In Young's double slit experiment,carried out | vedclass

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Question

(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5000 \ \mathring{A} = 5 	imes 10^{-7} \ 	ext{m}

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$10$

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(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5000 \ \mathring{A} = 5 	imes 10^{-7} \ 	ext{m}$,$D = 200 \ 	ext{cm} = 2 \ 	ext{m}$,and $d = 0.3 \ 	ext{mm} = 3 	imes 10^{-4} \ 	ext{m}$.Substituting these values,we get $\beta = \frac{5 	imes 10^{-7} 	imes 2}{3 	imes 10^{-4}} = \frac{10 	imes 10^{-3}}{3} \ 	ext{m}$.The position of the $n^{	ext{th}}$ maxima is given by $y_n = n \beta$.For the third maxima $(n=3)$,$y_3 = 3 	imes \left( \frac{10 	imes 10^{-3}}{3} ight) \ 	ext{m} = 10 	imes 10^{-3} \ 	ext{m} = 10 \ 	ext{mm}$.

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