Question

There is air on both sides of a soap film. What is the smallest thickness of the soap film \((n=1.420)\) that would appear dark if illuminated with \(500 .-\mathrm{nm}\) light?

Short answer

Answer: The smallest thickness of the soap film is approximately 175.88 nm.

Step-by-step solution

Write the formula for destructive interference of thin-film

In order to calculate the minimum thickness of the soap film, we need the formula for destructive interference in thin-films. The formula is: \(m \lambda = 2 n t \cos(\theta)\) Where \(m\) is an integer representing the order of interference, \(\lambda\) is the wavelength of light, \(n\) is the refractive index of the soap film, \(t\) is the thickness, and \(\theta\) is the angle of incidence.

Plug the given values and simplify

We are given the refractive index (\(n = 1.420\)), the wavelength of light (\(\lambda = 500\ \text{nm}\)), and the film appears dark (which means \(\theta = 0\)). Also, we are solving for the smallest thickness, so \(m=1\). Now plug these values into the equation: (1) \(500\ \mathrm{nm} = 2 \times 1.420 \times t \times \cos(0)\) Since \(\cos(0) = 1\), we can simplify the equation: \(500\ \mathrm{nm} = 2 \times 1.420 \times t\)

Solve for the thickness of the soap film (t)

Now, isolate the term 't' to solve for the thickness of the soap film: \(t = \frac{500\ \mathrm{nm}}{2 \times 1.420}\) Calculate the value: \(t = \frac{500}{2 \times 1.420} = 175.88\ \mathrm{nm}\) Thus, the smallest thickness of the soap film that would appear dark if illuminated with \(500\ \mathrm{nm}\) light is approximately \(175.88\ \mathrm{nm}\).

Key concepts

Thin-Film Interference

Thin-film interference is a phenomenon that occurs when light waves reflect off the upper and lower boundaries of a thin film, leading to a complex pattern of constructive and destructive interference. This interaction is contingent on several factors including the thickness of the film, the wavelength of the light, and the refractive index of the film material. When light meets a boundary at an angle, each reflected wave may travel a slightly different path length before recombining, leading to shifts in phase that cause interference.

For dark or 'destructive' interference to be observed, the recombined light waves need to be out of phase. This out-of-phase condition results in a reduction of light intensity, often seen as dark bands or areas when the film is viewed against a light source. Conversely, 'constructive' interference occurs when waves recombine in phase, leading to bright bands or areas. Understanding this concept helps to identify the conditions under which certain patterns will appear on thin films, such as soap bubbles or oil slicks on water.

Wavelength of Light

The wavelength of light, usually denoted as \(\lambda\), is the distance between two consecutive peaks (or troughs) in a wave. It is a key factor in the interference patterns that occur in thin films. Light is an electromagnetic wave, and different wavelengths within the visible spectrum are perceived by our eyes as different colors. In calculations related to thin-film interference, the wavelength of light determines the specifics of the phase shifts and path differences required for destructive or constructive interference.

In the context of textbook exercises, when light of a specific wavelength shines on a thin film, the exercise will often involve solving for the thickness of the film that results in a certain type of interference pattern, given that the color corresponding to that wavelength is known.

Refractive Index

The refractive index, symbolized by \(n\), is a dimensionless number that describes how fast light travels through a material compared to the speed of light in a vacuum. Materials with a higher refractive index slow down light to a greater degree than those with a lower refractive index. The concept of refractive index is essential for understanding thin-film interference because it affects how much light bends when entering or exiting the film and influences the phase shift during reflection.

When light reflects off a medium with a higher refractive index, it undergoes a half-wavelength phase shift. When a problem presents the refractive index of a film and the wavelength of light used, it becomes possible to compute the film thickness required for achieving the particular interference phenomenon, whether it be constructive or destructive interference.

Angle of Incidence

The angle of incidence is the angle at which incoming light rays approach and hit a surface. In thin-film interference, this angle is denoted by \(\theta\) and can affect the path length that the light travels within the film before reflecting. A larger angle of incidence generally means that the light travels a longer path within the thin film, which in turn, influences the conditions required for interference.

For perpendicular or near-perpendicular incidences (small angle of incidence), light travels almost directly through the film, which simplifies calculations regarding interference, as seen in the provided exercise where \(\theta = 0\). This simplification can be particularly useful for educational purposes, as it allows students to focus on understanding the core principles of interference without the added complexity of oblique angles.