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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
- The valence band and the conductance band of a solid overlap at low temperature
- A solid cylinder of mass 50kg and radius 0.5 m is, free to rotate about the horizontal axis
- Two identical cells of the same e.m.f. and same internal resistance
- A round disc of moment of inertia i₂ about its axis perpendicular to its plane
- Two charges each equal to 2µ C are 0.5m apart. If both of them exist inside vacuum
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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