| ⇦ | 
| ⇨ |
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: - The dimensions of latent heat are
- An electron in hydrogen atom makes a transition n₁ → n₂
- Which of the following is true for elastic potential energy density?
- A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage
- A boy playing on the roof of a 10m high building throws a ball with a speed
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dimensions of latent heat are
- An electron in hydrogen atom makes a transition n₁ → n₂
- Which of the following is true for elastic potential energy density?
- A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage
- A boy playing on the roof of a 10m high building throws a ball with a speed
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