convertion into energy will give
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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