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In the Newtons ring experiment, the diameter of the $10^{th}$ dark ring changes from 1.4 to 1.27 cm when a liquid is introduced between the lens & the plate. Calculate RI of liquid.
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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 $1.22$

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