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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 wheel of mass 10 kg has a moment of inertia of 10 kgm² about its own axis
- Two cells when connected in series are balanced on 8m on a potentiometer.
- The electric potential V at any point (x,y,z) all in meters in space is given by
- Initial angular velocity of a circular disc of mass M is ?￢ﾂﾁ.Then, two small spheres
- Steam at 100⁰C is passed into 20 g of water at 10°C. When water acquires a temperature
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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