The fringe widths are found to be omega_1 and | vedclass

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Question

(B) The fringe width $\omega$ in a Young's double slit experiment ($Y$.$D$.$S$.$E$.) when performed in a medium of refractive index $n$ is given by th

Vedclass · Accepted Answer

$\omega_1 > \omega_2$ if $n_1 < n_2$

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(B) The fringe width $\omega$ in a Young's double slit experiment ($Y$.$D$.$S$.$E$.) when performed in a medium of refractive index $n$ is given by the formula: $\omega = \frac{D \lambda}{d n}$.In this expression,$D$ is the distance between the slits and the screen,$d$ is the distance between the slits,and $\lambda$ is the wavelength of light in vacuum.Since $D, d,$ and $\lambda$ are constant for a given experimental setup,the fringe width is inversely proportional to the refractive index of the medium: $\omega \propto \frac{1}{n}$.Therefore,if $n_1  \omega_2$. Similarly,if $n_1 > n_2$,then $\omega_1 Thus,the statement $\omega_1 > \omega_2$ is correct if $n_1 < n_2$.

The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is:

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