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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 magnifying power of a telescope is 9. When it is adjusted for parallel rays
- A microscope is having objective of focal length 1 cm and eye-piece of focal length 6 cm.
- Work done by an external agent in separating the parallel plate capacitor is
- Rate of photoelectric emission is proportional to
- Two bodies A and B having temperature 327° C and 427° C are radiating heat
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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