Question

If prism angle $\alpha =1^o,\mu =1.54$, distance between screen and prism $\left(D\right)=0.7m$, distance between prism and source $a=0.3m,\lambda =180\pi nm$, then in Fresnal biprism find the value of $\beta$ (fringe width)

• A. ${10}^{-4}m$
• B. ${10}^{-3}mm$
• C. ${10}^{-4}\times \pi m$
• D. $\pi \times {10}^{-3}m$

Answer 1

The correct option is A ${10}^{-4}m$

Prism angle, $\alpha =1^o=\frac{\pi }{180}$ rad

Wavelength, $\lambda =180\pi nm=180\pi \times {10}^{-9}m$

Fringe width, $\beta =\frac{\lambda (a+D)}{2a\alpha (\mu -1)}$

$\therefore$ $\beta =\frac{(180\pi \times {10}^{-9})(0.3+0.7)}{2\left(0.3\right)\left(\frac{\pi }{180}\right)(1.54-1)}$

$⟹$ $\beta ={10}^{-4}m$