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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: - Two identical capacitors are first connected in series and then in parallel.
- A charged particle with a velocity 2×10³ ms⁻¹ passes undeflected through electric field
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Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two identical capacitors are first connected in series and then in parallel.
- A charged particle with a velocity 2×10³ ms⁻¹ passes undeflected through electric field
- When a mass m is attached to a spring, it normally extends by 0.2 m.
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- Magnetism of a magnet is due to
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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