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\(\bullet$$\bullet\) CDs and DVDs as diffraction gratings. A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the \(\mathrm{CD}\) are 1.60\(\mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740\(\mu\) mm apart. Repeat the calculation of part (a) for the DVD.

Short Answer

Expert verified
The maximum intensity angles are 23.3° for the CD and 64.5° for the DVD for m=1.

Step by step solution

01

Understand the Problem

When light shines on a CD or DVD, the tracks act as a diffraction grating. The formula for maxima in diffraction grating is given by \( d \sin \theta = m \lambda \) where \( d \) is the distance between the tracks (the grating spacing), \( \theta \) is the angle of diffraction, \( m \) is the order of the maximum (an integer), and \( \lambda \) is the wavelength of the light.
02

Calculate the Grating Spacing for CD

For part (a), the distance between tracks on a CD is given as 1.60 µm. First, convert this into meters: \( d = 1.60 \times 10^{-6} \) m. We can assume different integer values for \( m \) and solve for \( \theta \).
03

Calculate Angles for Maximum Intensity on CD

Use the equation \( d \sin \theta = m \lambda \) to find the angles. Assuming \( \lambda = 632.8 \times 10^{-9} \) m (convert nm to m). For \( m = 1 \), solve \( 1.60 \times 10^{-6} \sin \theta = 632.8 \times 10^{-9} \), find \( \theta = \arcsin( 632.8 \times 10^{-9} / 1.60 \times 10^{-6} ) \approx 23.3^\circ \). We only calculate for \( m \) until \( \sin \theta \) remains \( \leq 1 \).
04

Calculate the Grating Spacing for DVD

For part (b), the distance between tracks on a DVD is 0.740 µm. Convert this to meters: \( d = 0.740 \times 10^{-6} \) m. Repeat the process for different integer \( m \) values to find possible angles.
05

Calculate Angles for Maximum Intensity on DVD

Again using \( d \sin \theta = m \lambda \), for \( m = 1 \), solve \( 0.740 \times 10^{-6} \sin \theta = 632.8 \times 10^{-9} \), find \( \theta = \arcsin( 632.8 \times 10^{-9}/0.740 \times 10^{-6} ) \approx 64.5^\circ \). Ensure \( \sin \theta \leq 1 \) for valid \( m \) values.

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Key Concepts

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

Laser Wavelength
When examining how light interacts with surfaces like CDs or DVDs, the concept of laser wavelength is essential. Wavelength refers to the distance between successive peaks of a wave. In optics, it's especially significant because it dictates how light waves are diffracted by grating structures.
The wavelength of light used in optical devices is typically measured in nanometers (nm). For instance, in our exercise, we used a laser with a wavelength of 632.8 nm. Understanding this value is crucial because it is the basis for calculating diffraction patterns and angles of reflection.
Laser wavelength affects how light is split into various directions when passing over a grating. The longer the wavelength, the more spread out the diffraction pattern becomes. Always keep in mind that precision in measuring and converting these units matter for accurate results: for example, converting nanometers to meters by multiplying by \(10^{-9}\).
Angle of Reflection
The angle of reflection is a key concept in optics, particularly when discussing how light interacts with surfaces. It denotes the angle at which a beam of light, after hitting a reflective surface like a CD or DVD, emerges or reflects at a maximum intensity.
In this context, the angles are measured from the normal, which is an imaginary line perpendicular to the surface. The reflection angles are calculated using the diffraction grating formula \( d \sin \theta = m \lambda \), where \( d \) is the grating spacing, \( \theta \) is the angle, \( m \) is the order of maximums, and \( \lambda \) is the laser wavelength.
To find these angles, solve for \( \theta \) for successive orders \( m \) (i.e., 1, 2, 3,...). As laser light strikes the track spacings, integers \( m \) will dictate the particular directions where light will be most intense. It's crucial to ensure that the sine value calculated remains less than or equal to 1, as it represents the maximum possible sine value for any angle.
Grating Spacing
Grating spacing, denoted by \( d \), is the distance between the tracks on a CD or DVD that act as slits in a diffraction grating setup.
In CDs and DVDs, the grating spacing affects the range and intensity of reflected light. CDs have a larger grating spacing (1.60 µm in our problem) compared to DVDs (0.740 µm).
This difference in spacing is what produces distinct diffraction patterns for the same wavelength of light. Smaller spacings on DVDs lead to wider angular separations for maxima compared to CDs. To calculate these spacings in meters, remember the conversion from micrometers: multiply by \(10^{-6}\). Understanding and using this conversion allows you to accurately determine how light behaves across different media surfaces.
Diffraction Maxima
Diffraction maxima refer to the specific angles where the light intensity becomes greatest when it is reflected off a grating like a CD or DVD.
The concept is rooted in the diffraction grating equation: \( d \sin \theta = m \lambda \), which allows for calculations of these angles. In essence, these are positions where constructive interference occurs due to the light wave alignment.
Each integer \( m \) corresponds to a different order of maxima. The higher the value of \( m \), the further out from the center the maxima occur on either side of the central peak, provided that \( \sin \theta \) remains within its natural limits (i.e., less than or equal to 1).
  • The central peak (the primary maximum) is usually the most intense and occurs at \( m = 0 \).
  • First-order maxima are usually at \( m = ±1 \), second at \( m = ±2 \), and so on, leading to a spectrum of light intensities.
Understanding how these maxima form and vary with grating properties and laser wavelength is critical in applications like CDs, DVDs, and numerous optical systems.

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Most popular questions from this chapter

\(\bullet\) A thin film of polystyrene of refractive index 1.49 is used as a nonreflecting coating for Fabulite (strontium titanate) of refractive index \(2.409 .\) What is the minimum thickness of the film required? Assume that the wavelength of the light in air is 480 \(\mathrm{nm}\) .

\(\bullet\) Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge, so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546 nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

\(\bullet$$\bullet\) Doorway diffraction. Diffraction occurs for all types of waves, including sound waves. Suppose sound of frequency 1250 Hz leaves a room through a \(1.00-\) wide doorway. At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraun- hofer diffraction to apply. You can ignore effects of reflections.

\(\bullet\) Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interfer- ence pattern is at an angle of \(8.94^{\circ} .\) What is the angular posi- tion of the fourth-order maximum?

\(\bullet\) Conserving energy. You want to coat the inner surfaces of your windows (which have refractive index of 1.51\()\) with a film in order to enhance the reflection of light back into the room so that you can use bulbs of lower wattage than usual. You find that \(\mathrm{MgF}_{2},\) with \(n=1.38,\) is not too expensive, so you decide to use it. Since incandescent home lightbulbs emit reddish light with a peak wavelength of approximately 650 \(\mathrm{nm}\) , you decide that this wavelength is the one to enhance in the light reflected back into the room. (a) What is the minimum thickness of film that you will need? (b) If this layer seems too thin to be able to put on accurately, what other thicknesses would also work? Give only the three thinnest ones.

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