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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: - A certain metallic surface is illuminated with monochromatic light of wavelength
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Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A certain metallic surface is illuminated with monochromatic light of wavelength
- In the series resonance the value of impedence is 3 Ω and the resistance is 4Ω,
- In the elastic collision of objects
- A circular disc X of radius R is made from an iron plate of thickness t and another disc
- The moon’s radius is 1/ 4 that of the earth and its mass is 1 / 80 times that of the earth
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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