Question

In a Young’s double slit experiment, the fringe width is found to be 0.4 mm. To make things more interesting, if the whole apparatus is immersed in water of refractive index $\frac{4}{3}$ without disturbing the geometrical arrangement, the new fringe width will be:

• A. 0.30 mm
• B. 0.40 mm
• C. 0.53 mm
• D. 0.50 micron

Answer 1

The correct option is A 0.30 mm

$Inair:0.4=\frac{D{\lambda }_a}{d},InWater:\beta =\frac{D{\lambda }_w}{d}$
$\frac{0.4}{\beta }=\frac{{\lambda }_a}{{\lambda }_w},\beta =0.30mm$

Watch the next video for in-depth explanation.