Question

if the diameter of nth daRK Fringe in arrangement giving newtons rings change from 0.30 cm to 0.25 cm as a liquid is introduced between lens and plate, calculate refractive index of liquid and velocity of light in liquid

Answer 1

Answer:

for nth ring we have;

for air

Dn² =4nλR(air)

for liquid,

Dn²=4nλR(liq)/μ

μ=(0.30/0.25)²

=1.44

let v be the velocity of light in liquid

v=c/μ

= 3×10∧8/1.44

=2.09×10∧8.

Explanation:

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Answer 2

Answer:

Medium has a refractive index of 1.44 and medium allows the light to move with a speed of 2.08×10⁸m/s.

Explanation:

Arrangement of  Newton's ring experiment says that when a reflection of light takes place between 2 surfaces in which one is Spherical and other is flat an interference pattern is observed.

Given that:

Dₙ =0.30 cm, D'ₙ = 0.25 cm, µ=?

in presence of liquid the diameter of nth ring is

$D'n^{2} =$4nRλ/μ

diameter(in air) for the nth ring  is ,

$D'n^{2} =$4nRλ {µ=1 in case of air}

On division of equation (2) by (1), we get,

$u=(\frac{D_{n} ^{2} }{D'_{n} ^{2} } } )\\u= \frac{0.30^{2} }{0.25^{2} } \\u= 1.44.$

Refractive index of the liquid comes out to be 1.44.

Velocity of light in a medium whose refractive index is known is given by the standard relation,

$u=\frac{c}{v}$

where c is the speed of light and v is the velocity of light in the given medium.

v=c/μ

v=3*10⁸/1.44

v=2.08×10⁸m/s

Hence, Medium has a refractive index of 1.44 and medium allows the light to move with a speed of 2.08×10⁸m/s.