| ⇦ | 
| ⇨ |
In Young’s double slit experiment, the fringe width is 1×10⁻⁴ m, if the distance between the slit and screen is doubled and the distance between two slits is reduced to half and the wavelength is changed from 6.4×10⁻⁷ m to 4.0×10⁻⁷ m, then the value of new fringe width will be
Options
(a) 2.5×10⁻⁴ m
(b) 2.0×10⁻⁴ m
(c) 1×10⁻⁴ m
(d) 0.5×10⁻⁴ m
Correct Answer:
2.5×10⁻⁴ m
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The potential energy of a simple harmonic oscillator when the particle is half way
- In L-C-R series circuit, an alternating emf e and current i are given by the equations
- If Q, E and W denote respectively the heat added, change in internal energy
- If two charges +4e and +e are at a distance x apart, then at what distance charge q
- The frequency of fundamental tone in an open organ pipe of length
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The potential energy of a simple harmonic oscillator when the particle is half way
- In L-C-R series circuit, an alternating emf e and current i are given by the equations
- If Q, E and W denote respectively the heat added, change in internal energy
- If two charges +4e and +e are at a distance x apart, then at what distance charge q
- The frequency of fundamental tone in an open organ pipe of length
Topics: Wave Optics (101)
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