In Young's double slit experiment, $8^{ ext {th }}$ maximum with wavelength ' $\lambda_1$ ' is at a distance ' $\mathrm{d}_1$ ' from the central maximum and $6^{ ext {th }}$ maximum with wavelength ' $\lambda_2$ ' is at a distance ' $\mathrm{d}_2$ '. Then $\frac{\mathrm{d}_2}{\mathrm{~d}_1}$ is

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In Young's double slit experiment, $8^{	ext {th }}$ maximum with wavelength ' $\lambda_1$ ' is at a distance ' $\mathrm{d}_1$ ' from the central maximum and $6^{	ext {th }}$ maximum with wavelength ' $\lambda_2$ ' is at a distance ' $\mathrm{d}_2$ '. Then $\frac{\mathrm{d}_2}{\mathrm{~d}_1}$ is

Aicharya · Accepted Answer

$\begin{array}{ll} & \mathrm{d} \propto \mathrm{n} \lambda \ 	herefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{n}_2 \lambda_2}{\mathrm{n}_1 \lambda_1}=\frac{6 \lambda_2}{8 \lambda_1} \ 	herefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{3 \lambda_2}{4 \lambda_1}\end{array}$

In Young's double slit experiment, $8^{\text {th }}$ maximum with wavelength ' $\lambda_1$ ' is at a…

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