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?
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Answer 1

Answer:

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Answer 2

The refractive index of the liquid is $1.295$

Step by step explanation

Given data:

the number of the ring$=10th$

The diameter of the ring before liquid is introduced$=1.40cm$

The diameter of the ring after the liquid is introduced$=1.23cm$

To find:

When the liquid is introduced between the lens and glass plate we must find the refractive index of the liquid.

Formula to be used:

• $D_{l} =\sqrt{\frac{4NRlambda}{n} }$.........(a)

where: $D_{l} =$diameter of the ring when the liquid is introduced.

$N=$order of the rings

$R=$radius of curvature of the lens

lambda(λ)$=$wavelength of the light

n=refractive index

• $D_{N} =\sqrt{4NRlambda}$........(b)

Where:$D_{N} =$ diameter of the nth dark or bright fringe

$N=$order of the rings

$R=$radius of curvature of the lens

lambda(λ)$=$wavelength of the light

Solution:

Dividing the equation (b) by (a)

$\frac{D_{N} }{D_{l} } =\frac{\sqrt{4NRlambda} }{\sqrt{\frac{4NRlambda}{n} } }$

$\frac{D_{N} }{D_{l} } =\sqrt{n}$

$n=(\frac{D_{n} }{D_{l} } )^{2}$

$n=(\frac{1.40}{1.23} )^{2}$

$n=1.138^{2}$

$n=1.295$

Therefore the refractive index of the liquid is $1.295$

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