Question

Sometimes thin films are used as filters to prevent certain colors from entering a lens. Suppose an infrared filter is to be designed to prevent 800.0 -nm light from entering a lens. Find the minimum thickness for a film of \(\mathrm{MgF}_{2}(n=1.38)\) that will prevent this light from entering the lens.

Short answer

Answer: The minimum thickness required for the magnesium fluoride film to prevent 800 nm light from entering the lens is approximately 105.08 nm.

Step-by-step solution

Understand the concept of interference in thin films

In a thin film, interference occurs due to the reflection of light from the top and bottom surfaces of the film. This causes the light waves to superimpose and create a phenomenon called interference, which can be constructive or destructive. For a filter, we want to create destructive interference, which means we want the light waves to cancel each other out, so the specific wavelength of light doesn't enter the lens.

Find the destructive interference condition for the film

For destructive interference to occur, the path difference between the light waves should be an odd multiple of half-wavelength in the film medium. The formula to calculate the path difference is: Path difference = 2 * thickness * refractive index We want the path difference to be an odd multiple of half-wavelength (in the film), so the condition for destructive interference is: \(2nt = (2m + 1) \frac{\lambda_{film}}{2}\) where \(t\) is the thickness, \(n\) is the refractive index, \(\lambda_{film} = \frac{\lambda}{n}\) is the wavelength in the film medium, and \(m\) is an integer. For the given problem, the refractive index \(n = 1.38\) and the wavelength of light \(\lambda = 800 nm\).

Calculate the wavelength in the film medium

First, we need to find the wavelength of light in the film medium using the formula: \(\lambda_{film} = \frac{\lambda}{n}\) Substituting the given values, we get: \(\lambda_{film} = \frac{800\: nm}{1.38}\) \(\lambda_{film} \approx 579.71\: nm\)

Find the minimum film thickness

Now, we will use the condition of destructive interference to find the minimum thickness that will prevent 800 nm light from entering the lens. For minimum thickness, we will use \(m = 0\). \(2nt = (2m + 1) \frac{\lambda_{film}}{2}\) Substituting the values, we get: \(2(1.38)t = (2(0) + 1) \frac{579.71\: nm}{2}\) \(2.76t = 289.85\: nm\) Now, solve for the thickness \(t\): \(t = \frac{289.85\: nm}{2.76}\) \(t \approx 105.08\: nm\) So the minimum thickness required for the \(\mathrm{MgF}_2\) film to prevent 800 nm light from entering the lens is approximately 105.08 nm.

Key concepts

Interference in Thin Films

Interference in thin films is a captivating optical phenomenon where light is fractionally reflected and transmitted at the interfaces of materials with different refractive indexes. When light interacts with a thin film, such as a soap bubble or an oil slick on water, two reflected rays can result—one from the upper surface and one from the lower. If these two rays are in phase, they will amplify each other, causing constructive interference, and if out of phase, they cancel each other, leading to destructive interference.

Whether constructive or destructive interference occurs depends on several factors, including the thickness of the film, the wavelength of the light in the medium, and the refractive indexes of the materials involved. For destructive interference, the goal is for the reflected rays to be out of phase by half a wavelength. In a thin film of uniform thickness, this results in some wavelengths of light being cancelled out and not making it through the film, effectively filtering those colors from the transmitted or reflected light.

Thin Film Optical Filters

Thin film optical filters are precise tools in optics, employed for controlling the wavelengths of light that pass through or are reflected from them. By carefully designing thin films with specific thicknesses and refractive indices, engineers can create filters that selectively block certain colors or types of light.

For instance, an infrared filter designed to block 800.0 nm light functions through destructive interference. Destructive interference in this context means that the light wave reflected from the bottom surface has traveled a path equivalent to an odd multiple of the light wave's half-wavelength inside the film. This extra path causes it to be half a wavelength out of phase with the wave reflected from the top surface, which leads to the two waves cancelling each other out. Consequently, this specific wavelength is substantially diminished, if not entirely blocked, in the light that emerges from the filter. By leveraging this principle, thin film coatings can serve as notch filters, antireflection coatings, mirrors, and in numerous other optical applications.

Wavelength Calculation

Wavelength calculation in the context of thin films is crucial for designing filters that exploit destructive interference. The wavelength of light within a medium, \( \lambda_{film} \) is shorter than it is in a vacuum or air, as it is inversely proportional to the refractive index, \( n \) of the medium—represented by the equation \( \lambda_{film} = \frac{\lambda}{n} \) where \( \lambda \) is the wavelength of light in vacuum.

To determine the minimum thickness of a filter designed to prevent specific wavelengths of light from passing through, you can arrange the destructive interference condition to isolate the film thickness, \( t \) and solve accordingly. This is an essential consideration when creating thin film optical filters as it determines the exact point at which destructive interference will occur for a given wavelength of light, thus preventing that wavelength from entering a lens or being transmitted through the filter. Knowing how to calculate this thickness is fundamental for developing precisely engineered optical filters tailored for various technological applications.