Question

The distance between the slit and biprism and that between biprism and screen are each 50 cm. The obtuse angle of the briprism is 170${}^o$ and its refractive index is 1.5. If the fringe width is 0.0135 cm; calculate the wavelength of light in nm

Answer 1

$y_1=0.5m,y_2=0.5m,\alpha =5^{\circ }$

D = $y_1+y_2=0.5+.5=1m$

$\beta =0.0135cm$

d = $2(\mu -1)\ast \alpha \ast y_1$ = $2\ast 0.5\ast \left(5\pi /180\right)\ast 0.5=\pi /72$

$\beta =\lambda \ast D/d$

=> $\lambda =5890nm$

Answer. Wavelength $\lambda =5890nm$