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(D) The fringe width $\beta$ in a Fresnel's biprism experiment is given by $\beta = \frac{\lambda D}{d}$.In the medium (water),the wavelength becomes

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$3\beta_a$

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(D) The fringe width $\beta$ in a Fresnel's biprism experiment is given by $\beta = \frac{\lambda D}{d}$.In the medium (water),the wavelength becomes $\lambda_m = \frac{\lambda_a}{\mu_w}$,where $\mu_w$ is the refractive index of water.The distance $D$ between the source and the screen remains unchanged.The distance $d$ between the two virtual sources is given by $d = 2a(\mu - 1)\alpha$,where $a$ is the distance from the slit to the biprism,$\mu$ is the refractive index of the prism material,and $\alpha$ is the refracting angle.When the entire setup is immersed in water,the effective refractive index of the prism relative to water becomes $\mu' = \frac{\mu_g}{\mu_w}$.Thus,the new distance between virtual sources is $d' = 2a(\mu' - 1)\alpha = 2a\left(\frac{\mu_g}{\mu_w} - 1\right)\alpha = 2a\left(\frac{\mu_g - \mu_w}{\mu_w}\right)\alpha$.The new fringe width is $\beta_w = \frac{\lambda_m D}{d'} = \frac{(\lambda_a / \mu_w) D}{2a(\frac{\mu_g - \mu_w}{\mu_w})\alpha} = \frac{\lambda_a D}{2a(\mu_g - \mu_w)\alpha}$.Given $\beta_a = \frac{\lambda_a D}{2a(\mu_g - 1)\alpha}$,we have $\beta_w = \beta_a \frac{(\mu_g - 1)}{(\mu_g - \mu_w)}$.Substituting $\mu_g = 1.5$ and $\mu_w = 1.33$,we get $\beta_w = \beta_a \frac{1.5 - 1}{1.5 - 1.33} = \beta_a \frac{0.5}{0.17} \approx 2.94 \beta_a \approx 3\beta_a$.

$A$ Fresnel's biprism experiment is performed in water. If the fringe width in air is $\beta_a$,and the refractive indices of the biprism material and water are $1.5$ and $1.33$ respectively,find the new fringe width.

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