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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
- 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas
- A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm
- If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is
- In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V
- A series L-C-R circuit contains inductance 5 mH, capacitance 2 μF and resistance
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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