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Question

(A) The condition for the $n$-th bright fringe is given by $y = n \frac{\lambda D}{d}$.For the bright fringes of both wavelengths $\lambda_1 = 650 \ n

Vedclass · Accepted Answer

$429$

Vedclass · Answer

(A) The condition for the $n$-th bright fringe is given by $y = n \frac{\lambda D}{d}$.For the bright fringes of both wavelengths $\lambda_1 = 650 \ nm$ and $\lambda_2 = 550 \ nm$ to coincide,their positions must be equal: $y_1 = y_2$.$n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$.$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$.Since we want the least distance,we take the smallest integers $n_1 = 11$ and $n_2 = 13$.Substituting $n_1 = 11$ into the position formula:$y = 11 	imes \frac{650 	imes 10^{-9} 	imes 1.2}{2 	imes 10^{-3}}$.$y = 11 	imes 325 	imes 1.2 	imes 10^{-6} = 4290 	imes 10^{-6} \ m = 429 	imes 10^{-5} \ m$.

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