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 distance between imaginary sources will be

• A. $8.7mm$
• B. $4.36mm$
• C. $1.5mm$
• D. $3.5mm$

Answer 1

Answer: (D)

$3.5mm$

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=$\delta \times$ distance between source and prism$=0.4\times A(\mu -1)$

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

Hence $d=3.5mm$