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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 resistance in the two arms of the meter bridge are 5Ω and RΩ, respectively
- A body of mass 3 kg acted upon by a constant force is displaced by S meter
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- An electric charge 10⁻³ μC is placed at the origin (0,0) of X-Y co-ordinate system.
- The masses of three wires of copper are in the ratio 1:3:5 and lengths
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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