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Chapter 26: Problem 58

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 \(525 \mathrm{nm}\) 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 canceled?

Short Answer

Expert verified
It takes 23 years for the reflected light to be enhanced instead of canceled.

Step by step solution

01

Understand the Problem

This problem involves thin film interference, where the goal is to determine the minimum number of years required for the reflected light (initially canceled due to destructive interference) to become enhanced (constructive interference), given the rate at which the film wears away.
02

Calculate the Initial Film Thickness

For destructive interference of light, the effective path difference must be an odd multiple of half-wavelengths in the film. The condition for destructive interference is \(2n t = (m+\frac{1}{2})\frac{\lambda}{n_f}\), where \(n_f\) is the film's refractive index, \(t\) is the thickness, and \(\lambda\) is the wavelength in vacuum. Simplifying for the minimum thickness: \(t = \frac{(m+\frac{1}{2})\lambda}{2 n_f}\). Given \(n_f = 1.40\) and \(\lambda = 525 \text{ nm}\), calculate \(t\).
03

Determine the Wavelength in the Film

Calculate the effective wavelength in the film using \(\lambda_f = \frac{\lambda}{n_f} = \frac{525}{1.40}\approx 375 \text{ nm}.\)
04

Find the Thickness for Constructive Interference

For constructive interference, we need \(2nt = m\lambda_f\) for a thickness resulting from wearing. Using this equation, determine the thickness where constructive interference occurs instead of destructive.
05

Calculate Thickness Change Required

Since the film is initially set to cancel light at \((m+\frac{1}{2})\frac{\lambda}{2 n_f}\), the thickness change required to achieve constructive interference is \(\Delta t = \frac{\lambda_f}{4} = \frac{375}{4} = 93.75 \text{ nm}\).
06

Determine Minimum Number of Years

The film wears away at a rate of \(4.20 \text{ nm/year}\). To find the number of years required for this change in thickness, use the equation: \(\text{Number of years} = \frac{93.75}{4.20}\approx 22.32.\) Rounding up, it will take at least 23 years for the thickness change to enhance the reflected light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive Interference
When light waves encounter each other, they can interfere destructively if their peaks and troughs are perfectly misaligned. This results in a decrease in amplitude, thereby canceling each other out. In the context of thin films, destructive interference occurs when:
  • The path difference between the waves reflected from the top and bottom surfaces of the film equals an odd multiple of half the wavelength inside the film.
  • The mathematical expression for this condition is \(2n t = (m+\frac{1}{2})\frac{\lambda}{n_f}\), where \(n\) is the refractive index of the film, \(t\) is the film thickness, and \(\lambda\) is the wavelength of the light in the vacuum.
This results in the phase shift necessary for the waves to destructively interfere, effectively canceling the reflection of that specific wavelength.
Constructive Interference
Constructive interference happens when light waves overlap in such a way that their peaks and troughs are aligned, leading to an increase in amplitude and thus brighter reflected light. For a thin film:
  • This occurs when the path difference between the waves corresponds to an integral multiple of the wavelength inside the film.
  • The condition can be mathematically expressed as \(2nt = m\lambda_f\), where \(\lambda_f\) is the wavelength in the film.
In our scenario, wear down of the film thickness can shift the interference from destructive to constructive, leading the previously canceled light reflection to become enhanced.
Film Thickness Calculation
The thickness of the film plays a crucial role in determining the type of interference observed. To find the minimum thickness which provides destructive interference, utilize the formula \(t = \frac{(m+\frac{1}{2})\lambda}{2 n_f}\).
  • Here, \(n_f\) is the film's refractive index, \(\lambda\) is the wavelength in a vacuum, and \(m\) is an integer representing the interference order.
With the initial film thickness provided, changes such as wear can lead to variations in the interference pattern, shifting from destructive to constructive as demonstrated in the exercise.
Refractive Index
The refractive index is a fundamental property of materials that indicates how fast light travels through them compared to a vacuum. For thin films, this value is crucial as:
  • It affects the wavelength of light inside the film, changing it from its value in air or vacuum.
  • A higher refractive index means a slower light speed, leading to a shorter wavelength in the material.
  • In the problem, the film's refractive index is given as 1.40, influencing the calculation of effective wavelength and, subsequently, the interference pattern.
This showcases the importance of knowing the refractive index to accurately predict and understand interference within thin films.
Wavelength in Film
The concept of wavelength in the film differs from that in air or vacuum due to the refractive index. The wavelength inside a medium like a thin film is reduced and can be calculated using the formula:
  • \(\lambda_f = \frac{\lambda}{n_f}\), where \(\lambda_f\) is the wavelength in the film, \(\lambda\) is the original wavelength in air, and \(n_f\) is the refractive index of the film.
For the given exercise, the effective wavelength in the film was determined as approximately 375 nm. Understanding this adjusted wavelength is critical for correctly predicting when constructive and destructive interference occur.

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Most popular questions from this chapter

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{~Hz}\) passes through two thin slits and falls on a screen \(85.0 \mathrm{~cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{~cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and \(600 \mathrm{~Hz}\). The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\). You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved \(39.8 \mathrm{~cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by \(0.460 \mathrm{~mm}\). In the resulting interference pattern on a screen \(2.20 \mathrm{~m}\) away, adjacent bright fringes are separated by \(2.82 \mathrm{~mm}\). What is the wavelength of the light that falls on the slits?

If a diffraction grating produces a third-order bright spot for red light (of wavelength \(700 \mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength \(400 \mathrm{nm}\) )?
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