91Ó°ÊÓ

In a compact disc (CD), digital information is stored as a sequence of raised surfaces called "pits" and recessed surfaces called "lands." Both pits and lands are highly reflective and are embedded in a thick plastic material with index of refraction \(n=1.55\) (Fig. 34-34). As a 780-nm wavelength (in air) laser scans across the pit-land sequence, the transition between a neighboring pit and land is sensed by monitoring the intensity of reflected laser light from the CD. At the moment when half the width of the laser beam is reflected from the pit and the other half from the land, we want the two reflected halves of the beam to be \(180^{\circ}\) out of phase with each other. What should be the (minimum) height difference \(t\) between a pit and land? [When this light enters a detector, cancellation of the two out-of-phase halves of the beam produces a minimum detector output.]

Step by step solution

01

Understand the Problem and Given Data

The problem involves a laser beam reflecting from different surfaces on a CD. The beam needs to be out of phase by \(180^{\circ}\) for cancellation, which occurs when one beam undergoes an additional path difference of half the wavelength. The given data are: - Wavelength of laser in air: \(\lambda_0 = 780\, \text{nm}\)- Index of refraction of the CD material: \(n = 1.55\)

02

Convert Wavelength from Air to CD Material

To find the wavelength of the laser in the CD's material, use the formula \[ \lambda = \frac{\lambda_0}{n} = \frac{780\, \text{nm}}{1.55} \approx 503.23\, \text{nm} \] This is the new wavelength within the CD material.

03

Determine Path Difference for Destructive Interference

For two beams to be \(180^{\circ}\) out of phase, the path difference should be \(\frac{\lambda}{2}\). Thus:\[ \Delta x = \frac{\lambda}{2} = \frac{503.23\, \text{nm}}{2} \approx 251.615\, \text{nm} \]

04

Calculate Minimum Height Difference t

The actual height difference \(t\) should be such that it creates a path difference of \(\frac{\lambda}{2}\). Because the light travels down to the pit and back up, a height difference \(t\) results in a path difference of \(2t\). Thus:\[ 2t = \frac{\lambda}{2} \rightarrow t = \frac{\lambda}{4} \] Substituting the wavelength in the material, we get:\[ t = \frac{503.23\, \text{nm}}{4} \approx 125.81\, \text{nm} \]

05

Conclusion

The minimum height difference \(t\) between a pit and a land for the destructive interference condition is approximately \(125.81\, \text{nm}\).

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Key Concepts

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

Interference

When you shine a laser on a CD's surface, interesting things happen with light! This process connects to a concept in optics known as interference. Interference occurs when two or more light waves overlap and combine. This can happen constructively, making the light brighter, or destructively, canceling each other out and making it dimmer or even completely dark.

For a CD, when laser light reflects off the "pits" and "lands", these two reflected light waves can interfere with each other. If they are out of phase by 180 degrees, they will destructively interfere, meaning they cancel out. This is the kind of interference we're targeting in the CD problem.

• Constructive interference: Waves add up, intensity increases.
• Destructive interference: Waves cancel each other, intensity decreases.
• The phase difference can be influenced by the height difference between "pits" and "lands".

Understanding and applying these interference principles is essential for storing and reading data on optical media like CDs.

Wavelength

Wavelength is another crucial concept in understanding how light behaves when it interacts with materials. It is defined as the distance between successive peaks (or troughs) of a wave. In optics, when light enters a new medium—like the plastic of a CD—the wavelength changes due to the medium's index of refraction.

For instance, in the exercise, a laser with a wavelength of 780 nm in air becomes shorter inside the CD plastic. The reduced wavelength inside the CD influences how light waves interfere with one another. This change is calculated using the formula:\[ \lambda = \frac{\lambda_0}{n} \]where \( \lambda_0 \) is the original wavelength in vacuum (or air), and \( n \) is the index of refraction.

• Wavelength affects interference patterns.
• It's crucial for precisely determining the optical path differences between reflected light waves.
• Knowing the wavelength inside a medium helps calculate the geometrical height differences needed for the desired interference effect.

Understanding wavelength changes are a fundamental part of mastering optics.

Index of Refraction

The index of refraction, often denoted by \( n \), helps us understand how light behaves in different materials. It's a measure of how much the speed of light is reduced inside a medium compared to vacuum. For the CD material in the problem we discussed, it has an index of refraction of 1.55.

The higher the index of refraction, the slower light travels through the material, causing light to bend or change its direction. This slowing down also affects wavelength and is critical when calculating interference. The index of refraction is key for understanding how deep the pits on a CD should be to produce the correct interference needed to read information accurately from the disc.

• It directly affects light's speed and wavelength in materials.
• Helps determine the critical angle for total internal reflection.
• Understand how it influences interference by modifying optical path lengths.

In sum, the index of refraction is essential to the physics of optics, allowing us to precisely control and predict the behavior of light in different media.