| ⇦ | 
| ⇨ |
In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain 5th bright firnge at the same point?
Options
(a) 500 nm
(b) 630 nm
(c) 750 nm
(d) 420 nm
Correct Answer:
420 nm
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Which one of the following is a vector?
- We get Balmer series when electron comes from
- A battery having e.m.f. 4 volt and internal resistance 0.5 Ω is connected
- A p-n photodiode is made of a material with a band gap of 2.0 eV
- Ice is wrapped in black and white cloth. In which cases more ice will melt
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following is a vector?
- We get Balmer series when electron comes from
- A battery having e.m.f. 4 volt and internal resistance 0.5 Ω is connected
- A p-n photodiode is made of a material with a band gap of 2.0 eV
- Ice is wrapped in black and white cloth. In which cases more ice will melt
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