| ⇦ | 
| ⇨ |
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: - An unpolarised beam of intensity I₀ falls on a polaroid. The intensity
- Two identical thin plano-convex glass lenses (refractive index 1.5) each having
- In an orbital motion, the angular momentum vector is
- A nuclear reaction is given as 4 ₁H¹→₂He⁴+₀e¹+energy Mention the type of reaction.
- A wire having resistance 12Ω is bent in the form of an equilateral triangle.
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An unpolarised beam of intensity I₀ falls on a polaroid. The intensity
- Two identical thin plano-convex glass lenses (refractive index 1.5) each having
- In an orbital motion, the angular momentum vector is
- A nuclear reaction is given as 4 ₁H¹→₂He⁴+₀e¹+energy Mention the type of reaction.
- A wire having resistance 12Ω is bent in the form of an equilateral triangle.
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