Question

In Fresnels biprism $(\mu =1.5)$ experiment the distance between surface source and biprism is $0.3m$ and that between biprism and screen is $0.7m$ and angle of prism is $1^o$. The fringe width with light of wavelength $6000A^o$ will be:

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

Answer 1

The correct option is B $0.011cm$

Given that,

Distance between source and biprism $a=0.3$

Distance between biprism and screen $b=0.7$

Angle of prism $\alpha =1^0$

Index $\mu =1.5$

Wave length $\lambda =6000\times {10}^{-10}m$

Now, the fringe width is

$\beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }$

$\beta =\frac{(0.3+0.7)\times 6000\times {10}^{-10}}{2\times 0.3(1.5-1)\times (1\times \frac{\pi }{180})}$

$\beta =\frac{1.0\times 6\times {10}^{-7}\times 180}{0.3\times \pi }$

$\beta =1.146\times {10}^{-4}$

$\beta =0.011cm$

Hence, the fringe width is $0.011cm$