Problem 33 - University Physics with Modern Physics

Question

Compact Disc Player. A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 \(\mathrm{nm}\) 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. E35.33). 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.)

Step-by-Step Solution

Answer

The minimum pit depth required is approximately 219.45 nm.

Step 1: Understand the Problem and Concepts

The problem involves interference by reflection from two surfaces: the pit and the flat region of a CD. We need destructive interference, where the waves cancel each other. This occurs when the path difference between these two waves is half of the wavelength in the CD material. Remember, part of the light from the laser reflects from the flat surface, and part reflects from the pit.

Step 2: Identify the Relevant Wavelength

The laser emits light with a wavelength of 790 nm. However, this wavelength changes in the CD material due to the refractive index. Calculate the wavelength within the material using the formula \( \lambda_{material} = \frac{\lambda_{vacuum}}{n} \), where \( \lambda_{vacuum} = 790 \, \mathrm{nm} \) and \( n = 1.8 \).

Step 3: Calculate the Wavelength in the Material

Substitute the values into the formula: \( \lambda_{material} = \frac{790 \, \mathrm{nm}}{1.8} \approx 438.89 \, \mathrm{nm} \). This is the wavelength of the light within the CD's plastic substrate.

Step 4: Determine the Path Difference Requirement

For destructive interference, the reflected waves should differ in path length by half the wavelength:\( \Delta L = \frac{\lambda_{material}}{2} \). This difference occurs due to the extra depth of the pit.

Step 5: Calculate the Minimum Pit Depth

The path difference \( \Delta L \) equates to the pit depth \( t \): \( t = \frac{\lambda_{material}}{2} = \frac{438.89 \, \mathrm{nm}}{2} \approx 219.45 \, \mathrm{nm} \). Thus, this is the minimum pit depth needed to achieve destructive interference.

Key Concepts

Wavelength and Material Properties

When a laser emits light, like the semiconductor laser used in a compact disc player, it sends light in the form of waves. These waves have a distinct distance between two similar points, known as the wavelength. For example, in a vacuum, the wavelength of the laser light used here is 790 nm. However, when light waves travel through different materials, this distance changes.

In our scenario, the light passes through a plastic substrate. To find how the wavelength is affected, we use the refractive index of the material. The equation is:

• \( \lambda_{material} = \frac{\lambda_{vacuum}}{n} \)

where \( \lambda_{vacuum} \) is the original wavelength and \( n \) is the refractive index. Here, with a refractive index of 1.8, the wavelength inside the CD becomes approximately 438.89 nm. This concept is crucial because it determines how light behaves when interacting with the CD's structures.

Destructive Interference

Light behaves in fascinating ways when it encounters different surfaces, such as those on a CD. It can reflect off both pits and flat regions, with a phenomenon called interference occurring when these reflected waves meet. If they align perfectly in peaks and troughs, they amplify each other — this is constructive interference.

However, if they meet in such a way that a peak and a trough coincide, they cancel each other out, known as destructive interference. For this to happen, the path difference between waves should be half the wavelength within the medium. When it comes to CDs, this precise cancellation helps in recognizing data, set by the specific depth of the pits.

Therefore, to achieve destructive interference, the pit's depth must be such that the path difference between the reflected wave from the pit and the flat region equals half the wavelength (in the material). This plays a key role in how CDs function, as it is this cancellation that allows the CD player to interpret the start and end of data bits.

Refractive Index

The refractive index is a measure that describes how light travels through a material. It's denoted as \( n \) and is a ratio comparing the speed of light in a vacuum to how fast it travels in the material. A higher refractive index indicates that light travels slower in that material compared to a lower index. For CDs, this number is essential in determining the wavelength of light within.

By using the refractive index, one can calculate the shifted wavelength when transitioning from air to the CD material. For example, with the original laser light at 790 nm and a CD's refractive index of 1.8, the wavelength within the CD becomes 438.89 nm. This index profoundly impacts the laser's ability to interact with the pits and flat areas, necessary for data reading.

Keep in mind:

• A larger refractive index reduces the wavelength in the material.
• This can also affect other properties such as the velocity of light through the material.

Compact Disc (CD) Technology

Compact Discs are a marvel of technology that utilize tiny pits and flat surfaces to encode data. This data is read by a laser beam that interacts precisely with the CD's structure, allowing it to register the information encoded. Light from the laser, which is initially 790 nm in a vacuum, navigates through the plastic substrate of the CD, changing due to the refractive index to 438.89 nm.

CD players detect changes in light patterns created by the pits' depths and flat areas through interference. For instance, when the beam reflects onto these differing surfaces and achieves destructive interference, it signals the device to recognize data's beginning or end, much like switching between binary 0s and 1s.

In addition to understanding interference, some terms related to CDs are:

• Pit — A depression that creates a difference in path length for light.
• Flat region (land) — The elevated part where light reflects differently.

These operations are a critical part of data retrieval in what was once cutting-edge technology!