Problem 48 - University Physics with Modern Physics

Question

A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 nm that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

Step-by-Step Solution

Answer

23 years

Step 1: Understanding Thin Film Interference

Thin film interference occurs when light waves reflect off the top and bottom surfaces of a thin film. The light can constructively or destructively interfere depending on the optical path difference and phase changes upon reflection.

Step 2: Formula for First Destructive Interference

For destructive interference, the condition is: \ \(2n imes d = \left(m + \frac{1}{2}\right)\lambda\), where \(n\) is the refractive index of the film, \(d\) is the thickness, \(\lambda\) is the wavelength in a vacuum, and \(m\) is an integer.

Step 3: Initial Thickness Calculation

Given \(n = 1.40\), \(\lambda = 525 \text{ nm}\), and we seek the first-order cancellation \((m=0)\). Solving for thickness: \ \(2 \times 1.40 \times d = \frac{1}{2} \times 525\), \ \(d = \frac{525}{2 \times 1.40 \times 2} = 93.75 \text{ nm}\).

Step 4: Subsequent Order for Enhancement

For the light to be enhanced after losing thickness, it must reach the next order of constructive interference: \ \(2n \times d' = m \lambda\). The first constructive interference above 93.75 nm is for \(m=1\): \ \ \(2 \times 1.40 \times d' = 525\), \ \(d' = \frac{525}{2 \times 1.40} = 187.5 \text{ nm}\).

Step 5: Determining Thickness Change

Determine the difference between the initial destructive thickness and the first constructive thickness: \ \(d' - d = 187.5 \text{ nm} - 93.75 \text{ nm} = 93.75 \text{ nm}\).

Step 6: Calculating Time for Thickness Change

Given the film wears away at 4.20 nm per year, calculate the years for this change: \ \(\text{Years} = \frac{93.75}{4.20} \approx 22.32 \text{ years}\).

Step 7: Rounding to Whole Years

Since time must be a whole number of years, round up to the nearest integer. \(\text{Minimum Years} = 23\).

Key Concepts

Destructive Interference

Destructive interference is a phenomenon that occurs when two waves combine in such a way that they cancel each other out. In the context of thin film interference, this happens when light waves reflect off the top and bottom surfaces of a film. The condition for destructive interference is met based on the path difference of the waves and the phase change upon reflection. This is mathematically represented by the formula: \[2n \times d = \left(m + \frac{1}{2}\right)\lambda\]where:

• \(n\) is the refractive index of the film,
• \(d\) is the thickness of the film,
• \(\lambda\) is the wavelength of light in a vacuum,
• and \(m\) is an integer indicating the interference order (commonly \(m=0\) for the first instance).

For destructive interference, any light that manages to travel through the film and reflects back interferes with light reflected from the top surface, thereby cancelling each other's effect. This means that if we observe a thin film in this configuration, certain wavelengths will "disappear" from the reflected light, resulting in a dark spot in the spectrum for those wavelengths.

Constructive Interference

Constructive interference is the process where two waves combine to make a wave with a larger amplitude. For light waves reflecting off the surfaces of a thin film, constructive interference occurs when the waves reinforce each other, making the light appear brighter. This can happen when the path difference between the waves is an integer multiple of the wavelength. The formula for this is:\[2n \times d = m\lambda\]where:

• \(n\) is the refractive index of the film,
• \(d\) is the thickness of the film,
• \(\lambda\) is the wavelength in vacuum,
• and \(m\) is an integer (starting from 1 for the first constructive interference).

When light waves meet the conditions for constructive interference, they synchronously align, essentially "adding" together to boost brightness at a specific wavelength. As thin films lose thickness, conditions shift and at certain points, previously cancelled out light can become visible, appearing as bright bands or colors where previously there were none.

Refractive Index

The refractive index, often symbolized as \(n\), is a measure of how much the speed of light is reduced inside a medium compared to its speed in vacuum. In thin film interference, the refractive index critically affects both destructive and constructive interference. It is an essential factor in determining how light shifts in phase as it moves through or reflects off different surfaces of a thin film. For a thin film with a refractive index \(n\):

• When light moves from a medium of lower to higher refractive index, it causes phase shifts which affect interference.
• The refractive index determines how much bending, or refraction, the light experiences entering the new medium.
• It also influences the optical path length of the light, changing how we perceive interference patterns.

Understanding the refractive index helps predict and explain patterns seen in everyday life, such as the vibrant colors in soap bubbles or the sheen on oil slicks. It allows us to calculate precise conditions for destructive and constructive interference necessary to solve problems on thin film interactions.