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(D) The fringe width in air is given by $\beta_a = \frac{\lambda D}{d}$,where $d = 2a(\mu - 1)\alpha$. When immersed in a medium of refractive index $

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$3 \beta_a$

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(D) The fringe width in air is given by $\beta_a = \frac{\lambda D}{d}$,where $d = 2a(\mu - 1)\alpha$. When immersed in a medium of refractive index $\mu_m$,the wavelength changes to $\lambda' = \frac{\lambda}{\mu_m}$ and the refractive index of the prism relative to the medium becomes $\mu' = \frac{\mu_g}{\mu_m}$. The new fringe width $\beta_w$ is given by $\beta_w = \frac{\lambda' D}{d'} = \frac{\lambda D / \mu_m}{2a(\frac{\mu_g}{\mu_m} - 1)\alpha} = \frac{\lambda D}{2a(\mu_g - \mu_m)\alpha}$. Comparing this to $\beta_a = \frac{\lambda D}{2a(\mu_g - 1)\alpha}$,we get $\beta_w = \beta_a 	imes \frac{\mu_g - 1}{\mu_g - \mu_m}$. Given $\mu_g = 1.5 = 3/2$ and $\mu_w = 1.33 \approx 4/3$. $\beta_w = \beta_a 	imes \frac{1.5 - 1}{1.5 - 1.333} = \beta_a 	imes \frac{0.5}{0.166} \approx 3 \beta_a$.

If a biprism experiment is immersed in water,and the fringe width in air is $\beta_a$,find the new fringe width if the refractive index of the biprism material is $1.5$ and the refractive index of water is $1.33$.

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