| ⇦ | 
| ⇨ |
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 process of adding impurities to the pure semiconductor is called
- CO₂ laser uses
- Magnetic flux φ (in weber) linked with a closed circuit of resistance 10 Ω varies
- A proton and an alpha particle both enter a region of uniform magnetic field B,
- If a small sphere is let fall vertically in a large quantity of still liquid of density
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The process of adding impurities to the pure semiconductor is called
- CO₂ laser uses
- Magnetic flux φ (in weber) linked with a closed circuit of resistance 10 Ω varies
- A proton and an alpha particle both enter a region of uniform magnetic field B,
- If a small sphere is let fall vertically in a large quantity of still liquid of density
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