Question

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.

Answer 1

$R=100 cm$

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

For the maxima

$r^2=\frac{R\lambda(2k-1)}{2}$

For the minima

$r^2=k\lambda R$

So,

$k\lambda_1R=\frac{R\lambda_2(2k-1)}{2}\to k\lambda_1=\frac{\lambda_2(2k-1)}{2}$ , $\lambda_1=\lambda,\lambda_2=\lambda_1+\Delta\lambda=\lambda+\Delta\lambda$

$k\lambda=\frac {(\lambda+\Delta\lambda)(2k-1)}{2}\to k \Delta \lambda \approx\frac{\lambda}{2}$

$k=\frac{\lambda}{2\Delta\lambda}=\frac{\lambda_1}{2(\lambda_2-\lambda_1)}=\frac{589}{2(589.6-589)}\approx491$

$l=\sqrt{R^2-(R-\frac{k\lambda}{2})^2}=\sqrt{1^2-(1-\frac{491\cdot589\cdot10^{-9}}{2})^2}\approx0.017m=17mm$