| ⇦ | 
| ⇨ |
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: - A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of 10⁻⁵ Wb.
- A solid cylinder of mass M and radius R rolls without slipping down and inclined plane
- Masses of three wires of copper are in the ratio of 1:3:5 and their lengths
- The most important characteristic of electron in the production of X-rays is
- If alternating source of primary coil on transformer is replaced by a Laclanche cell
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of 10⁻⁵ Wb.
- A solid cylinder of mass M and radius R rolls without slipping down and inclined plane
- Masses of three wires of copper are in the ratio of 1:3:5 and their lengths
- The most important characteristic of electron in the production of X-rays is
- If alternating source of primary coil on transformer is replaced by a Laclanche cell
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