Q. 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 $0.7m$ and angle of prism is $1{^\circ}$. The fringe width with light of wavelength $6000\ \mathring{\mathrm{A}}$ will be
A) $3\ cm$
B) $0.011\ cm$
C) $2\ cm$
D) $4\ cm$
Explanation
(b) $\beta = \frac{(a + b)\lambda}{2a(\mu - 1)\alpha}$
Where $a =$ distance between source and biprism $= 0.3\ m$
$b =$ distance between biprism and screen $= 0.7\ m$
$\alpha =$ Angle of prism $= 1{^\circ},\ \mu = 1.5,\ \lambda = 6000 \times 10^{- 10}m$
Hence, $\beta = \frac{(0.3 + 0.7) \times 6 \times 10^{- 7}}{2 \times 0.3(1.5 - 1) \times \left( 1{^\circ} \times \frac{\pi}{180} \right)}$
$= 1.14 \times 10^{- 4}m = 0.0114\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.