In a double-slit interference experiment, the | vedclass

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(B) In Young's double-slit interference experiment, the fringe width $\beta$ is given by the formula:$\beta = \frac{\lambda D}{d}$where $\lambda$ is t

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$0.8$

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(B) In Young's double-slit interference experiment, the fringe width $\beta$ is given by the formula:$\beta = \frac{\lambda D}{d}$where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the separation between the slits.Given:$\beta_1 = 1.2 \ 	ext{mm}$$d_2 = 1.5 	imes d_1$Since $\beta \propto \frac{1}{d}$ (assuming $\lambda$ and $D$ remain constant),$\frac{\beta_1}{\beta_2} = \frac{d_2}{d_1} = 1.5$Substituting the values:$\frac{1.2}{\beta_2} = 1.5$$\beta_2 = \frac{1.2}{1.5} = 0.8 \ 	ext{mm}$Thus, the new fringe width is $0.8 \ 	ext{mm}$.

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