Two light waves of wavelengths 800 , nm and 6 | vedclass

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Question

(C) The fringe width for a wavelength $\lambda$ is given by $\omega = \frac{\lambda D}{d}$.Given: $\lambda_1 = 800 \, nm = 8 	imes 10^{-7} \, m$,$\la

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) The fringe width for a wavelength $\lambda$ is given by $\omega = \frac{\lambda D}{d}$.Given: $\lambda_1 = 800 \, nm = 8 	imes 10^{-7} \, m$,$\lambda_2 = 600 \, nm = 6 	imes 10^{-7} \, m$,$D = 7 \, m$,and $d = 0.35 \, mm = 3.5 	imes 10^{-4} \, m$.Calculating fringe widths:$\omega_1 = \frac{8 	imes 10^{-7} 	imes 7}{3.5 	imes 10^{-4}} = 16 	imes 10^{-3} \, m = 16 \, mm$.$\omega_2 = \frac{6 	imes 10^{-7} 	imes 7}{3.5 	imes 10^{-4}} = 12 	imes 10^{-3} \, m = 12 \, mm$.The bright fringes coincide at a distance $y$ from the central maximum where $y = n_1 \omega_1 = n_2 \omega_2$,where $n_1$ and $n_2$ are integers.To find the shortest distance,we calculate the Least Common Multiple $(LCM)$ of $\omega_1$ and $\omega_2$.$LCM(16, 12) = 48 \, mm$.Thus,the shortest distance is $48 \, mm$.

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