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

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Question

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.Given initial values: $\beta_1 = 1 	imes 10^{-4}

Vedclass · Accepted Answer

$2.5 	imes 10^{-4} \ m$

Vedclass · Answer

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.Given initial values: $\beta_1 = 1 	imes 10^{-4} \ m$,$\lambda_1 = 6.4 	imes 10^{-7} \ m$.New parameters: $D_2 = 2D_1$,$d_2 = \frac{d_1}{2}$,$\lambda_2 = 4.0 	imes 10^{-7} \ m$.Using the ratio: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} 	imes \frac{D_2}{D_1} 	imes \frac{d_1}{d_2}$.Substituting the values: $\frac{\beta_2}{1 	imes 10^{-4}} = \left( \frac{4.0 	imes 10^{-7}}{6.4 	imes 10^{-7}} ight) 	imes \left( \frac{2D_1}{D_1} ight) 	imes \left( \frac{d_1}{d_1/2} ight)$.$\frac{\beta_2}{1 	imes 10^{-4}} = \left( \frac{4.0}{6.4} ight) 	imes 2 	imes 2 = \frac{16}{6.4} = 2.5$.Therefore,$\beta_2 = 2.5 	imes 10^{-4} \ m$.

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