Answer to Question #222815 in Optics for CHETU

Question

Newton's rings are formed in reflected light of wavelength 6000Å with aliquid between the plane and curved surfaces. If the diameter of the 6th bright ring be 3.1 mm and the radius of curvature of the curved surface be 1 m, the refractive index of the liquid will be

Answer 1

The Refractive index of Liquid can be found by using the equation,

$\mu= \frac{4Rnλ}{D^2_n}$

Here $\mu$ is the refractive index of the liquid.

R is the radius of curvature = 1 m

n is the count of bright ring.

$\lambda$ is the wavelength of refracted light.

D is the diameter of $n^{th}$ ring $= 3.1 \; mm = 3.1 \times 10^{-3} \;m$

$\mu= \frac{4 \times 1 \times 6 \times 6000 \times 10^{-10}}{(3.1 \times 10^{-3})^2} \\ = \frac{14.4 \times 10^{-6}}{9.61 \times 10^{-6}} \\ = 1.498$

Therefore refractive index of the liquid is 1.498.