| ⇦ | 
| ⇨ |
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: - In a uniform circular motion,
- The Rutherford scattering experiment proves that an atom consists of
- ln a surrounding medium of temperature 10⁰C a body takes 7 min for a fall
- A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum
- A coil of N turns and mean cross-sectional area A is rotating with uniform angular
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a uniform circular motion,
- The Rutherford scattering experiment proves that an atom consists of
- ln a surrounding medium of temperature 10⁰C a body takes 7 min for a fall
- A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum
- A coil of N turns and mean cross-sectional area A is rotating with uniform angular
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