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Chapter 35: Problem 31

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E35.31). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

Short Answer

Expert verified
The minimum pit depth is approximately 110 nm.

Step by step solution

01

Understand the Interference Condition

For destructive interference to occur, the difference in the path length of the two beams should be half of the wavelength in the medium. This condition is due to the phase difference of 180 degrees (or \(\pi\) radians) required for the cancellation of the two beams.
02

Calculate Wavelength in the Medium

The wavelength \(\lambda'\) in the medium is calculated by dividing the wavelength in vacuum \(\lambda = 790 \text{ nm}\) by the refractive index \(n = 1.8\). So, \[\lambda' = \frac{790 \text{ nm}}{1.8}\] This results in \[\lambda' \approx 438.89 \text{ nm}\]
03

Determine Path Difference for Destructive Interference

For destructive interference, the path difference should be \(\frac{\lambda'}{2}\). Thus, the path difference is \[\frac{438.89 \text{ nm}}{2} \approx 219.45 \text{ nm}\]
04

Calculate Minimum Pit Depth

The path difference is twice the depth of the pit because the light travels down to the pit and back up again. Therefore, the minimum pit depth \(d\) is:\[2d = 219.45 \text{ nm}\]Solving for \(d\), we get:\[d = \frac{219.45 \text{ nm}}{2} \approx 109.72 \text{ nm}\]

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Key Concepts

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

Wave Interference
Wave interference is a fascinating phenomenon where two or more waves superpose to form a resultant wave. This can result in either amplification or cancellation of the waves.

In the case of a compact disc (CD), two laser beams reflect off different surfaces; one from the pit and another from the flat region. These beams combine to form an interference pattern.
  • If the waves align in a way where their peaks and troughs coincide, it leads to constructive interference, which amplifies the wave.
  • Conversely, if the waves align such that the peak of one wave meets the trough of another, it causes destructive interference, canceling the resultant wave.
Refractive Index
In optics, the refractive index denotes how much light slows down when it enters a different medium.

For our exercise, a laser beam passes through a plastic substrate with a refractive index of 1.8. This means:
  • Light travels 1.8 times slower in this medium compared to a vacuum.
  • As a result, the wavelength of the light changes within the medium.

To find the new wavelength, divide the original wavelength by the refractive index. This adjustment helps us understand how light behaves during its journey through the CD's material.
Laser Wavelength
The wavelength of a laser is the distance between repeating points of the light wave. In the original problem, the laser has a wavelength of 790 nm in a vacuum.

However, since the laser beam travels through a material with a refractive index of 1.8, we must adjust this wavelength.

The formula to find the new wavelength inside the medium is: \[\lambda' = \frac{\lambda}{n}\]

where \(\lambda\) is the original wavelength and \(n\) is the refractive index. After calculation, we find the wavelength decreases to roughly 438.89 nm, which is essential for determining interference outcomes.
Destructive Interference
Destructive interference is crucial for recognizing the pit's beginning and end on a CD. It occurs when two waves combine in such a way that they cancel each other out.

For destructive interference, the waves must have a path difference that equals half the wavelength of the light in the medium.

Calculating this path difference in the problem involves:
  • Determining the wavelength in the medium, calculated previously as approximately 438.89 nm.
  • Finding the desired path difference for destructive interference, which is half of this modified wavelength.

For our scenario, that path difference equals about 219.45 nm. This value helps find the pit depth needed to ensure the wave peaks cancel the troughs effectively.

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Most popular questions from this chapter

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 nm. How many interference fringes are observed per centimeter in the reflected light?

Two speakers \(A\) and \(B\) are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\). For points far from the speakers, find all the angles relative to the centerline (Fig. P35.44) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 \(\mu\)m apart, and in line with an observer, so that one source is 2.04 \(\mu\)m farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04 \(\mu\)m farther away from the observer than the other? (c) For what visible wavelengths will there be \(destructive\) interference at the location of the observer?

A researcher measures the thickness of a layer of benzene (\(n\) = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength 575 nm is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?
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