Question

pod VAJ
5. In Newton's rings experiment, the diameter of 20th dark ring changes
from 1.50 cm to 1.39 cm, when a liquid is introduced between the lens
and the glass plate. Calculate the refractive index of the liquid.​

Answer 1

The refractive index of the liquid is 1.14

Given that;

The diameter of the 20th dark ring changes from 1.50 cm to 1.39 cm, when a liquid is introduced between the lens and the glass plate.

To find;

The refractive index of the liquid.

​

Solution;

It is given that,

$D_{20}$ (for air film formed between lens and glass plate) = 1.50

$D_{20}$ (for the liquid film formed between the lens and glass plate)= 1.39

Since, the expression for the diameter of the  nth dark  ring after the introduction of a liquid of refractive index μ is given by,

$D_{n} (l)$ = $\sqrt{4nR lambda}$/μ ... (1)

Similarly, for air film, the expression for the diameter of the nth dark ring is given by,

$D_{n}$ = $\sqrt{4nlambdaR}$...(2)

Dividing equation (2) by equation (1) we get,

$D_{n}$  / $D_{n} (l)$ = sqrt μ

μ = $(D_{n} /D_{n}({l}) )^{2}$...(3)

on substituting the given values in equation (3) we get,

μ = $(\frac{1.50}{1.39} )^{2}$ = 1.14

Hence, the refractive index of the liquid is 1.14

#SPJ1

Answer 2

Answer:

The wavelength of light used is 5646A∘.

Explanation:

From the above question,

They have given :

The diameter of 20th dark ring changes from 1.50 cm to 1.39 cm, when a liquid is introduced between the lens and the glass plate.

Circular interference fringes produced by enclosing a thin air film of varying thickness between the surface of a convex lens of a large radius of curvature and a plane glass plate are known as Newton’s rings. The monochromatic light which produces these rings is given by the following mentioned formula.

Formula Used:

       λ=r2n+m−r2nmR

where R is the radius of curvature of the surfaces of the lens in contact with the glass plate.

rn and rn+m are the radiuses of the nth and (n+m)th dark fringe.

Circular interference fringes produced by enclosing a very thin air film of varying thickness between the surface of a convex lens of a large radius of curvature and a plane glass plate are known as Newton’s rings.

The radius of dark fringes:

For a minimum intensity or a dark fringe, the mathematical condition is:

                                2μt=nλ

Diameter of the 20th dark ring, D20=5.8mm=5.82×10−3m

Radius of the 20th dark ring, r20=2.91×10−3m

Diameter of the 10th dark ring, D10=3.6mm=3.6×10−3m

Radius of the 10th dark ring, r10=1.68×10−3m

Radius of plano convex lens, R=1m

                              λ = 5646A∘.

Therefore, the wavelength of light used is 5646A∘.

For more such related questions : https://brainly.in/question/40450248

#SPJ1