In Young's double slit experiment, the ' nth ' maximum of wavelength… | MHT CET 2021 (20 Sep, Shift 1)

Q50MHT CET 2021 · 20 SEP · SHIFT 1Easy

YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment, the ' $n^{\text{th}}$ ' maximum of wavelength ' $\lambda_1$ ' is at a distance ' $y_1$ ' from the central maximum. When the wavelength of the source is changed to ' $\lambda_2$ ', $\left(\frac{n}{2}\right)$ th maximum is at a distance of ' $y_2$ ' from its central maximum. The ratio $\frac{y_1}{y_2}$ is

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Answer: (b)

Bright fringe: $y_n = \frac{n\lambda D}{d}$
$y_1 = \frac{n\lambda_1 D}{d}$ , $y_2 = \frac{(n/2)\lambda_2 D}{d}$
$\frac{y_1}{y_2} = \frac{n\lambda_1}{(n/2)\lambda_2} = \frac{2\lambda_1}{\lambda_2}$
So the answer is (b) $\frac{2\lambda_1}{\lambda_2}$ .

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YDSE: Fringe Width & Fringe Positions question from MHT CET 2021 (20 Sep, Shift 1)

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