Question

In Fresnel's biprism $(\mu=1.5)$ experiment the distance between source and biprism is $0.3 \,m$ and that between biprism and screen is $0.7 \,m$ 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)

$\beta=\frac{(a+b) \lambda}{2 a(\mu-1) \alpha}$
where $a=$ distance between source and biprism $=0.3 \,m$
$b=$ distance between biprism and screen $=0.7 \,m$.
$a=$ 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 =0.0114 \,cm $