In a biprism experiment,21 fringes are observ | vedclass

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(A) The width of the region $L$ is constant and is given by the product of the number of fringes $n$ and the fringe width $\beta$. $L = n_1 \beta_1 =

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$18$

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(A) The width of the region $L$ is constant and is given by the product of the number of fringes $n$ and the fringe width $\beta$. $L = n_1 \beta_1 = n_2 \beta_2$ Since the fringe width $\beta = \frac{\lambda D}{d}$,we have: $n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$ $n_1 \lambda_1 = n_2 \lambda_2$ Given $n_1 = 21$,$\lambda_1 = 4800 	ext{ Å}$,and $\lambda_2 = 5600 	ext{ Å}$: $21 	imes 4800 = n_2 	imes 5600$ $n_2 = \frac{21 	imes 4800}{5600} = \frac{21 	imes 48}{56} = \frac{21 	imes 6}{7} = 3 	imes 6 = 18$ Thus,the number of fringes observed is $18$.

In a biprism experiment,$21$ fringes are observed in a given region using light of wavelength $4800 \text{ Å}$. If light of wavelength $5600 \text{ Å}$ is used,the number of fringes in the same region will be:

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