In a Young's double-slit experiment,the wavel | vedclass

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(C) The formula for fringe width in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since the experimental setup ($D$ and $

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$1.2$

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(C) The formula for fringe width in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since the experimental setup ($D$ and $d$) remains unchanged,the fringe width is directly proportional to the wavelength of light: $\beta \propto \lambda$.Therefore,$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.Given: $\lambda_1 = 5000 \, \mathring{A}$,$\beta_1 = 1 \, mm$,and $\lambda_2 = 6000 \, \mathring{A}$.Substituting the values: $\beta_2 = \beta_1 	imes \frac{\lambda_2}{\lambda_1} = 1 	imes \frac{6000}{5000} = 1.2 \, mm$.Thus,the new fringe width is $1.2 \, mm$.

In a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.

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