The width of the fringe is 2,mm on the screen | vedclass

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(C) The formula for fringe width $(\beta)$ in a Young's double-slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance betw

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(C) The formula for fringe width $(\beta)$ in a Young's double-slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance between the slits and the screen,$\lambda$ is the wavelength of light,and $d$ is the distance between the slits.Since $D$ and $d$ remain constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.Therefore,we can write the ratio as $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.Given $\beta_1 = 2\,mm$,$\lambda_1 = 400\,nm$,and $\lambda_2 = 600\,nm$.Substituting these values: $\frac{\beta_2}{2\,mm} = \frac{600\,nm}{400\,nm} = \frac{3}{2}$.Solving for $\beta_2$: $\beta_2 = 2\,mm 	imes 1.5 = 3\,mm$.

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

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