In a biprism experiment, monochromatic light of wavelength ' λ ' is used. The distance between 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 Z₁, Z₂, Z₃ and Z₄ then (a) Z₁D₁ = Z₂D₂ = Z₃D₃ = Z₄D₄ (b) (Z₁)/(D₁) = (Z₂)/(D₂) = (Z₃)/(D₃) = (Z₄)/(D₄) (c) D₁√(Z₁) = D₂√(Z₂) = D₃√(Z₃) = D₄√(Z₄) (d) Z₁√(D₁) = Z₂√(D₂) = Z₃√(D₃) = Z₄√(D₄)

Answer: (b). Z = (λ D)/(d) with λ, d fixed, so Z ∝ D. Hence Z/D = (λ)/(d) is constant. So the answer is (b) (Z₁)/(D₁) = (Z₂)/(D₂) = (Z₃)/(D₃) = (Z₄)/(D₄).

In a biprism experiment, monochromatic light of wavelength ' λ ' is… | MHT CET 2023 (13 May, Shift 1)

Q17MHT CET 2023 · 13 MAY · SHIFT 1Easy

Biprism, Thin Films & Other Devices

In a biprism experiment, monochromatic light of wavelength ' $\lambda$ ' is used. The distance between 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 $Z_1, Z_2, Z_3$ and $Z_4$ then

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Answer: (b)

$Z = \frac{\lambda D}{d}$ with $\lambda$ , $d$ fixed, so $Z \propto D$ .
Hence $\frac{Z}{D} = \frac{\lambda}{d}$ is constant.
So the answer is (b) $\frac{Z_1}{D_1} = \frac{Z_2}{D_2} = \frac{Z_3}{D_3} = \frac{Z_4}{D_4}$ .

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