Question

The fringe width in air, in Young's experiment is 3.6mm for light of $\lambda =6000\dot{A}$. If the arrangement is immersed in a transparent liquid of $\mu =1.2$ the fringe width would be

• A. 4.32mm
• B. 3mm
• C. 0.33mm
• D. 1.8mm

Answer 1

The correct option is B 3mm

${\beta }_a=\frac{D\lambda }{d}$; $\beta l=\frac{D{\lambda }_l}{d}$
$\frac{{\beta }_a}{{\beta }_l}=\frac{\lambda }{{\lambda }_l}={\mu }_l$
${\beta }_l=\frac{{\beta }_a}{\mu l}$
${\beta }_l=\frac{3.6}{1.2}=3mm$