Newton's Ring

NEWTON' S RING

Newton's ring are the circular interference fringes which can be produced by enclosing a very thin film of air or any other transparent medium with variable thickness between a plane convex lens of large radius of curvature and thin glass plate. This kind of interference fringes were firstly produced by Newton's so that they are known as Newton's ring.

NEWTONS RONG DUE TO REFLECTED OF LIGHT WAVE;

The schematic diagram to produce circular interference fringes due to reflected light wave is shown in the figure below.

The light waves from point source S are made parallel with the help of lens \(L_1\) and the parallel beams are allowed to incident on the glass plate \(G_1\) at 4\(5^\circ\) so that the light waves moves downwards to the plano convex lens. The plano convex lens \(L_2\) and thin glass plate \(G_2\), enclosed a very thin film of air in which interference fringes are produced between the reflected light waves and the fringes are observed with the help of eye piece as shown in figure (1). The production of reflected light wave as shown in figure (2) and figure (3) is used to determine the relationship between thickness of the medium (t) and radius of Newton's ring (\(\sigma_n\)). Now, from figure (2);

$$(OM)^2=(OC)^2+(CM)^2$$

$$or,\;\;\;R^2=(R-t)^2+(r_n)^2$$

$$or,\;\;\;R^2=R^2-2Rt+t^2+(r_n)^2$$

$$or,\;\;\;2Rt=t^2+(r_n)^2$$

$$\therefore\;\;\;t=\frac{(r_n)^2}{2R}\dotsm(1)$$

Here \(t^2\) is neglected being very small. We know optical path difference between reflected light waves is

2\(\mu tcos r +\frac{\lambda}{2}\)

In this case, cos r = 1 ( \(\therefore\) being the very small angle of incident )

And \(\mu\) = 1 ( \(\therefore being the air medium)

\(\therefore\) Optical path difference = 2t + \(\frac{\lambda}{2}\)

= 2 \(\frac{r^2_n}{2R} + \frac{\lambda}{2}\)

= \(\frac{r^2_n}{R} + \frac{\lambda}{2}\dotsm(2) \)

Special Cases:

• Condition of bright fringes: The bright ringes are produced when,

\(\frac{r^2_n}{R} + \frac{\lambda}{2} = n\lambda\)

$$or,\;\;\;r_n=\sqrt{\frac{(2n-1)}{2}\lambda R}\dotsm(3)$$

Here, n = 1, 2, 3, 4, 5\(\dotsm\)

• Condition for dark fringes: The dark rings are produced when,

\(\frac{r^2_n}{R} + \frac{\lambda}{2} = (2n+1)\frac{\lambda}{2}\dotsm(4)\)

Which implies that; \(r_n = \sqrt{n\lambda R}\) ; where, n = 0, 1, 2, 3, \(\dotsm\)

The Newton's ring's due to reflected light waves is shown in the figure above.

NEWTON'S RING DUE TO TRANSMITTED LIGHT WAVE

The schematic diagram of newton's ring due to transmitted light wave are same as shown in the diagram (1) & (2) above due to reflected light wave.

The light wave coming from the source S are made parallel with the help of the lens \(L_1\) and the parallel beams are allowed to incident on the glass plate \(G_1\) and are reflected towards the plano-convex lens. The plano-convex lens \(L_2\) and thin glass plate \(G_2\) enclose a very thin film of air in which the interference fringes are developed between the transmitted light wave. The production of transmitted light wave is shown in the diagram above, and the other figure are used to determine the radius of Newton's ring and the thickness of the medium.

We have from figure (ii),

$$(OM)^2=(OC)^2+(CM)^2$$

$$or,\;\;\;R^2=(R-t)^2+(r_n)^2$$

$$or,\;\;\;R^2=R^2-2Rt+t^2+(r_n)^2$$

$$or,\;\;\;2Rt=t^2+(r_n)^2$$

$$\therefore\;\;\;t=\frac{(r_n)^2}{2R}\dotsm(1)$$

Here \(t^2\) is neglected being very small. We know optical path difference between transmitted light waves is, 2\(\mu tcos r\)

Here, , cos r = 1 [ \(\therefore\) being the very small angle of incident]

And \(\mu\) = 1 [ \(\therefore\) being the air medium]

\(\therefore\) Optical path difference = 2t

= 2 \(\frac{r^2_n}{2R}\)

= \(\frac{r^2_n}{R}\dotsm(2)\)

SPECIAL CASES;

• Condition of bright fringes:

The bright ringes are produced when,

\(\frac{r^2_n}{R} = n\lambda\)

$$or,\;\;\;r_n=\sqrt{n\lambda R}\dotsm(3)$$

Here, n = 0, 1, 2, 3, 4, \(\dotsm\)

• Condition for dark fringes:

The dark rings are produced when,

\(\frac{r^2_n}{R} = (2n-1)\frac{\lambda}{2}\dotsm(4)\)

Which implies that; $$or,\;\;\;r_n= \sqrt{\frac{(2n-1)}{2}\lambda R}\dotsm(4)$$ ; where, n = 1, 2, 3, 4, \(\dotsm\)

The Newton's ring's due to transmitted light waves is shown in the figure below.

fig

CENTRAL BRIGHT POINT DUE TO REFLECTED LIGHT WAVE

In the Newton's ring experiment, the central point appears dark due to superposition of reflected light wave when there is an air film between the lens and the glass plate. However, it can not be made bright by introducing suitable transparent medium between plano convex lens and thin glass plate.

Let, \(\mu_1\) = refractive index of plano convex lens,

\(\mu\) = refractive index of transparent medium enclosed between plano convex lens and thin glass plate (sassafras oil ).

\(\mu_2\) = refractive index of thin glass plate

FIGURE HERE

Initially, let us adjust \(\mu_1\) > \(\mu\) > \(\mu_2\)

Let us consider, a transparent medium of refractive index \(\mu\) trapped between the two surfaces in contact. This is possible if a little oil of sassafras is placed between a convex lens of crown glass and a plate of flint glass. The reflection in both the case will be from denser to rarer medium and the two interfering rays are reflected under the same conditions. Therefore, in this case the central spot will be bright i.e. there exist no extra path difference, So, central point becomes bright. The diameter of the \(n^{th}\) bright ring is \(D_n = 2\sqrt{\frac{n\lambda R}{\mu}}\).

If refractive index indices are adjusted in such a way that \(\mu_1\) < \(\mu\) < \(\mu_2\), the central spot will also be bright. Here, extra path difference exist due to the reflection of light waves from denser surface to rarer medium in both beams. Again, the two interfering beams are reflected under the similar conditions. So, the extra term is canceled out and the central point again appears bright.