| ⇦ | 
| ⇨ |
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: - A block of mass m1 rests on a horizontal table. A light string connected
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks
- A point charge q is placed at a distance x/2 directly above the centre of a cube
- The potential energy of a 1 kg particle free to move along the x-axis is given by
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A block of mass m1 rests on a horizontal table. A light string connected
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks
- A point charge q is placed at a distance x/2 directly above the centre of a cube
- The potential energy of a 1 kg particle free to move along the x-axis is given by
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
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