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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

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550.0 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of wavelength 585 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 u 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?

Jan first uses a Michelson interferometer with the 606-nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502-nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

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