A beam of light consisting of two wavelengths | vedclass

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(C) Given: $\lambda_1 = 7000 \; \mathring{A} = 7000 	imes 10^{-10} \; m$,$\lambda_2 = 5500 \; \mathring{A} = 5500 	imes 10^{-10} \; m$,$d = 2.5 \; m

Vedclass · Accepted Answer

$462$

Vedclass · Answer

(C) Given: $\lambda_1 = 7000 \; \mathring{A} = 7000 	imes 10^{-10} \; m$,$\lambda_2 = 5500 \; \mathring{A} = 5500 	imes 10^{-10} \; m$,$d = 2.5 \; mm = 2.5 	imes 10^{-3} \; m$,$D = 150 \; cm = 1.5 \; m$.The condition for the coincidence of bright fringes is $y_n = y_m$,where $y_n = \frac{n \lambda_1 D}{d}$ and $y_m = \frac{m \lambda_2 D}{d}$.Equating the two,we get $n \lambda_1 = m \lambda_2$,which implies $\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14}$.For the least distance,we take the smallest integer values $n = 11$ and $m = 14$.The distance $y$ is given by $y = \frac{n \lambda_1 D}{d} = \frac{11 	imes 7000 	imes 10^{-10} 	imes 1.5}{2.5 	imes 10^{-3}}$.$y = \frac{11 	imes 7000 	imes 1.5}{2.5} 	imes 10^{-7} = \frac{115500}{2.5} 	imes 10^{-7} = 46200 	imes 10^{-7} = 462 	imes 10^{-5} \; m$.Comparing this with $n 	imes 10^{-5} \; m$,we get $n = 462$.

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