Problem 48 - University Physics with Modern Physics

Question

Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has \(n\) = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

Step-by-Step Solution

Answer

(a) 154.55 nm; (b) 77.275 nm.

Step 1: Understanding the Problem

We are given a laser light with a wavelength \(\lambda = 510 \, \text{nm}\), a plastic rod with refractive index \(n = 1.30\) and a coating with refractive index 1.65. We need to find the minimum nonzero thickness of this coating to maximize and minimize light transmission.

Step 2: Calculate Wavelength in Coating

First, find the wavelength of light in the coating: \[ \lambda_{coating} = \frac{\lambda_{air}}{n_{coating}} \] Substitute the values:\[ \lambda_{coating} = \frac{510 \, \text{nm}}{1.65} = 309.1 \, \text{nm} \]

Step 3: Condition for Maximum Transmission (Constructive Interference)

For maximum transmission, the optical path difference should be an integer multiple of the wavelength in the coating. This can be simplified to:\[ 2t = m \lambda_{coating} \] where \(m = 1, 2, 3, \ldots\), as we want the minimum thickness, use \(m = 1\).

Step 4: Solve for Maximum Transmission Thickness

Plug \(m = 1\) into the constructive interference equation to find the minimum thickness:\[ 2t = 1 \times 309.1 \, \text{nm} \]Solving for \(t\):\[ t = \frac{309.1 \, \text{nm}}{2} = 154.55 \, \text{nm} \]

Step 5: Condition for Minimum Transmission (Destructive Interference)

For minimum transmission, the optical path difference should be an odd multiple of half the wavelength:\[ 2t = \left(m + \frac{1}{2}\right) \lambda_{coating} \]For minimum thickness, choose \(m = 0\).

Step 6: Solve for Minimum Transmission Thickness

Plug \(m = 0\) into the destructive interference equation:\[ 2t = \left(0 + \frac{1}{2}\right) \times 309.1 \, \text{nm} \] Solving for \(t\):\[ t = \frac{309.1 \, \text{nm} \times 0.5}{2} = 77.275 \, \text{nm} \]

Key Concepts

Constructive Interference

When light waves meet, they can combine in a way that increases their overall effect. This is called constructive interference. To imagine this, think of ocean waves, where two crests meet and form a bigger wave.
Constructive interference occurs when the phase difference between the waves is an integer multiple of their wavelength. In simpler terms, when two waves with the same frequency meet in such a way that their peaks add up, they reinforce each other.

• This results in a bright, intensified light.
• For constructive interference, the path difference between two waves is equal to a whole number of wavelengths (m = 0, 1, 2, ...).

In the example, to achieve maximum transmission, the coating's thickness causes the light wave inside to maintain this condition. This is calculated with the formula: \[ 2t = m \lambda_{coating} \] where m is a positive integer, and \(\lambda_{coating}\) is the wavelength of the light in the coating.

Destructive Interference

Destructive interference happens when two waves meet and cancel each other out, similar to when a crest and a trough in ocean waves meet, reducing the overall effect. In the context of light, this results in darkness rather than bright light.
This occurs when the path difference between the waves leads to a phase difference of half a wavelength, making one wave peak where the other wave trough is.

• Destructive interference leads to minimized light transmission.
• The path difference is an odd multiple of half the wavelength.

For the coating, this means finding a thickness that aligns with the formula: \[ 2t = \left( m + \frac{1}{2} \right) \lambda_{coating} \]This arrangement ensures minimal transmission, using the smallest non-zero m value.

Refractive Index

The refractive index (n) is a measure of how much a material slows down light. It’s like running through different types of surfaces: you run slower on sand than on concrete.
The refractive index tells us the ratio between the speed of light in a vacuum and the speed of light in a medium. A higher refractive index indicates greater slowing of light.

• Air has a refractive index close to 1, while denser materials like glass or plastic have higher values.
• Refractive index affects how light bends, or refracts, entering a new medium.

In our exercise, the plastic rod's refractive index is 1.30, and the coating is 1.65. This impacts how much the light's wavelength changes as it travels through these materials.

Wavelength in Medium

Light travels differently within various materials because of refraction, affecting its wavelength. When light passes from one medium into another with a different refractive index, both its speed and wavelength change.
The wavelength in a medium is calculated using the relation:\[ \lambda_{medium} = \frac{\lambda_{air}}{n_{medium}} \]where \(\lambda_{air}\) is the wavelength of light in a vacuum or air, and \(n_{medium}\) is the refractive index of the material.

• Higher refractive indices shorten the wavelength more compared to lower ones.
• Changing wavelengths is crucial in applications like anti-reflective coatings and other optical devices.

In our problem, the wavelength of light in the coating becomes 309.1 nm, calculated using the coating's refractive index of 1.65.