Answer to Question #189345 in Optics for Parul

Question

Newton’s rings are formed in reflected light of wavelength 5890 *10^-8 cm with a

liquid between the plane and curved surfaces. The diameter of the fifth ring is

0.3 cm and the radius of curvature of the curved surface is 100 cm. Calculate

the refractive index of the liquid, when the ring is bright.

Answer 1

Given,

Wavelength $\lambda= 5890\times 10^{-8}\ cm=589\times10^{-9}\ m$

Radius of curvature R = 100cm = 1 m

n = refractive index of liquid

So,

Diameter of k th ring

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

$\dfrac{d_5}{2}=\sqrt{\frac{9}{2}\cdot \frac{589\times 10^{-9}\cdot 1}{n}}$

$\frac{3\times 10^{-3}}{2}=\sqrt{\frac{5301\times 10^{-9}}{2n}}\\n=\dfrac{5301\times 10^{-9}\cdot 2}{9\times 10^{-6}}\\\Rightarrow n=1.178$

Hence, refractive index of liquid is 1.178