| ⇦ | 
| ⇨ |
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:
- Which of the following is a vector quantity?
- In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected
- Pick out the correct statements from the following
- The input resistance of a silicon transistor is 100 W. Base current is changed
- A projectile is fired at 30°. Neglecting friction, the change in kinetic energy when
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