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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
- The order of distance of an electron revolving in an atom from nucleus is
- A very small circular loop of radius a is initially (at t=0) coplanar and concentric
- A radiotransmitter radiates 1 kW power at a wavelength 198.6 m. How many photons
- A body of mass 10 kg is moving on an inclined plane of inclination 30°with an acceleration
- A body of mass m is thrown upwards at an angle θ with the horizontal with velocity
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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