In an interference experiment, the mth bright fringe for light of wavelength λ₁ coincides with the nth dark fringe for light of wavelength λ₂. The ratio (λ₂)/(λ₁) is (a) (m)/(n - 1) (b) (m)/((2n - 1)) (c) (2m)/((2n - 1)) (d) (2m)/((2n + 1))

Answer: (c). m th bright: y = (mλ₁ D)/(d); n th dark: y = ((2n-1)λ₂ D)/(2d). mλ₁ = ((2n-1)λ₂)/(2) ⇒ (λ₂)/(λ₁) = (2m)/(2n-1) So the answer is (c) (2m)/((2n - 1)).

In an interference experiment, the mth bright fringe for light of… | MHT CET 2026 (20 Apr, Shift 1)

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YDSE: Film, Medium & Two Wavelengths

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$ . The ratio $\frac{\lambda_2}{\lambda_1}$ is

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Answer: (c)

$m$ th bright: $y = \dfrac{m\lambda_1 D}{d}$ ; $n$ th dark: $y = \dfrac{(2n-1)\lambda_2 D}{2d}$ .
$m\lambda_1 = \dfrac{(2n-1)\lambda_2}{2} \Rightarrow \dfrac{\lambda_2}{\lambda_1} = \dfrac{2m}{2n-1}$
So the answer is (c) $\frac{2m}{(2n - 1)}$ .

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