Question

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

The minimum pit depth is approximately 110 nm.

Step-by-step solution

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.

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 \[\lambda' \approx 438.89 \text{ nm}\]

Determine Path Difference for Destructive Interference

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

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:\[d = \frac{219.45 \text{ nm}}{2} \approx 109.72 \text{ nm}\]

Key concepts

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.