Problem 27 - University Physics with Modern Physics

Question

A uniform film of TiO\(_2\) , 1036 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) wavelengths of the light in the TiO\(_2\) film.

Step-by-Step Solution

Answer

Add 63.24 nm thickness. Path difference: 198.47 nm or 1 wavelength.

Step 1: Identify the condition for destructive interference

In order for the reflected light to cancel, the path difference should be an odd multiple of half-wavelengths in the film. This condition can be expressed as \( m \cdot \lambda_{film} + \frac{1}{2}\lambda_{film} \) where \( m \) is an integer, and \( \lambda_{film} \) is the wavelength of light in the TiO\(_2\) film.

Step 2: Calculate the wavelength in the TiO2 film

The wavelength of light inside a medium is different from its wavelength in vacuum or air. It is given by \( \lambda_{film} = \frac{\lambda_0}{n} \), where \( \lambda_0 = 520 \) nm (wavelength in air) and \( n = 2.62 \) (index of refraction of TiO\(_2\)). Thus, \( \lambda_{film} = \frac{520 \text{ nm}}{2.62} \approx 198.47 \text{ nm} \).

Step 3: Determine the additional thickness required

The film should be adjusted so that the round-trip path difference equals an odd half-multiple of the new wavelength in the film. Given the existing thickness \( t_0 = 1036 \text{ nm} \), calculate the additional thickness \( t \) needed: solve \( 2(t_0 + t) = (2m + 1) \cdot \lambda_{film} \) for the smallest non-zero \( m \). After substitution and simplification, \( t = m \cdot \lambda_{film}/2 - t_0\). We solve for \( m = 1\), yielding \( t \approx 63.24 \text{ nm} \).

Step 4: Calculate the path difference after adjustment

After the additional thickness is added, the round-trip path difference should equal one wavelength in the film. Based on \( 2(t_0 + t) = (2m + 1) \cdot \lambda_{film} \), for \( m = 1 \), the path difference \( \Delta d = \lambda_{film} = 198.47 \text{ nm} \).

Step 5: Express the path difference in terms of wavelengths

Convert the path difference \( \Delta d \) into wavelengths of light in the TiO\(_2\) film. Since the path difference is equal to the wavelength \( \lambda_{film} \) itself, it corresponds to 1 wavelength.

Key Concepts

Destructive Interference

Destructive interference occurs when two waves combine to produce a smaller amplitude than either of the individual waves. This happens because the crest of one wave aligns with the trough of the other, causing them to cancel each other out.
In optics, destructive interference is used to diminish the intensity of light waves, as seen in thin film interference situations. For complete cancellation, the path difference between the two interfering waves should be an odd multiple of half the wavelength of light within the medium. Specifically, the path difference can be expressed as \( (2m + 1)\cdot\frac{\lambda}{2} \), where \( m \) is an integer indicating the number of half-wavelengths.
This principle is essential in applications like anti-reflective coatings where specific thicknesses are designed to ensure destructive interference of reflected light waves.

Thin Film Interference

Thin film interference is an optical phenomenon where light waves reflected from the top and bottom surfaces of a thin film, such as oil on water or a soap bubble, interfere with each other.
This interference occurs because light waves penetrate the thin film and are partially reflected both at the top and bottom interfaces. Thin film interference can result in either constructive interference, where the waves amplify each other, or destructive interference, where they cancel each other out.
For a thin film of uniform thickness, the path difference between the two waves is primarily responsible for the interference pattern. The condition for destructive interference is when the path difference matches an odd multiple of half the wavelength of light in the film. Adjusting the thickness of the film affects the interference pattern, thereby allowing for the control of reflected light, which is useful in creating coatings that minimize reflection.

Refractive Index

The refractive index, or index of refraction, is a measure of how much light slows down as it travels through a medium, compared to its speed in a vacuum.
The refractive index \( n \) is defined as the ratio of the speed of light in a vacuum \( c \) to the speed of light in the medium \( v \): \( n = \frac{c}{v} \). A higher refractive index indicates that light travels more slowly through the medium.
In the context of thin film interference, the refractive index affects the wavelength of light within the film. The wavelength in the film is given by \( \lambda_{film} = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the wavelength of light in vacuum. This change in wavelength is crucial for determining the conditions for constructive or destructive interference, making the refractive index a key factor in designing optical devices such as anti-reflective coatings.