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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 monoatomic gas at a pressure P, having a volume V expands isothermally
- The internal energy change in a system that has absorbed 2 kcals of heat
- A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg.
- A body of mass 0.25 kg is projected with muzzle velocity 100 m/s from a tank
- A gas is suddenly expanded such that its final volume becomes 3 times
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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