The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is: - Physics MCQ AIEEE 2004 2026 | ExamTest.live | ExamTest.live

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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is:

A $Infinite$
B $Five$
C $Three$
D $Zero$

Solution & Step-by-step Explanation

The condition for interference maxima is d sin θ = nλ .Given d = 2 λ : 2 λ sin θ = nλ ⟹ sin θ = \frac{n}{2} Since the maximum value of ∣ sin θ ∣ is 1 : \frac{n}{2} \leq 1 ⟹ ∣ n ∣ \leq 2 The possible integer values for n are {- 2, - 1, 0, 1, 2} .Total number of maxima = 5 .

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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is:

A

Infinite

B

Five

C

Three

D

Zero

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