Question

In Newton's ring experiment, the diameter of
the 10th ring changes from 1.40 to 1.23 cm
when a liquid is introduced between the lens
and glass plate. What is the refractive index of
the liquid?
(a) 1.05
(c) 1.25
(b) 1.15
(d) 1.35​

Answer 1

Given:

n = 10

l1 = 1.40 cm

l2 = 1.23 cm

To find:

n2 =?

Explanation:

The radii of Newton's light ring with a constant radius of curvature of the lens is given by

$\displaystyle r_k = \sqrt{\frac{( k - 0.5) \lambda R}{n}$

where

r = radius of the ring

k = newton's constant

λ = wavelength of a ray

R = radius of curvature

n = refractive index

Now, the radius of the 10th ring having refractive index n1 is given by

$\displaystyle r_1 = \sqrt{\frac{(10 - 0.5) \lambda R}{n_1}$

$\displaystyle 1.40 = \sqrt{\frac{( 10 - 0.5) \lambda R}{n_1}$    ....(1)

Now, the radius of the 10th ring having refractive index n2 is given by

$\displaystyle r_2 = \sqrt{\frac{( k - 0.5) \lambda R}{n_2}$

$\displaystyle 1.23 = \sqrt{\frac{( 10 - 0.5) \lambda R}{n_2}$     ...(2)

Divide eq (1) by eq (2)

$\displaystyle \frac{1.40}{1.23} = \sqrt{\frac{n_2}{n_1}}$

but n1 = 1

∴ n2 = 1.22

∴ The refractive index of the liquid is 1.22

Answer 2

Answer:

Given: ( D ₁₀) air = 1.40 cm

(D ₁₀ ) liquid= 1.27 cm

To find: μ= refractive index of liquid=?

Formula:

(D n ²)=4nλR/μ

Solution:

without liquid in air ,for 10th ring,μ=1 and n=10

(D ₁₀ ²) air= 40λR             -------(1)

with liquid ,for 10th ring, n=10

(D ₁₀ ²) liquid= 40λR/μ     -------(2)

Dividing Eq (1) by Eq (2),

(D ₁₀ ²) air/(D ₁₀ ²) liquid= 40λR x μ /40λR

μ=(D ₁₀ ²) air/(D ₁₀ ²) liquid

 = (1.40/1.23)²

(C) μ  = 1.25

Explanation:

use the concept of  application of newton's ring:

determination of refractive index of liquid