Answer to Question #107517 in Optics for navaz

Question #107517
In Newton's ring experiment, sodium light is used (A1=589 nm, 12=589.6nm). Radius of curvature of plano-convex lens R=100 nm.

Calculate the order number of the ring at which they fades away. Find out the distance from the point of contact from where the rings fades away.
1
Expert's answer
2020-04-02T10:56:33-0400

R=100cmR=100 cm


Sharpness of the rings pattern is the worst when the maxima and minima intermingle.

For the maxima


r2=Rλ(2k1)2r^2=\frac{R\lambda(2k-1)}{2}


For the minima


r2=kλRr^2=k\lambda R


So,


kλ1R=Rλ2(2k1)2kλ1=λ2(2k1)2k\lambda_1R=\frac{R\lambda_2(2k-1)}{2}\to k\lambda_1=\frac{\lambda_2(2k-1)}{2} , λ1=λ,λ2=λ1+Δλ=λ+Δλ\lambda_1=\lambda,\lambda_2=\lambda_1+\Delta\lambda=\lambda+\Delta\lambda


kλ=(λ+Δλ)(2k1)2kΔλλ2k\lambda=\frac {(\lambda+\Delta\lambda)(2k-1)}{2}\to k \Delta \lambda \approx\frac{\lambda}{2}


k=λ2Δλ=λ12(λ2λ1)=5892(589.6589)491k=\frac{\lambda}{2\Delta\lambda}=\frac{\lambda_1}{2(\lambda_2-\lambda_1)}=\frac{589}{2(589.6-589)}\approx491


l=R2(Rkλ2)2=12(14915891092)20.017m=17mml=\sqrt{R^2-(R-\frac{k\lambda}{2})^2}=\sqrt{1^2-(1-\frac{491\cdot589\cdot10^{-9}}{2})^2}\approx0.017m=17mm














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