Question #211553

In Newton's ring experiment the diameter of 16 ^ (th) bright ring changes from 1.8 cm to 1.5cm when a liquid is introduced the plate and the lens. Find the refractive index of the liquid and the diameter of the 5 ^ (th) dark ring.


1
Expert's answer
2021-06-29T10:00:25-0400

The radius of the k-th bright Newton ring:\text {The radius of the k-th bright Newton ring:}

rk=(k12)λRn(1)r_k=\sqrt{(k-\frac{1}{2})\frac{\lambda R}{n}}(1)

measurement results for air:\text{measurement results for air:}

r16=1.82=0.9cm=0.009mr_{16}=\frac{1.8}{2}=0.9cm = 0.009m

nE1n_E\approx1

r16=15.5λRr_{16}=\sqrt{15.5*\lambda R}

measurement results for liquid:\text{measurement results for liquid:}

r16=1.52=0.75cm=0.0075mr'_{16}=\frac{1.5}{2}=0.75cm = 0.0075m

r16=15.5λRnLr'_{16}=\sqrt{15.5*\frac{\lambda R}{n_L}}


r16r16=nL\frac{r_{16}}{r'_{16}}=\sqrt{n_L}

nL=0.0090.0075=1.2\sqrt{n_L} = \frac{0.009}{0.0075}=1.2

nL=1.44n_L=1.44


Radius of Newton’s k-th dark ring:\text{Radius of Newton's k-th dark ring:}

rk=kλRnr_k=\sqrt{k\frac{\lambda R}{n}}

r5=5λRnr_5=\sqrt{5*\frac{\lambda R}{n}}

From equality (1)\text{From equality (1)}

r16=15.5λRnr_{16}=\sqrt{15.5*\frac{\lambda R}{n}}

r5=r16515.5=0.005mr_5 = r_{16}*\sqrt{\frac{5}{15.5}}=0.005m

d5=2r5=0.01md_5 = 2*r_5=0.01m


Answer: refractive index of the liquid =1.44;\text {Answer: refractive index of the liquid =} 1.44;

 the diameter of the 5- th dark ring= 0.01m\text{ the diameter of the 5- th dark ring= }0.01m



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