| ⇦ | 
| ⇨ |
The maximum number of possible interference maxima for slit-seperation equal to twice the wavelength, in young’s double slit experiment is
Options
(a) Infinite
(b) five
(c) three
(d) zero
Correct Answer:
five
Explanation:
For interference maxima, d sin θ = nλ
⇒ 2λ sin θ = nλ ⇒ sin θ = n / 2
sin θ can have values between 0 and ±1.
Hence n can be (-2, -1, 0, +1, +2) or five values.
The possible maxima are five.
Related Questions: - If h is the height of the capillary rise and r be the radius of capillary tube,
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
- Fusion reaction takes place at high temperature because
- A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface
- When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If h is the height of the capillary rise and r be the radius of capillary tube,
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
- Fusion reaction takes place at high temperature because
- A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface
- When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid
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