Problem 29 - University Physics with Modern Physics

Question

A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of TiO \(_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO\(_{2}\) film

Step-by-Step Solution

Answer

(a) Add 49.625 nm of TiO₂. (b) Path difference: (i) 198.5 nm, (ii) 1 wavelength.

Step 1: Understand the Interference Condition

To achieve the cancellation of reflected light (destructive interference) in thin films, the condition is that the path difference should be an odd multiple of half the wavelength of the light within the film. The formula for destructive interference is: \(2nt = (m + \frac{1}{2})\lambda_{film}\) where \(m\) is an integer, \(n\) is the refractive index of the film, \(t\) is the thickness, and \(\lambda_{film}\) is the wavelength of light in the film.

Step 2: Calculate Wavelength in TiO₂ Film

First, find the wavelength of the light within the \(\mathrm{TiO}_2\) film. This is obtained by dividing the wavelength in air by the refractive index of the film: \(\lambda_{film} = \frac{\lambda_0}{n} = \frac{520}{2.62} \approx 198.5 \text{ nm}\).

Step 3: Find Additional Thickness for Destructive Interference

We need to determine the additional thickness, \(\Delta t\), that will result in the reflected light being canceled. Initially, the thickness \(t = 1036 \text{ nm}\). The condition \(2n(t + \Delta t) = (2m+1)\frac{\lambda_{film}}{2}\) must hold, ensuring an odd number of half-wavelengths. Since we start with an even multiple (resulting in constructive interference), adding a quarter-wavelength will produce destructive interference: \(\Delta t = \frac{\lambda_{film}}{4} = \frac{198.5}{4} \approx 49.625 \text{ nm}\).

Step 4: Calculate Path Difference After Adding Thickness

The path difference after adding the minimum thickness to cause cancellation is one additional full wavelength: \(2n\Delta t = \lambda_{film}\), which calculates to \(2(2.62)(49.625) = 198.5 \text{ nm}\). In terms of wavelengths of the film, the path difference is exactly one wavelength, \(\lambda_{film}\).

Key Concepts

Destructive Interference

In the world of optics, interference refers to the phenomenon where waves overlap, resulting in a new wave pattern. For light waves, this often means seeing bright and dark patterns. Destructive interference happens when two light waves combine in a way that they cancel each other out. This is crucial for creating specific reflective effects, such as making a light beam disappear from view.
In thin films, achieving destructive interference means ensuring that the path difference—between light waves bouncing off different layers—results in zero light. This requires the path difference to be an odd multiple of half-wavelengths of the light within the film. The formula used to express this is:

• \(2nt = (m + \frac{1}{2})\lambda_{film}\)

Here:

• \(n\) is the film's refractive index,
• \(t\) is its thickness,
• \(\lambda_{film}\) is the light's wavelength in the film, and
• \(m\) represents an integer indicating the order of interference.

Choosing the correct thickness, as shown in the example, involves adjusting the film to add or subtract specific wavelengths, ensuring the reflected light waves negate each other.

Refractive Index

The refractive index is a key player in understanding how light behaves when traveling through different materials. This number describes how much slower light travels in a material compared to a vacuum. For instance, in our exercise,

• the refractive index of \(\mathrm{TiO}_2\) is 2.62,
• while that of crown glass is 1.52.

These values tell us how much more a light beam will bend or slow down as it enters the film.
A higher refractive index means the light speed is reduced significantly, altering the light's path and phase within the film. Since

• \(\mathrm{TiO}_2\)'s refractive index is greater than that of air (which is approximately 1),

light traveling from air into the film slows down, thereby changing the light's behavior at interfaces. The refractive index not only affects the speed but also the wavelength of light within that medium.

Wavelength in Film

Understanding the wavelength of light within a film is vital when analyzing and solving optical interference problems. The wavelength of light changes as it passes through materials, such as thin films, due to the refractive index. This change is calculated by dividing the wavelength of light in a vacuum (or air) by the refractive index of the material.

• For the given problem, the wavelength of 520 nm in air becomes \(\frac{520}{2.62} \approx 198.5\) nm in the \(\mathrm{TiO}_2\) film.

This adjusted wavelength, known as "wavelength in the film," is shorter than the original because light travels slower and thus has a shorter wavelength in the denser medium.Ensuring that these calculations are accurate is crucial for preparing specific optical effects, such as the destructive interference in this exercise. Manipulating the thickness of the film to align with these wavelengths ensures that we achieve the desired light cancellation effect.