In Young's double slit experiment,the seventh | vedclass

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(A) The position of the $n^{th}$ maximum in Young's double slit experiment is given by the formula $x_n = \frac{n \lambda D}{d}$,where $n$ is the orde

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$\frac{\lambda_1}{\lambda_2}$

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(A) The position of the $n^{th}$ maximum in Young's double slit experiment is given by the formula $x_n = \frac{n \lambda D}{d}$,where $n$ is the order of the maximum,$\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.For the seventh maximum $(n=7)$ with wavelength $\lambda_1$,the distance from the central maximum is $d_1 = \frac{7 \lambda_1 D}{d}$.For the seventh maximum $(n=7)$ with wavelength $\lambda_2$,the distance from the central maximum is $d_2 = \frac{7 \lambda_2 D}{d}$.Taking the ratio of the two distances:$\frac{d_1}{d_2} = \frac{\frac{7 \lambda_1 D}{d}}{\frac{7 \lambda_2 D}{d}} = \frac{\lambda_1}{\lambda_2}$.

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ is:

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