Question #232467
In Newton's ring experiment,the diameter of 10th ring changes from 1.40cm to 1.27cm when a liquid is introduced between the lens and the plate.to calculate the refractive index of the liquid.
1
Expert's answer
2021-09-03T09:26:48-0400

The radii of Newton's light rings with a constant radius of curvature of the lens can be found using the formula:

rk=(k0.5)λRnr_k=\sqrt {\frac{(k-0.5)\cdot λ\cdot R}{n}}

if,

r1=1.4cm,r2=1.27cmr_1 = 1.4 cm, r_2 = 1.27 cm

r1=(100.5)λRn1,n1=1r_1=\sqrt {\frac{(10-0.5)\cdot λ\cdot R}{n_1}}, n_1=1

r2=(100.5)λRn2r_2=\sqrt {\frac{(10-0.5)\cdot λ\cdot R}{n_2}}

Next, divide r1 by r2 and get:

r1r2=n2\frac{r_1}{r_2}=\sqrt{n_2}

n2=r12r22=1.421.272=1.215n_2=\frac{r_{1}^2}{r_{2}^2}=\frac{1.4^2}{1.27^2}=1.215

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