Question

Interference fringes are formed by a biprism of angle $1^o$ and refractive index $1.5$ . The slit is fixed at a distance of $10cm$ from the biprism and is illuminated by the light of wavelength $600nm$. Calculate the fringe width at a distance of $1m$ from the biprism in mm.

• A. $3.78$
• B. $37.8$
• C. $0.0378$
• D. $0.378$

Answer 1

The correct option is D $0.378$

For a fresnal biprism having refractive index $\mu$ and prism angle $\alpha ,$
Fringe width $\beta$ is given by, $\beta =\frac{\lambda \times D}{2\times a\times (\mu -1)\times \alpha }$
Where, a $=$ distance between slit & biprism $=0.1$ $m$
D $=$ distance between slit & screen $=1.1$ $m$
$\alpha =1=\frac{\pi }{180}rad$
$\therefore \beta =\frac{600\times 1.1}{2\times 0.1\times (1.5-1)\times \left(\frac{\pi }{180}\right)}nm=378152nm=0.378mm$