Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000\stackrel{0}{A}$. When the slit is illuminated by light of another wavelength, then the angular width decreases by 30% . The same decrease in angular width of central maximum is obtained when the original apparatus is immersed in a liquid. The refractive of the liquid will be 

Angular width of central maxima is   $\omega =2\theta =\frac{2\lambda }{a}$$\Rightarrow \omega =\frac{2\times 6000\times {10}^{-10}}{a}$ 
When the wavelength is changed the angular width of central maxima is reduced by 30%. Thus new angular width is given by  $w^{\text{'}}=w-0.3w=0.7w$
  $\Rightarrow 0.7w=\frac{2\lambda \text{'}}{a} \Rightarrow \lambda \text{'}=\left(0.7w\right)\left(\frac{a}{2}\right)=0.7\left(\frac{2\times 6000\times {10}^{-10}}{a}\right)\left(\frac{a}{2}\right)$
$\Rightarrow \lambda \text{'}=0.7\times 6000\times {10}^{-10}=4200\times {10}^{-10}m \Rightarrow \lambda \text{'}=4200\stackrel{0}{\text{A}}$

When the setup is submerged in a liquid, then also the angular width of central maxima decreases by 30%, which indicates that the wavelength of light decreases by 30% , so

 $\lambda \text{'}=\frac{\lambda }{\mu }\Rightarrow 4200=\frac{6000}{\mu }\Rightarrow \mu =\frac{6000}{4200}=\frac{10}{7}=1.43$