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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.
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Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A copper disc of radius 0.1 m is rotated about its centre, with 10 revolutions
- If R is the radius of earth, the height at which the weight of a body 1/4 of its weight
- The intensity of magnetisation of a bar magnet is 5×10⁴ Am⁻¹. The magnetic length
- A body of mass 120 kg and density 600 kg/m³ floats in water.
- Tangent galvanometer is used to measure
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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