In Young's double slit experiment,we get 60 f | vedclass

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(C) The width of the field of view $W$ is constant.In Young's double slit experiment,the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}

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$40$

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(C) The width of the field of view $W$ is constant.In Young's double slit experiment,the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.The total number of fringes $n$ in a field of view $W$ is given by $n = \frac{W}{\beta} = \frac{Wd}{\lambda D}$.Since $W$,$d$,and $D$ are constant,we have $n \lambda = 	ext{constant}$.Therefore,$n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 60$,$\lambda_1 = 4000 \; \mathring{A}$,and $\lambda_2 = 6000 \; \mathring{A}$.Substituting the values: $60 	imes 4000 = n_2 	imes 6000$.$n_2 = \frac{60 	imes 4000}{6000} = 40$.Thus,the number of fringes obtained is $40$.

In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

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