Question

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

• A. 3 cm
• B. 0.011 cm
• C. 2 cm
• D. 4 cm

Answer 1

Answer: (B)

$\beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }$
where $a=$ distance between source and biprism $=0.3m$
$b=$ distance between biprism and screen $=0.7m$.
$a=$ Angle of prism $=1^{\circ },\mu =1.5,\lambda =6000\times 10^{-10}m$
Hence, $\beta =\frac{(0.3+0.7)\times 6\times 10^{-7}}{2\times 0.3(1.5-1)\times \left(1^{\circ }\times \frac{\pi }{180}\right)}$
$=1.14\times 10^{-4}m=0.0114cm$