The maxium number of possible interference maxima for slit – separation equal to twice the wavelength in Young’ s double slit experiment is
Text Solution
Verified by ExpertsThe correct answer is:
B
For possible interference maxima on the screen the condition is
d sin θ = n λ
Given : d = sit - whidth = 2 λ
∴ 2λ sin θ = n λ ⇒ 2 sin θ = n
The maximum value of sin θ is 1 hence,
n = 2 ×1 = 2
Thus, Eg. (i) must be satisfied by 5 integer values ie, – 2 – 1,– 1,2. Hence the maximum number of possible interference maxima is 5.
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