In a Young's double slit experiment,12 fringe | vedclass

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(B) The width of the segment on the screen is given by $W = n \beta$,where $n$ is the number of fringes and $\beta$ is the fringe width.Since $\beta =

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$18$

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(B) The width of the segment on the screen is given by $W = n \beta$,where $n$ is the number of fringes and $\beta$ is the fringe width.Since $\beta = \frac{\lambda D}{d}$,the width of the segment is $W = n \frac{\lambda D}{d}$.For a fixed segment $W$,the product $n \lambda$ remains constant because $D$ and $d$ are constant.Therefore,$n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 12$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.Substituting the values: $12 	imes 600 = n_2 	imes 400$.$n_2 = \frac{12 	imes 600}{400} = 12 	imes 1.5 = 18$.Thus,$18$ fringes will be observed.

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

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