| ⇦ | 
| ⇨ |
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: - A black hole is an object whose gravitational field is so strong that even light
- An explosion blows a rock into three parts. Two parts go off at right angles to each other
- Two capacitors of capacitances 2 μF and 3 μF are joined in series.
- Identify the paramagnetic substance
- In a series resonant LCR circuit, the voltage across R is 100 volts and R =1kΩ
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A black hole is an object whose gravitational field is so strong that even light
- An explosion blows a rock into three parts. Two parts go off at right angles to each other
- Two capacitors of capacitances 2 μF and 3 μF are joined in series.
- Identify the paramagnetic substance
- In a series resonant LCR circuit, the voltage across R is 100 volts and R =1kΩ
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