Question

In a Young's double slit experiment,the fringe width is found to be $0.4mm$. What will be the new fringe width, if the whole apparatus is immersed in a water of refractive index 4/3 without disturbing the geometrical arrangement?

• A. 0.30 $mm$
• B. 0.40 $mm$
• C. 0.53 $mm$
• D. 450 $mm$

Answer 1

The correct option is A 0.30 $mm$

$\beta =\frac{\lambda D}{d};\beta \propto \lambda \mathrm{Gives}\mathrm{us},\frac{\beta \text{'}}{\beta }=\frac{0.4}{4/3}\Rightarrow \beta \text{'}=0.30\mathrm{mm}$