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प्रश्न
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.
पर्याय
three
five
infinite
zero
MCQ
रिकाम्या जागा भरा
उत्तर
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is five.
Explanation:
For constructive interference path difference (As sin θ ≤ 1)
d sin θ = nλ
Given d = 2λ
∴ 2λ sin θ = nλ ⇒ sin θ = `"n"/2`
n = 0, 1, -1, 2, -2 hence five maxima are possible.
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