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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 = \f

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$X_{blue} < X_{green}$

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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 $X_{blue} < X_{green}$.

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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