Problem 48 - Physics

Question

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{~nm}\) ), determine the minimum nonzero thickness that the coating can have.

Step-by-Step Solution

Answer

The minimum thickness is approximately 102.9 nm.

Step 1: Identify Key Information

We need to find the minimum thickness of a nonreflective coating. The refractive indices are given as \( n_1 = 1.38 \) for magnesium fluoride and \( n_2 = 1.52 \) for glass. The wavelength of light in vacuum is \( \lambda_0 = 565 \text{ nm} \).

Step 2: Understand the Concept of Minimum Thickness

For destructive interference (which prevents reflection), the path difference should be \( \frac{\lambda}{2} \). The coating thickness \( t \) must satisfy \( 2nt = m \frac{\lambda_0}{n_1} \), where \( m \) is an odd integer.

Step 3: Substitute and Solve for Thickness

Assume \( m = 1 \) for minimum nonzero thickness.Substituting into the equation:\[ 2t \cdot 1.38 = \frac{565}{1.38} \]Simplifying gives:\[ t = \frac{565}{2 \times 1.38 \times 1.38} \]

Step 4: Compute the Thickness

Calculate the value:\[ t = \frac{565}{3.8} \approx 102.9 \text{ nm} \]

Key Concepts

Destructive Interference

Destructive interference is a fascinating phenomenon that occurs when waves interact in such a way that they cancel each other out. In the context of thin films, such as the magnesium fluoride coating on a camera lens, it can prevent unwanted reflections, making images sharper and clearer.
In our example, we want to cancel out the reflection of specific light, in this case, yellow-green light. The necessary condition for destructive interference in thin films is that the path difference of the light waves should be half of the wavelength of the light inside the film. This can be expressed mathematically as a path difference of \( \frac{\lambda}{2} \).

• The path difference includes the distance that light travels through the film twice (there and back).
• For no reflection to occur, the interference between the two reflected paths (at the front and back of the film) must be destructive.

This is achieved when the thickness of the film is such that it leads to a path difference that results in destructive interference, effectively "canceling" the reflection of the target wavelength.

Refractive Index

The refractive index is a measure of how much the speed of light reduces when it enters a medium. In the case of the camera lens, two materials are involved: magnesium fluoride and glass, with refractive indices of 1.38 and 1.52, respectively.
Understanding the refractive index is crucial because when light enters a medium with a different refractive index, its speed and wavelength change, though its frequency remains constant.

• The refractive index \( n \) is defined as \( n = \frac{c}{v} \), where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.
• A higher refractive index means light travels slower in that medium compared to air or vacuum.
• This change affects the wavelength of light inside the film, which is why the formula to calculate minimum thickness incorporates the refractive index \( n_1 \) of magnesium fluoride.

The refractive index is thus integral to determining how light splits within the film, impacting the resulting interference pattern.

Wavelength

Wavelength is the distance between successive peaks of a wave. For the problem at hand, the light's wavelength is given as 565 nm in a vacuum, which is crucial for determining the behavior of light in a medium.
Inside a different medium, the wavelength of light changes, and this change is given by the formula \( \lambda = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the wavelength in the vacuum and \( n \) is the refractive index of the medium.

• Understanding how the wavelength changes inside materials allows for precise engineering applications, such as designing coatings to utilize interference effectively.
• In our example, determining the exact wavelength inside the magnesium fluoride coating ensures that we can calculate the exact film thickness for destructive interference.

By calculating the wavelength of light as it travels through different media, we can tailor materials to enhance the overall performance of optical systems, such as camera lenses.