In Young's double-slit experiment,we get 10 f | vedclass

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(A) The fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since the field of view $L$ is constant,the t

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$8$

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(A) The fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since the field of view $L$ is constant,the total number of fringes $n$ is given by $n = \frac{L}{\beta}$.Therefore,$n \propto \frac{1}{\beta} \propto \frac{1}{\lambda}$.Given $n_1 = 10$ for $\lambda_1 = 4000 	ext{ } \mathring{A}$ and we need to find $n_2$ for $\lambda_2 = 5000 	ext{ } \mathring{A}$.Using the relation $n_1 \lambda_1 = n_2 \lambda_2$,we get:$10 	imes 4000 = n_2 	imes 5000$.$n_2 = \frac{10 	imes 4000}{5000} = 8$.Thus,the number of fringes obtained is $8$.

In Young's double-slit experiment,we get $10$ fringes in the field of view using monochromatic light of wavelength $4000 \text{ } \mathring{A}$. If we use monochromatic light of wavelength $5000 \text{ } \mathring{A}$,then the number of fringes obtained in the same field of view is

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