(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe (maximum) from the central maximum is given by the formula: $x_n = \f

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(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe (maximum) from the central maximum is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 4$ (for the fourth maximum),$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.Since $n$,$D$,and $d$ are constant for both experiments,the position $x$ is directly proportional to the wavelength $\lambda$ $(x \propto \lambda)$.Given $\lambda_{blue} = 4360 \, \mathring{A}$ and $\lambda_{green} = 5460 \, \mathring{A}$.Since $\lambda_{blue} < \lambda_{green}$,it follows that $X_{blue} < X_{green}$.

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

Medium

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

A
$X_{blue} > X_{green}$
B
$X_{blue} < X_{green}$
C
$X_{blue} = X_{green}$
D
$\frac{X_{blue}}{X_{green}} = \frac{5460}{4360}$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

$A$ mixture of light of wavelength $590 \, nm$ and an unknown wavelength is incident on the two slits of a Young's double-slit experiment. The central bright fringes of both lights coincide. The $3^{rd}$ bright fringe of the known wavelength coincides with the $4^{th}$ bright fringe of the unknown wavelength. Find the unknown wavelength in $nm$.

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In two separate set-ups of the Young's double slit experiment,fringes of equal width are observed when lights of wavelengths in the ratio $1:2$ are used. If the ratio of the slit separation in the two cases is $2:1$,the ratio of the distances between the plane of the slits and the screen in the two set-ups is

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In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

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In a Young's double slit experiment,the intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1 S_2$ on screen $S$. Find the minimum,maximum,and the intensity averaged over the entire screen,respectively.

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$A$ beam of light consisting of two wavelengths $6300 \, \mathring{A}$ and $\lambda \, \mathring{A}$ is used to obtain interference fringes in a Young's double-slit experiment. If the $4^{th}$ bright fringe of $6300 \, \mathring{A}$ coincides with the $5^{th}$ dark fringe of $\lambda \, \mathring{A}$,the value of $\lambda$ (in $\mathring{A}$) is:

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The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

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In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

In a Young's double slit experiment,the intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1 S_2$ on screen $S$. Find the minimum,maximum,and the intensity averaged over the entire screen,respectively.

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