Question

The distance between the slit and the bi-prism and that of between bi-prism and screen each is $0.4m$. The obtuse angle of bi-prism is ${179}^0$ and refractive index is $1.5$. If the fringe width is $1.8\times {10}^{-4}m$ then the wavelength of light will be :

• A. $7850A^o$
• B. $6930A^o$
• C. $5890A^o$
• D. $3750A^o$

Answer 1

Answer: (A)

$7850A^o$

Obtuse angle in the bi-prism is ${179}^{\circ }$

Thus each acute angle of the prism,

$A=\frac{{180}^{\circ }-{179}^{\circ }}{2}={0.5}^{\circ }=\frac{0.5\times \pi }{180}$radians

Deviation of the source beam=$\delta =A(\mu -1)$

Total deviation of each beam=$\delta \times$δ×distance between source and prism=$0.4\times A(\mu -1)$

Distance between imaginary sources, $d=2\times$ total deviation of each beam

$\omega =\frac{D\lambda }{d}$

Hence, $\lambda =\frac{\omega d}{D}$

D=0.8m

Thus $\lambda =7850A^{\circ }$

d=3.5mmA=180∘−179∘2=0.5∘=0.5×π180