| ⇦ | 
| ⇨ |
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
Related Questions: - The moon’s radius is 1/ 4 that of the earth and its mass is 1 / 80 times that of the earth
- Potentiometer measures the potential difference more accurately than a voltmeter
- Which of the following phenomenon exhibited the particle nature of light?
- Pascal-second has the dimensions of
- A current I ampere flows along the inner conductor of a co-axial cable and returns
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moon’s radius is 1/ 4 that of the earth and its mass is 1 / 80 times that of the earth
- Potentiometer measures the potential difference more accurately than a voltmeter
- Which of the following phenomenon exhibited the particle nature of light?
- Pascal-second has the dimensions of
- A current I ampere flows along the inner conductor of a co-axial cable and returns
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply