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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: - The unit of reactance is
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Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The unit of reactance is
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- A long straight wire is carrying a current of 12 A. The magnetic field at a distance
- The drive shaft of an automobile rotates at 3600 rpm and transmits 80 HP
- Two parallel long wires carry currents i₁ and i₂ with i₁ > i₂ .
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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