Physics Bootcamp

Subsection 50.3.1 Derivation of Interference Conditions for Dielectric Film

In this section we will work out the formulas I have given above for the most common case of use, viz., \(n_1=n_3\lt n_2\text{,}\) such as air/glass/air. Since we only wish to find the interference conditions and not actual values of intensities, we do not need to do a full calcualtion using the electric field wave. We can actually derive the conditions using a few simple rules.

1. 1. When a wave in a medium \(n_{first}\) is incident on a medium \(n_{second}\text{,}\) then reflected wave will have additional phase change by \(\pi\) radians if \(n_{first} \lt n_{second}\text{,}\) otherwise no additional phase change occurs.
2. 1. Wave length in a medium depends on refractive index. Thus, if wavelength in vacuum is \(\lambda_0\text{,}\) then refractive index in a medium of refractive index \(n\) will be
   \begin{equation*} \lambda = \frac{\lambda_0}{n}. \end{equation*}
3. 3. For each wavelength traveled, phase changes by \(2\pi\) radians.

Now, let us redraw the figure for reflections from a thin film (see Figure 50.3.4) so that we can use it here rather than switch back and forth from above. We wish to calculate the phase difference between the wave going in paths PADD'Q and PABCC'Q. Note that phase changes on DD'Q and CC'Q are same, so they cancel out in the difference and we need to only consider paths PAD and PABC.

To find the phase difference in the two paths, we trace them so that we can locate phase changing reflections and appropriate path lengths starting from their common point P before any reflection occurs. Note that although PA part is common to both, but there is a difference between immediately before A and immediately after A in the two. So, you can imagine P to be as close to A as possible, but just before the reflection. This will give us the following phase difference \(\Delta_{12}\text{.}\)

In the present case, the reflection at A causes a phase change of \(\pi\) radians since \(n_1\lt n_2\text{,}\) but the reflection at B does not have any phase change since \(n_2 \gt n_3\) - make sure you understand these comparisions in terms of first medium and second medium for each reflection event.

Now, we can find the phase change for traveling a distance AD in the first medium by noting that phase of a wave changes by \(2\pi\) radians when it travels one wavelength in the medium. We must be careful here and use the correct wavelength for each path since waves are traveling in different media. We are using \(\lambda_1\) and \(\lambda_2\) for the wavelength in the two media. Note that although the frequency of light is same in the two media, the wavelength will be different in the two media because speed of the wave depends on the refractive index.

Therefore, we will have the following relation between the wavelengths.

Hence, phase changes over the two paths are as follows.

Therefore phase difference between the two waves will be

We can replace the distances AB, BC, and AD in terms of the thickness \(d\) of the film, angle of incidence \(\theta\) and angle of refraction \(\theta_r\text{.}\)

where I have made use of the Snell's law for the refraction at A, \(n_1\sin\theta = n_2 \sin\theta_r\text{.}\) Substituting for AB, BC and AD in the expression for \(\Delta_{12}\) gives

The interference will be constructive if this is integer multiple of \(2\pi\) and destructive if odd multiple of \(\pi\text{.}\)

We can send \(\pi\) on the other side, simplify, and replace \(n\) and \(n^\prime\) by appropriate \(m\) and \(m^\prime\text{.}\) Due to moving \(\pi\) to the right side, the choice of \(m\) is different than that of \(n\text{.}\)

Subsection 50.3.2 Reflected Waves from a Wedge-shaped Film

A wedge is a simple system of linearly varying thickness. The phase difference between the rays reflected from the top and the bottom of a wedge-shaped dielectric film will depend upon where on the wedge this happens. A set-up that makes use of the reflections from a wedge-shaped film is shown in Figure 50.3.5. An extended source is used to strike at different parts of the wedge, and observations are made on the reflections of nearly normally incident rays.

We again consider the case \(n_1=n_3\text{.}\) If \(n_1\lt n_2\text{,}\) then the reflection at A will cause a phase shift of \(\pi\) radians and no phase shift for the reflection at B. If \(n_1=n_3>n_2\text{,}\) then the reflection at A will not cause phase shift but the reflection at B will cause phase shift of \(\pi\) radians. Therefore, as far as the phase shift contributions in the two waves labeled \(E_{1r}\) and \(E_{2r}\) are concerned, the situation here is similar to the plane plate worked out in the last section.

Since we are looking at near normal directions, the angles of incidence and refraction would be nearly zero. Therefore we will set \(\cos\theta_r \approx 1\) in the condition for interference worked out in the last section. Let \(x\) be the distance from the corner of the wedge and \(\alpha\) be the angle at the corner of the wedge, then the thickness \(d\) of the dielectric material in the wedge at the place shown in the figure will be

where \(\alpha\) is in radian and I have assumed that the angle at the wedge is very small, making \(\alpha \ll 1.0\text{ rad}\text{.}\) Depending on the values of \(d\) and \(x\text{,}\) we will have constructive or destructive interferences on the screen as given by the following conditions. With \(2d \approx 2\alpha x\) we get

Since different places on the wedge have different \(d\text{,}\) the constructive interference condition for \(m=1,\ 2, \cdots\text{,}\) is satisfied dor different wavelengths. this gives rise to bands of colors when white light is shone on the wedge. These bands are also called Fizeau fringes.

Subsubsection 50.3.2.1 Newton's Rings

Figure 50.3.6 shows a hemispherical surface on a flat surface. This arrangment provides an interesting way to create air gap of varying heights. Reflections from top of glass/air and then air/glass interfaces interfere and one sees fringes. These fringes are called Newton's rings. Let us analyze this setup and find a formula for normal viewing.

The thickness d at a horizontal distance r from the center is given by the following relation from the Pythagoras theorem.

\begin{equation*} (R-d)^2 + r^2 = R^2. \end{equation*}

Solve for \(d\) keeping the root with d\lt R.

\begin{equation*} d = R - \sqrt{R^2 - r^2}. \end{equation*}

The condition for constructive interference in case of near normal incidence is as follows.

\begin{equation*} d = R - \sqrt{R^2 - r^2} = \frac{m}{2} \frac{\lambda_0}{2}. \end{equation*}

where \(m\) is an odd number and \(\lambda_0\) is the wavelength in air, the medium of the wedge. The bright fringes form at following radial values from the center.

\begin{equation*} r = \sqrt{ \frac{m}{2} R \lambda_0 }, \end{equation*}

where we have assumed \(\lambda_0 \ll R\) and dropped a term. Since bright and dark fringes form in circles, they look like rings as illustrated in Figure 50.3.7. Isaac Newton reported these rings in his book Optiks: or, a Treatise on the Reflexions, Refractions, Inflexions and Colours of Light. We now call these rings Newton's rings. Newton pressed a double convex lens against the curved sides of the planar convex lens to create wedge that produced colored rings.