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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
- Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear
- A thin prism of angle 15⁰ made of glass of refractive index µ₁ = 1.5 is combined
- Which series of hydrogen spectrum corresponds to ultraviolet region?
- The radius of the first orbit of the hydrogen atom is a₀. The radius of the second orbit
- The molar specific heats of an ideal gas at constant pressure and volume are denoted
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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