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

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(B) The distance of the $n^{th}$ bright fringe from the central maxima is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 3$,$\lambda = 6000\ \mathring A = 6000 \times 10^{-10}\ m$,$D = 1.0\ m$,and $d = 0.5\ mm = 0.5 \times 10^{-3}\ m$.Substituting these values into the formula:$x_3 = \frac{3 \times 6000 \times 10^{-10} \times 1.0}{0.5 \times 10^{-3}}$$x_3 = \frac{18000 \times 10^{-10}}{0.5 \times 10^{-3}}$$x_3 = 36000 \times 10^{-7}\ m = 3.6 \times 10^{-3}\ m = 3.6\ mm$.Therefore,the separation is $3.6\ mm$.

Class 12 Physics Wave Optics Young's Double Slit Experiment (YDSE)

Medium

In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.

A
$4$
B
$3.6$
C
$3$
D
$2.5$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

Two coherent light sources $S_1$ and $S_2$ $(\lambda = 6000 \ \mathring{A})$ are $1 \, mm$ apart from each other. The screen is placed at a distance of $25 \, cm$ from the sources. The width of the fringes on the screen should be (in $, cm$)

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In Young's double slit experiment,a mica sheet of thickness $t$ and refractive index $\mu$ is introduced in the path of the ray from the first source $S_1$. By what distance will the fringe pattern be displaced?

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Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

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In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

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The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

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Two coherent light sources $S_1$ and $S_2$ $(\lambda = 6000 \ \mathring{A})$ are $1 \, mm$ apart from each other. The screen is placed at a distance of $25 \, cm$ from the sources. The width of the fringes on the screen should be (in $, cm$)

In Young's double slit experiment,a mica sheet of thickness $t$ and refractive index $\mu$ is introduced in the path of the ray from the first source $S_1$. By what distance will the fringe pattern be displaced?

Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

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