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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 comb run through one’s dry hair attracts small bits of paper. This is due to
- The ratio of longest wavelength corresponding to Lyman and Blamer series
- Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
- The half life of radium is 1620 yr and its atomic weight is 226 kg per kilo mol.
- A thin rod of length L and mass M is bent at its midpoint into two halves
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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