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Mechanical Properties of Solids
Question
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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

  • infinite

  • five

  • three

  • zero

Answer

B.

five

For possible interference maxima on the screen, the condition is

d sinθ = n λ

Given:- d = slit - width = 2 λ

∴  2 λ sinθ = n λ

⇒ 2 sinθ = n

The maximum value of sinθ is 1, hence,

n = 2 × 1 = 2

Thus, Eq. (i) must be satisfied by 5 integer values ie, -2, -1, 0, 1, 2. Hence, the maximum
number of possible interference maxima is 5.

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