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(B) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the source and th

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$5400$

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(B) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the source and the screen,and $d$ is the distance between the two virtual sources.For the first case: $20\beta_1 = 2\, cm$,so $\beta_1 = \frac{2}{20} = 0.1\, cm = 1\, mm$.Thus,$\beta_1 = \frac{\lambda_1 D}{d} = 1\, mm$.For the second case: $30\beta_2 = 2.7\, cm$,so $\beta_2 = \frac{2.7}{30} = 0.09\, cm = 0.9\, mm$.Thus,$\beta_2 = \frac{\lambda_2 D}{d} = 0.9\, mm$.Taking the ratio: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} = \frac{0.9}{1} = 0.9$.Therefore,$\lambda_2 = 0.9 	imes \lambda_1 = 0.9 	imes 6000\, \mathring{A} = 5400\, \mathring{A}$.

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