Answer to Question #263545 in Optics for Arpit

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

$d_{b16} =1.8cm$

$d'_{b16} =1.5cm$

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

$\frac{r_{b16}}{r'_{b16}}=\frac{d_{b16}}{d'_{b16}}=\frac{1.8}{1.5}= 1.2$

$r_{b16}=\sqrt{(16-\frac{1}{2})\frac{\lambda R}{n_1}}$

$r'_{b16}=\sqrt{(16-\frac{1}{2})\frac{\lambda R}{n_2}}\ (1)$

$\frac{r_{b16}}{r'_{b16}}=\sqrt{\frac{n_2}{n_1}}$

$n_1\approx 1 \text{ suppose the expiration is in the air}$

$n_2 =1.2^2=1.44$

$r_{dk}=\sqrt{k\frac{\lambda R}{n}}$

$\text{From the formula (1)}$

$\sqrt{\frac{\lambda R}{n_2}}=r'_{b16}*\sqrt{\frac{1}{15.5}}\text{ then}$

$r'_{d5}=\sqrt{5}*\frac{d'_{b16}}{2}*\sqrt{\frac{1}{15.5}}=0.75*\sqrt{\frac{5}{15.5}}=0.43cm$

$\text{Answer:}$

$1.44 -\text{refractive index of a liquid}$

$0.43cm - \text{radius 5 of the dark ring when immersed in liquid}$