Question

Obtain the expression for the radius of the $n^{th}$ dark ring in Newton's rings experiment.

Answer 1

Expression for the radius of the $n^{th}$ dark ring can be obtained as follows :
Let us consider the vertical section $SOP$ of the plano convex lens through its centre of curvature $C$. Let $R$ be the radius of curvature of the planoconvex lens and $O$ be the point of contact of the lens with the plane surface. Let $t$ be the thickness of the air film at $S$ and $P$. Draw $ST$ and $PQ$ perpendiculars to the plane surface of the glass plate. Then $ST=AO=PQ=t$.
Let $r_n$ be the radius of the $n^{th}$ dark ring which passes through the points $S$ and $P$.
Then $SA=AP=r_n$
If $ON$ is the vertical diameter of the circle, then by the law of segments.
$SA\cdot AP=OA\cdot AN$
$r^2=t(2R-t)$
$r_n^2=2Rt$ (neglecting $t^2$ comparing with $2R$) $2t=\frac{r_n^2}{R}$
According to the condition for darkness
$2t=n\lambda$
$\therefore \frac{r_n^2}{R}=n\lambda \Rightarrow r_n^2=nR\lambda$ or $r_n=\sqrt{nR\lambda }$
Since $R$ and $\lambda$ are constants, we find that the radius of the dark king is directly proportional to square root of its order.
(ie) $r_1\propto \sqrt{1}$, $r_2\propto \sqrt{2}$, $r_3\propto \sqrt{3}$ and so on.