In an interference experiment, the n th bright fringe for light of wavelength λ₁ (n = 0, 1, 2, 3) coincides with the m th dark fringe for light of wavelength λ₂ (m = 1, 2, 3). The ratio (λ₁)/(λ₂) is (a) (m - 1)/(n) (b) (2m - 1)/(n) (c) (2m - 1)/(2n) (d) (2m + 1)/(2n)

Answer: (c). n th bright: y = (nλ₁ D)/(d); m th dark: y = ((2m-1)λ₂ D)/(2d). nλ₁ = ((2m-1)λ₂)/(2) ⇒ (λ₁)/(λ₂) = (2m-1)/(2n) So the answer is (c) (2m - 1)/(2n).

In an interference experiment, the n th bright fringe for light of… | MHT CET 2024 (3 May, Shift 1)

Q34MHT CET 2024 · 3 MAY · SHIFT 1Medium

YDSE: Film, Medium & Two Wavelengths

In an interference experiment, the $n$ th bright fringe for light of wavelength $\lambda_1 (n = 0, 1, 2, 3 \ldots)$ coincides with the $m$ th dark fringe for light of wavelength $\lambda_2 (m = 1, 2, 3 \ldots)$ . The ratio $\frac{\lambda_1}{\lambda_2}$ is

Show answer and solution

Answer: (c)

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

Report a mistake

YDSE: Film, Medium & Two Wavelengths question from MHT CET 2024 (3 May, Shift 1)

More questions on YDSE: Film, Medium & Two Wavelengths