In Young's double-slit experiment,the distanc | vedclass

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(B) The distance of the $n^{th}$ bright fringe from the central fringe is given by the formula: $x_n = n \frac{\lambda D}{d}$.Given values are:$n = 3$

Vedclass · Accepted Answer

$1.5$

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(B) The distance of the $n^{th}$ bright fringe from the central fringe is given by the formula: $x_n = n \frac{\lambda D}{d}$.Given values are:$n = 3$$\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$$D = 200 \, cm = 2 \, m$$d = 0.2 \, mm = 0.2 	imes 10^{-3} \, m = 2 	imes 10^{-4} \, m$Substituting these values into the formula:$x_3 = 3 	imes \frac{(5 	imes 10^{-7} \, m) 	imes (2 \, m)}{2 	imes 10^{-4} \, m}$$x_3 = 3 	imes \frac{10 	imes 10^{-7}}{2 	imes 10^{-4}}$$x_3 = 3 	imes 5 	imes 10^{-3} \, m$$x_3 = 15 	imes 10^{-3} \, m = 1.5 	imes 10^{-2} \, m = 1.5 \, cm$.Thus,the distance is $1.5 \, cm$.

In Young's double-slit experiment,the distance between the two slits is $0.2 \, mm$ and the screen is $200 \, cm$ away from the slits. If the wavelength of light used is $5000 \, \mathring{A}$,find the distance of the third bright fringe from the central bright fringe in $cm$.

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Two slits are separated by a distance of $0.5\, mm$ and illuminated with light of $\lambda = 6000\ \mathring A$. If the screen is placed $2.5\, m$ from the slits,the distance of the third bright fringe from the centre will be........$mm$.

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