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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Light Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Light Waves

For an optical arrangement as depicted in the figure, the distance...

For an optical arrangement as depicted in the figure, the distance DD is significantly greater than the slit separation dd. Assume dd is set to the minimum value required to produce a dark fringe at point OO. Determine the fringe width of the interference pattern.
For an optical arrangement as depicted in the figure, the distance D is significantly greater than the slit separation

Options

  1. A2d2d
  2. B4d4d
  3. Cd2\dfrac{d}{2}
  4. Ddd

Correct answer

B. 4d4d

Step-by-step solution

Path difference at point OO is ΔxO=(S0S1+S1O)−(S0S2+S2O)\Delta x_O = (S_0 S_1 + S_1 O) - (S_0 S_2 + S_2 O)

For D≫dD \gg d, the distances are:

S0S1=S1O=D2+d2≈D+d22DS_0 S_1 = S_1 O = \sqrt{D^2 + d^2} \approx D + \dfrac{d^2}{2D}

S0S2=S2O=DS_0 S_2 = S_2 O = D

ΔxO=2(D+d22D)−2D=d2D\Delta x_O = 2\left(D + \dfrac{d^2}{2D}\right) - 2D = \dfrac{d^2}{D}

For a dark fringe at OO with minimum dd, ΔxO=λ2\Delta x_O = \dfrac{\lambda}{2}

d2D=λ2  ⟹  λ=2d2D\dfrac{d^2}{D} = \dfrac{\lambda}{2} \implies \lambda = \dfrac{2d^2}{D}

The fringe width β\beta on the screen for sources separated by dd at distance DD is β=λDd\beta = \dfrac{\lambda D}{d}

β=(2d2D)Dd=2d\beta = \left(\dfrac{2d^2}{D}\right) \dfrac{D}{d} = 2d
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