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(D) The fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since $D$ and $d$ are constant for the experi

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$560$

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(D) The fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.Since $D$ and $d$ are constant for the experimental setup,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.Therefore,we can write the ratio: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.Given: $\lambda_1 = 630 \, nm$,$\beta_1 = 8.1 \, mm$,and $\beta_2 = 7.2 \, mm$.Substituting the values: $\lambda_2 = \frac{\beta_2}{\beta_1} 	imes \lambda_1$.$\lambda_2 = \frac{7.2}{8.1} 	imes 630$.$\lambda_2 = \frac{8}{9} 	imes 630 = 8 	imes 70 = 560 \, nm$.

$A$ laser light of wavelength $630 \, nm$ is incident on a pair of slits,and the fringe width of the interference pattern produced is $8.1 \, mm$. In another interference pattern produced by a different light,the fringe width is $7.2 \, mm$. Find the wavelength of this second light in $nm$.

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