In Fresnel's biprism (mu = 1.5) experiment,th | vedclass

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Question

(B) The formula for fringe width $\beta$ in a Fresnel's biprism experiment is given by $\beta = \frac{(a + b)\lambda}{2a(\mu - 1)\alpha}$,where $\alph

Vedclass · Accepted Answer

$0.011$

Vedclass · Answer

(B) The formula for fringe width $\beta$ in a Fresnel's biprism experiment is given by $\beta = \frac{(a + b)\lambda}{2a(\mu - 1)\alpha}$,where $\alpha$ is in radians.Given:$a = 0.3\, m$$b = 0.7\, m$$\mu = 1.5$$\lambda = 6000\ \mathring{A} = 6 	imes 10^{-7}\, m$$\alpha = 1^\circ = \frac{\pi}{180}\, 	ext{radians}$Substituting the values:$\beta = \frac{(0.3 + 0.7) 	imes 6 	imes 10^{-7}}{2 	imes 0.3 	imes (1.5 - 1) 	imes (\frac{\pi}{180})}$$\beta = \frac{1.0 	imes 6 	imes 10^{-7}}{0.6 	imes 0.5 	imes 0.01745}$$\beta = \frac{6 	imes 10^{-7}}{0.3 	imes 0.01745} \approx 1.146 	imes 10^{-4}\, m$$\beta \approx 0.0114\, cm$.Thus,the correct option is $B$.

In Fresnel's biprism $(\mu = 1.5)$ experiment,the distance between the source and the biprism is $0.3\, m$,the distance between the biprism and the screen is $0.7\, m$,and the angle of the prism is $1^\circ$. The fringe width with light of wavelength $6000\ \mathring{A}$ will be: (in $, cm$)

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