In a YDSE,light of two different wavelengths | vedclass

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(B) The position of the $n^{th}$ maxima for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.The position of the $m^{th}$ maxima for

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$5, 6, 6000 \ \mathring{A}$

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(B) The position of the $n^{th}$ maxima for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.The position of the $m^{th}$ maxima for wavelength $\lambda_2$ is given by $y_m = \frac{m \lambda_2 D}{d}$.Since the maxima coincide exactly in front of one of the slits,the distance from the central fringe is $y = \frac{d}{2}$.Thus,$\frac{n \lambda_1 D}{d} = \frac{d}{2} \implies n \lambda_1 = \frac{d^2}{2D} = \frac{(3 	imes 10^{-3})^2}{2 	imes 1.5} = \frac{9 	imes 10^{-6}}{3} = 3 	imes 10^{-6} \ m = 30000 \ \mathring{A}$.Similarly,$m \lambda_2 = 30000 \ \mathring{A}$.For option $B$: $n=5, m=6$. Then $\lambda_1 = \frac{30000}{5} = 6000 \ \mathring{A}$ and $\lambda_2 = \frac{30000}{6} = 5000 \ \mathring{A}$.Both wavelengths $6000 \ \mathring{A}$ and $5000 \ \mathring{A}$ lie within the given range $4500 \ \mathring{A} Therefore,the correct values are $n=5, m=6, \lambda_1 = 6000 \ \mathring{A}$.

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