In a biprism experiment, monochromatic light of wavelength ' λ ' is used. The distance between the two coherent sources ' d ' is kept constant. If the distance between slit and eyepiece ' D ' is varied as D₁, D₂, D₃, D₄ and corresponding measured fringe widths are W₁, W₂, W₃, W₄ then (a) W₁D₁ = W₂D₂ = W₃D₃ = W₄D₄ (b) (W₁)/(D₁) = (W₂)/(D₂) = (W₃)/(D₃) = (W₄)/(D₄) (c) W₁√(D₁) = W₂√(D₂) = W₃√(D₃) = W₄√(D₄) (d) D₁√(W₁) = D₂√(W₂) = D₃√(W₃) = D₄√(W₄)

Answer: (b). W = (λ D)/(d) with λ, d fixed, so W ∝ D. Hence W/D = (λ)/(d) is constant. So the answer is (b) (W₁)/(D₁) = (W₂)/(D₂) = (W₃)/(D₃) = (W₄)/(D₄).

In a biprism experiment, monochromatic light of wavelength ' λ ' is… | MHT CET 2024 (4 May, Shift 2)

Q14MHT CET 2024 · 4 MAY · SHIFT 2Easy

Biprism, Thin Films & Other Devices

In a biprism experiment, monochromatic light of wavelength ' $\lambda$ ' is used. The distance between the two coherent sources ' $d$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $D_1, D_2, D_3, D_4$ and corresponding measured fringe widths are $W_1,\ W_2,\ W_3,\ W_4$ then

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Answer: (b)

$W = \frac{\lambda D}{d}$ with $\lambda$ , $d$ fixed, so $W \propto D$ .
Hence $\frac{W}{D} = \frac{\lambda}{d}$ is constant.
So the answer is (b) $\frac{W_1}{D_1} = \frac{W_2}{D_2} = \frac{W_3}{D_3} = \frac{W_4}{D_4}$ .

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