In Young's double slit experiment, 8 th maximum with wavelength ' λ₁ ' is at a distance ' d₁ ' from the central maximum and 6 th maximum with wavelength ' λ₂ ' is at a distance ' d₂ '. Then (d₂)/(d₁) is (a) (3λ₁)/(4λ₂) (b) (3λ₂)/(4λ₁) (c) (4λ₁)/(3λ₂) (d) (4λ₂)/(3λ₁)

Answer: (b). y = (nλ D)/(d) d₁ = (8λ₁D)/(d), d₂ = (6λ₂D)/(d) (d₂)/(d₁) = (6λ₂)/(8λ₁) = (3λ₂)/(4λ₁), so the answer is (b).

In Young's double slit experiment, 8 th maximum with wavelength ' λ₁… | MHT CET 2023 (14 May, Shift 1)

Q32MHT CET 2023 · 14 MAY · SHIFT 1Easy

YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment, $8$ th maximum with wavelength ' $\lambda_1$ ' is at a distance ' $d_1$ ' from the central maximum and $6$ th maximum with wavelength ' $\lambda_2$ ' is at a distance ' $d_2$ '. Then $\frac{d_2}{d_1}$ is

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Answer: (b)

$y = \frac{n\lambda D}{d}$
$d_1 = \frac{8\lambda_1D}{d}$ , $d_2 = \frac{6\lambda_2D}{d}$
$\frac{d_2}{d_1} = \frac{6\lambda_2}{8\lambda_1} = \frac{3\lambda_2}{4\lambda_1}$ , so the answer is (b).

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YDSE: Fringe Width & Fringe Positions question from MHT CET 2023 (14 May, Shift 1)

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