Interference of Light

# Newton’s Rings

Newton’s Rings is a special case of interference of light waves reflected from the opposite surfaces of a thin film. It was discovered by Newton but it was Young who first gave a satisfactory explanation of the formation of the rings in terms of waves, being due to interference.

When a plano convex lens of long focal length (about one metre) is placed on a plane glass plate, a thin film of air is enclosed between the lower surface of the lens and the upper surface of the glass plate.

The thickness of the air film gradually increases from the centre outwards and it is zero at the point of contact. The two interfering beams, derived from a monochromatic source, satisfy the coherence condition for interference. A series of bright and dark circular rings is seen when a travelling microscope is focused on the air film.

The fringes produced are concentric circles. With monochromatic light, bright and dark fringes are produced and with white light, the fringes are coloured. As the radii of the rings increase, the separation between them decreases. At the centre, where there is no air film, there is a dark spot.

## Determination of Wavelength of Monochromatic Light

Let $R$ be the radius of curvature of the lower surface of the lens. The complete circular section of a sphere of radius $R$ is shown in the figure given below.

Let $r_n$ be the radius $OL$ of the nth ring at $L$ and let the thickness $LM$ of the air film be $t$. If $BO$ is produced to meet the circle at $Q$ then $BOQ$ is the diameter $(BOQ=2R).$

By the theorem of intersecting chords, $SO×OL=QO×OB$ $r_n×r_n=(2R-t)×t$ $r_n^2=2Rt-t^2$

$t^2$ is very small compared to $2Rt$ since $R$ is large. Hence, neglecting $t^2$, we get, $r_n^2=2Rt\;\;\text{__(1)}$

The path difference between the two interfering waves at $L$ is $2t$ for light incident normally on the air film. If a dark ring is formed at $L$, then from the condition of interference, we have, $\text{Path Difference}=n\lambda$ $\therefore 2t=n\lambda\;\;\text{__(2)}$ where, $n=1,2,3,…$ and $\lambda$ is the wavelength of light. From equations $(1)$ and $(2)$, $r_n^2=Rn\lambda$ $\therefore r_n=\sqrt{n\lambda R}$

This gives the expression for the radius of nth dark ring. Hence, the radius of nth dark ring is proportional to the square root of number of order of rings $\text{i.e.}\;\;r_n∝\sqrt{n}$

Similarly, a bright ring would be formed if path difference will be $(2n+1)\frac{\lambda}{2}$ $\text{i.e}\;\; 2t=(2n+1)\frac{\lambda}{2}$ $r_n^2=R(2n+1)\frac{\lambda}{2}$ $\therefore r_n=\sqrt{R(2n+1)\frac{\lambda}{2}}$

This gives the expression of nth bright ring. Hence, the radius of nth bright ring is proportional to the odd number of order of the rings $\text{i.e.}\;\;r_n∝\sqrt{2n+1}$

Now, let $D_n$ be the diameter of the nth dark ring, then, $r_n=\frac{D_n}{2}$ $\therefore \frac{D_n}{2}=\sqrt{n\lambda R}$ $D_n^2=4n\lambda R\;\;\text{__(a)}$

Similarly, if $D_{n+m}$ is the diameter of the (n+m)th dark ring, then, $D_{n+m}^2=4(n+m)\lambda R\;\;\text{__(b)}$

From equations (a) and (b), we have, $D_{n+m}^2-D_n^2=4(n+?)\lambda R-4n\lambda R$ $D_{n+m}^2-D_n^2=4m\lambda R$ $\therefore\lambda=\frac{D_{n+m}^2-D_n^2}{4mR}$