Answer to Question #211553 in Optics for Jigyasa Karakoti

Question

In Newton's ring experiment the diameter of 16 ^ (th) bright ring changes from 1.8 cm to 1.5cm when a liquid is introduced the plate and the lens. Find the refractive index of the liquid and the diameter of the 5 ^ (th) dark ring.

Answer 1

$\text {The radius of the k-th bright Newton ring:}$

$r_k=\sqrt{(k-\frac{1}{2})\frac{\lambda R}{n}}(1)$

$\text{measurement results for air:}$

$r_{16}=\frac{1.8}{2}=0.9cm = 0.009m$

$n_E\approx1$

$r_{16}=\sqrt{15.5*\lambda R}$

$\text{measurement results for liquid:}$

$r'_{16}=\frac{1.5}{2}=0.75cm = 0.0075m$

$r'_{16}=\sqrt{15.5*\frac{\lambda R}{n_L}}$

$\frac{r_{16}}{r'_{16}}=\sqrt{n_L}$

$\sqrt{n_L} = \frac{0.009}{0.0075}=1.2$

$n_L=1.44$

$\text{Radius of Newton's k-th dark ring:}$

$r_k=\sqrt{k\frac{\lambda R}{n}}$

$r_5=\sqrt{5*\frac{\lambda R}{n}}$

$\text{From equality (1)}$

$r_{16}=\sqrt{15.5*\frac{\lambda R}{n}}$

$r_5 = r_{16}*\sqrt{\frac{5}{15.5}}=0.005m$

$d_5 = 2*r_5=0.01m$

$\text {Answer: refractive index of the liquid =} 1.44;$

$\text{ the diameter of the 5- th dark ring= }0.01m$