In a Fresnel's biprism experiment,the distanc | vedclass

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(C) Given: Distance between source and screen $D = 1 \, m = 100 \, cm$. Distance between source and biprism $a = 10 \, cm$. Distance between biprism a

Vedclass · Accepted Answer

$1.573$

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(C) Given: Distance between source and screen $D = 1 \, m = 100 \, cm$. Distance between source and biprism $a = 10 \, cm$. Distance between biprism and screen $b = D - a = 90 \, cm$. Fringe width $\beta = 0.03 \, cm$. Wavelength $\lambda = 6000 \, \mathring{A} = 6 	imes 10^{-5} \, cm$. Refracting angle $\alpha = 1^\circ = \frac{\pi}{180} \, 	ext{radians}$.The fringe width formula is $\beta = \frac{\lambda D}{2d}$,where $2d$ is the distance between the two virtual sources.$2d = \frac{\lambda D}{\beta} = \frac{6 	imes 10^{-5} 	imes 100}{0.03} = 0.2 \, cm$.The distance between virtual sources is given by $2d = 2a(\mu - 1)\alpha$.Substituting the values: $0.2 = 2 	imes 10 	imes (\mu - 1) 	imes \frac{\pi}{180}$.$0.2 = 20 	imes (\mu - 1) 	imes 0.01745$.$0.2 = 0.349 	imes (\mu - 1)$.$\mu - 1 = \frac{0.2}{0.349} \approx 0.573$.$\mu = 1.573$.

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