Notes on Newton’s Rings and Determination of the Wavelength of Light | Grade 12 > Physics > Interference | KULLABS.COM

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#### Newton’s Rings

When a plano-convex lens is placed on the top of a flat glass surface a thin air film is formed between the two surfaces as shown in the figure. The thickness of air film varies from zero at the point of contact to some value t at point P. If the radius of curvature R of the lens is much greater than the distance r, and if the system is viewed from above, a pattern of bright and dark rings is observed as shown in a figure. The rings in the fringes are called Newton’s rings.

##### Theory

Let R be the radius of curvature of the lens, AOB be the vertical section of the lens surface through its centre of curvature C as shown in the figure. The lens is in contact at O with the plane glass plate MON, in such a manner that the points B and A are equidistant from O. The circle AOBE is completed and the diameter ODCE is drawn. BN and AM are drawn perpendicular to the plate MN and let AM = BN = t where t is the maximum thickness of the air film.

When light is incident on a thin film, a part of it is reflected at the upper surface and the rest is transmitted into the medium. It is again partly reflected at the lower surface and emerges out of the upper surface. The two beams of light, one reflected at the upper surface and the other reflected at the lower surface M give rise to the phenomenon of interference. The path difference is given by 2 µt cos r where µ is the refractive index of the medium i.e. at M or N, a further phase change of p or path difference λ/2 takes place. Hence,

$$\text {Total path difference} = 2\mu t \cos r + \frac {\lambda }{2}$$

In a case of a thin air film enclosed between two glass surfaces and for normal incidence i = r = 0 and hence cos r = 1.

Also for air, the refractive index is unity.

$$\text {Path difference} = 2t + \frac {\lambda }{2}$$

For maximum intensity of bright rings

\begin{align*} 2t + \frac {\lambda }{2} = n\lambda \\ \text {or,} \: 2t = (2n-1) \frac {\lambda }{2} \dots (i) \\ \end{align*}

For minimum intensity of dark rings

$$2t = n\lambda \dots (ii)$$

From figure (iii), we have

\begin{align*} \: AD \times DB &= OD \times DE \\ &= OD \times (2R – OD) \\ &= 2Rt – OD^2 \\ AD \times BD &= 2RT \dots (iii) \\ \end{align*}

Since OD, the thickness of air film is very small, OD2 can be neglected as compared to 2Rt. If r is the radius of the nth ring, then

$$AD = BD = r$$

\begin{align*} \text {Then equation} \: (iii) \text {can be written as} \\ r\times r &= 2Rt \\ \text {i.e.} \: r^2 &= 2Rt \\ \therefore 2t &= \frac {r^2}{R} \dots (iv) \\ \text {Substituting the value of 2t in equation} \: (i) \text {for bright rings we have} \\ \frac {r^2}{R} = (2n-1) \frac {\lambda }{2} \\ \text {radius of}\: n^{th} \text {bright ring}\\ r =\sqrt {\frac {(2n-1)\lambda R}{2}} \dots (v)\\ \text {Substituting the value of 2t in equation} \: (ii) \text {for dark rings, we have} \\ \frac {r^2}{R} = n\lambda \\ \text {or,} \: r^2 = n\lambda R \\ \therefore \text {Radius of}\: n^{th} \: \text {dark ring} \\ r = \sqrt {n\lambda R} \\\end{align*}

From equation (v), we find that the radius of the nth dark ring proportional to the square root of natural numbers $$\text {i.e.} \: R \propto \sqrt {n}$$.

##### Determination of the Wavelength of Light

Figure shown is an experimental arrangement for the determination of wavelength of sodium light. S is a source of sodium light. A parallel beam of light from the lens L1 is reflected by the glass plate G inclined at an angle of 45o to the horizontal. L is a plano-convex lens of large focal length. Newton’s strings are viewed through the eyepiece of the travelling microscope M focused on the sir film. Circular bright and dark rings are seen with the dark central fringe. With the help of travelling microscope measure the diameter of the nth dark ring.

\begin{align*} \text {let} \: D_n \text {be the diameter of the} \: n^{th} \text {ring and} \\ r_n^2 = n\lambda R \\ \text {But} \: r_n = \frac {D_n}{2}, \\ \frac {D_n^2}{4} = n \lambda R \\ \text {or,} \: D_n^2 = 4n \lambda R \dots (i) \\ \text {The diameter of the} \: (n+m)^{th} \: \text {dark ring is measured. Let it be}\: D_{n+m} \\ \therefore D^2_{n\times m} = 4(n + m) \lambda R \dots (ii) \\ \text {Subtracting} \: (i) \text {from} \: (ii), \text {we get} \\ \therefore (D_{n+m})^2 – (D_n)^2 = 4m \lambda R \dots (ii) \\ \text {or,} \: \lambda = \frac {( D_{n+m})^2 – (D_n)^2} {4mR} \\ \end{align*}

Hence, the wavelength λ of the sodium light can be measured.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.

If the radius of curvature R of the lens is much greater than the distance r, and if the system is viewed from above, a pattern of bright and dark rings which are called Newton’s rings.

Circular bright and dark rings are seen with the dark central fringe.

Wavelength of light is $$\lambda = \frac {( D_{n+m})^2 – (D_n)^2} {4mR}$$

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