In a biprism experiment,a steady interference | vedclass

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(D) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is th

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increase by $28 \%$

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(D) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{\lambda D}{d}$,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.Since $D$ and $d$ remain constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.Given $\lambda_1 = 5000 \ Å$ and $\lambda_2 = 6400 \ Å$.The ratio of the new fringe width $\beta_2$ to the initial fringe width $\beta_1$ is $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} = \frac{6400}{5000} = 1.28$.This means $\beta_2 = 1.28 \beta_1$.The percentage increase in fringe width is given by $\frac{\beta_2 - \beta_1}{\beta_1} 	imes 100\% = (1.28 - 1) 	imes 100\% = 0.28 	imes 100\% = 28\%$.Therefore,the fringe width increases by $28\%$.

In a biprism experiment,a steady interference pattern is observed on the screen using light of wavelength $\lambda_1 = 5000 \ Å$. Without disturbing the setup of the experiment,the source of light is replaced by a source of wavelength $\lambda_2 = 6400 \ Å$. The fringe width will:

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