Laser light (630 , nm) incident on a pair of | vedclass

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

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$576.86$

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(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.For the first light source: $\beta_1 = \frac{\lambda_1 D}{d} = 8.3 \, mm$.For the second light source: $\beta_2 = \frac{\lambda_2 D}{d} = 7.6 \, mm$.Taking the ratio of the two fringe widths: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.Rearranging for $\lambda_2$: $\lambda_2 = \lambda_1 	imes \left( \frac{\beta_2}{\beta_1} ight)$.Substituting the given values: $\lambda_2 = 630 \, nm 	imes \left( \frac{7.6 \, mm}{8.3 \, mm} ight) = 630 	imes 0.91566 = 576.86 \, nm$.

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