In Young’s double slit experiment, the $8th$ maximum with wavelength $\lambda_{1}$ is at a distance $d_{1}$ from the central maximum and the $6th$ maximum with a wavelength $\lambda_{2}$ is at a distance $d_{2}$. Then $(d_{1}/d_{2})$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b) Position of $n^{th}$ maxima from central maxima is given by $x_{n} = \frac{n\lambda D}{d}$
$\Rightarrow x_{n} \propto n\lambda \Rightarrow \frac{d_{1}}{d_{2}} = \frac{n_{1}\lambda_{1}}{n_{2}\lambda_{2}} = \frac{8\lambda_{1}}{6\lambda_{2}} = \frac{4}{3}\left( \frac{\lambda_{1}}{\lambda_{2}} \right)$

Q. In Young’s double slit experiment, the $8th$ maximum with wavelength $\lambda_{1}$ is at a distance $d_{1}$ from the central maximum and the $6th$ maximum with a wavelength $\lambda_{2}$ is at a distance $d_{2}$. Then $(d_{1}/d_{2})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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