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Chapter 35: Problem 46

35.46. 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 cancelled?

Short Answer

Expert verified
The minimum number of years is 23.

Step by step solution

01

Understanding Thin Film Interference

In thin film interference, when light is incident on a film, some of it is reflected from the top surface and some from the bottom surface. These reflections can interfere constructively or destructively. The condition for destructive interference for reflected light is: \[ 2nt = (m+\frac{1}{2})\lambda \] where \(n\) is the refractive index of the film, \(t\) is the thickness of the film, \(\lambda\) is the wavelength of light in vacuum or air, and \(m\) is an integer.
02

Calculate Current Film Thickness

Given that the film perfectly cancels the 525 nm wavelength, it's known that the current thickness is greater than the minimum required for cancellation. Using the destructive interference condition with \(m+1\), we calculate the thickness as \[ t = \frac{(m+\frac{1}{2})\lambda}{2n} \] Let us assume it currently cancels at a value of \(m+1 = 2\). Thus, the thickness \(t\) is: \[ t = \frac{(2+\frac{1}{2})\times 525}{2 \times 1.40} = 468.75 \, \text{nm}. \]
03

Determine Minimum Thickness Change

To change from destructive to constructive interference, the condition for constructive interference is required: \[ 2nt = m\lambda \] This happens when the thickness reduces to satisfy the condition \(m=2\) in the current case. Calculate the required thickness for \(m=1\): \[ t = \frac{2\times 525}{2 \times 1.40} = 375 \, \text{nm}. \] Thus, the film thickness needs to change from 468.75 nm to 375 nm.
04

Calculate Time for Thickness Reduction

The difference in thickness needed for constructive interference is \[ 468.75 \text{ nm} - 375 \text{ nm} = 93.75 \text{ nm} \] Given that the film wears away at 4.20 nm per year, the time needed to reach the required thickness is: \[ \text{years} = \frac{93.75 \text{ nm}}{4.20 \text{ nm/year}} \approx 22.32 \text{ years}. \] Rounding up gives a minimum of 23 years.

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Key Concepts

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

Refractive Index
The refractive index is a fundamental concept in optics and is crucial to understanding thin film interference. It is denoted by the symbol \( n \), and it's a measure of how much light slows down when passing through a medium compared to its speed in a vacuum. Every material has its own unique refractive index, which affects how light behaves as it enters the material.

  • The higher the refractive index, the more the light bends away from its original path.
  • In the context of thin films, the refractive index determines how light waves travel through the film and how they interfere with each other.
In the given exercise, a film with a refractive index of 1.40 coats a glass plate with a refractive index of 1.55. This difference plays a key role in how the light waves reflected from the two surfaces of the film either cancel each other out (destructive interference) or enhance each other (constructive interference). Understanding the refractive index helps us predict and explain these interference effects.
Destructive Interference
Destructive interference is a phenomenon where two or more light waves combine to form a wave with reduced or zero amplitude. This can result in darkness or a reduction in light intensity, and it occurs under specific conditions in thin film interference.

  • For destructive interference to occur in a thin film, the path difference between the waves reflected from the top and bottom surfaces should be an odd multiple of half of the wavelength (\( \lambda \)).
  • The condition is mathematically given by \( 2nt = (m+\frac{1}{2})\lambda \), where \( m \) is an integer.
In the exercise, this principle explains the cancellation of the 525 nm wavelength light, leading to no reflection visible at that specific thickness. As the film wears away, achieving destructive interference again depends on adjustments in thickness to continue meeting this condition.
Constructive Interference
Constructive interference occurs when two or more light waves combine to form a wave with greater amplitude, resulting in bright or enhanced light intensity. In thin film interference, it happens under different conditions compared to destructive interference.

  • The conditions for constructive interference in a thin film are met when the path difference between the reflected waves from the top and bottom surfaces is an integer multiple of the wavelength.
  • This is described by the formula \( 2nt = m\lambda \), where \( m \) is again an integer.
In the scenario from our exercise, after a certain period as the film thins, constructive interference occurs for the 525 nm light. Over time, as the film's thickness decreases steadily, the change from destructive to constructive interference is observed. This change is predictable and can be calculated based on the rate of wear and the refractive index.

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Most popular questions from this chapter

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ? )\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

35.31. A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 \(\mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO_ 2 film.

35.9. Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)

35.25. Points \(A\) and \(B\) are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a \(12.5-\mathrm{MHz}\) signal horizontally. These transmitters are in phase with other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C\) . directly opposite the midpoint of \(A B\) . The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?
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