Question

Determine the minimum thickness of a soap film \((n=1.32)\) that would produce constructive interference when illuminated by light of wavelength \(550 . \mathrm{nm}\)

Short answer

Answer: The minimum thickness of the soap film is 208 nanometers.

Step-by-step solution

Write down the formula for constructive interference in thin films

The condition for constructive interference in thin films can be expressed by the formula: \(2 \cdot t \cdot n= m\lambda\) Where: - \(t\) is the thickness of the soap film - \(n\) is the refractive index of the soap film - \(m\) is an integer representing the order of interference (\(0\) for minimum thickness, \(1\) for next minimum thickness, and so on) - \(\lambda\) is the wavelength of the light source

Convert wavelength into meters

The wavelength should be in meters to match the units for thickness. The given wavelength is \(550\) nm. You can convert nm to meters by multiplying it by \(10^{-9}\) meters: \(\lambda = 550 \cdot 10^{-9} \mathrm{m} \)

Find the minimum thickness by plugging in the values in the formula

Since we're interested in the minimum thickness, let's find the thickness for the \(m=0\) case: \(2 \cdot t \cdot 1.32 = 0 \cdot (550 \cdot 10^{-9})\) \(t = \frac{0 \cdot (550 \cdot 10^{-9})}{2 \cdot 1.32}\) \(t = 0\) The thickness for the \(m=0\) case turns out to be \(0\) which is not a meaningful solution for existence of soap film. So we need to move on to the next case, which is \(m = 1\).

Find the thickness for the next minimum case, m = 1

Now let's find the thickness for the \(m=1\) case: \(2 \cdot t \cdot 1.32 = 1 \cdot (550 \cdot 10^{-9})\) \(t = \frac{1 \cdot (550 \cdot 10^{-9})}{2 \cdot 1.32}\) \(t = \frac{550 \cdot 10^{-9}}{2 \cdot 1.32}\) \(t = 2.08 \cdot 10^{-7} \mathrm{m}\)

Express the result in nanometers

To express the thickness in nanometers, multiply it by \(10^9\): \(t_{nm} = 2.08 \cdot 10^{-7} \cdot 10^9\) \(t_{nm} = 208 \mathrm{nm}\) The minimum thickness of the soap film that would produce constructive interference is \(208\) nanometers.

Key concepts

Thin Film Interference

Imagine you blow a soap bubble and notice the array of colors painting its surface. This is a beautiful example of thin film interference, a phenomenon that occurs when light waves reflected off the top surface of a thin film, like the bubble, combine with waves that have travelled through and reflected off the bottom surface.

Understanding this requires a basic appreciation of how waves superimpose. When the crest of one wave aligns with the crest of another, they add up to create a brighter, or more intense, light. This is called constructive interference, and it's precisely what's happening to generate those vibrant hues in soap films. Otherwise, if the crest of one wave aligns with the trough of another, they cancel each other out, causing destructive interference—resulting in places where the film appears darker.

In our exercise, the task is to find the minimum thickness of a film that would lead to constructive interference, specifically for light of a certain wavelength. The appropriate interference pattern is critical, as it determines how much light is reflected versus how much is transmitted, influencing the visible color or brightness.

Optical Path Difference

To grasp how thin film interference works, let's dissect a key term: optical path difference (OPD). This is the difference in the path lengths that light takes when reflecting off different layers within a material. In thin films, OPD can result in different types of interference depending on the thickness of the film and the wavelength of light.

In mathematical terms, OPD is calculated based on the refractive index of the film multiplied by its thickness, accounting for the extra distance the wave travels inside the film compared to the air. If the OPD is an integer multiple of the wavelength, it leads to constructive interference, while a half-integer multiple typically results in destructive interference.

The exercise provided perfectly illustrates OPD. When light of a specific wavelength strikes a soap film, the waves that reflect off the top interface are out of phase with those reflecting from the bottom. By finding when this path difference equals a multiple of the wavelength, we can determine when constructive interference will occur, and this helps us to measure the minimum thickness of the soap film.

Soap Film Interference

Soap film interference might appear simple at first glance but forms a fundamental part of optics. The vibrant patterns we see on soap bubbles are due to varying film thicknesses, cause a spectrum of colors due to interference effects. Here's how we apply our knowledge practically as shown in the exercise solution.

By applying the interference formula and considering light of 550 nm wavelength (a part of the visible spectrum), we find out how thin a film needs to be for it to amplify this light via constructive interference. It's worth noting that the 'm' in our formula represents different 'orders' of thickness that could work, similar to musical notes in chord strings resonating at fundamental frequencies and their harmonics.

The 'zero-order' (m=0) does not apply as it suggests a film of zero thickness, which isn't practical. The first order, however, gives us a meaningful value: a film thickness that precisely meets the condition for constructive interference for the given wavelength. In this case, it's 208 nm for the first order, and that would appear as a specific color depending on the viewing angle, illuminating the mystery behind the bubble's shimmering facade.