Physics - Wave Optics Question with Solution | TestHub

Given:
Refractive index of water, $\mu =1.33$
Slit separation, $d=1\times 10^{-3}$ m
Distance to screen, $D=1.33$ m
Wavelength of light in vacuum, $\lambda =6300\times 10^{-10}$ m $=6.3\times 10^{-7}$ m

The wavelength of light in water is given by:
${\lambda }^{\prime }=\frac{\lambda }{\mu }$

The fringe width in the medium is:
$\beta =\frac{{\lambda }^{\prime }D}{d}=\frac{\lambda D}{\mu d}$

Substituting the given values:
$\beta =\frac{6.3\times 10^{-7}\times 1.33}{1.33\times 10^{-3}}$

$\beta =6.3\times 10^{-4}$ m

Comparing with $x\times 10^{-4}$ m, we get $x=6.3$.

Answer: $6.3$