Solution:

The radii of Newton's light ring with a constant radius of curvature of the lens is given by

$$ r_{k}=\sqrt{\frac{(k-0.5) \lambda R}{n}} $$

where,

$r=$ radius of the ring

$k=$ newton's constant

$\lambda=$ wavelength of a ray

$\mathrm{R}=$ radius of curvature

$\mathrm{n}=$ refractive index

Now, the radius of the 10 th ring having refractive index $n 1$ is given by,

$ r_{1}=\sqrt{\frac{(10-0.5) \lambda R}{n_{1}}} \\ $

$ 1.40=\sqrt{\frac{(10-0.5) \lambda R}{n_{1}}} \\ $

Now, the radius of the 10 th ring having refractive index $n 2$ is given by,

$$ \begin{aligned} &r_{2}=\sqrt{\frac{(k-0.5) \lambda R}{n_{2}}} \\\\ &1.23=\sqrt{\frac{(10-0.5) \lambda R}{n_{2}}} \end{aligned} \\ $$

Divide eq (1) by eq (2),

$$ \frac{1.40}{1.23}=\sqrt{\frac{n_{2}}{n_{1}}} \\ $$

but $\mathrm{n} 1=1 \\$ $$ \therefore n 2=1.22 $$

$ \therefore$ The refractive index of the liquid is $1.22 $