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Chapter 36: Problem 30

A laser beam of wavelength \(\lambda\) = 632.8 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 CD are 1.60 \(\mu\)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 \(\mu\)m apart. Repeat the calculation of part (a) for the DVD.

Short Answer

Expert verified
(a) CD: 胃 鈮� 23.3掳 for m=1, 胃 鈮� 52.0掳 for m=2; (b) DVD: 胃 鈮� 58.7掳 for m=1. Higher orders not possible for DVD.

Step by step solution

01

Understanding the Condition for Intensity Maximum

The condition for maximum intensity in reflection, i.e., constructive interference, is given by the diffraction grating formula:\[m \lambda = d \sin \theta\]where \(m\) is the order of the maximum, \( \lambda \) is the wavelength of the incident light (632.8 nm), \(d\) is the distance between tracks (1.60 \mu m for the CD and 0.740 \mu m for the DVD), and \(\theta\) is the angle of reflection from the normal.
02

Calculating the Angle for Maximum Intensity for CD

For the CD, use the distance \(d = 1.60 \mu m = 1600 \ nm\). Now rearrange the equation for \(\sin \theta\):\[sin \theta = \frac{m \lambda}{d}\]For the first order maximum (\(m = 1\)), calculate \(\sin \theta\):\[\sin \theta = \frac{1 \times 632.8 \ nm}{1600 \ nm} = 0.3955\]Calculate \(\theta\) using \(\theta = \sin^{-1}(0.3955)\).
03

Calculating Possible Orders for CD

Check possible orders by considering \(m = 2, 3, ...\) as long as \(\sin \theta \leq 1\). For example, for \(m = 2\):\[\sin \theta = \frac{2 \times 632.8}{1600} = 0.7910\]Calculate \(\theta = \sin^{-1}(0.7910)\) and continue finding \(\theta\) for higher \(m\) values.
04

Repeat the Calculation for the DVD

Repeat the process for the DVD with track distance \(d = 0.740 \mu m = 740 \ nm\) and the same wavelength (632.8 nm). Start with \(m = 1\):\[\sin \theta = \frac{632.8}{740} = 0.8549\]Then, calculate \(\theta = \sin^{-1}(0.8549)\).
05

Calculating Possible Orders for DVD

Similar to the CD, calculate possible orders for the DVD by checking higher\(m\) values until \(\sin \theta > 1\). For example, for \(m = 2\):\[\sin \theta = \frac{2 \times 632.8}{740} = 1.7098\]Since \(\sin \theta > 1\), only \(m = 1\) is valid for the DVD.

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Key Concepts

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

Constructive Interference
When light waves meet at a point, they can combine to form a new wave pattern. This phenomenon is called interference. Constructive interference happens when the crests of two waves align together and amplify the wave. This results in a brighter light spot.
In the case of a diffraction grating, like the surface of a CD or a DVD, constructive interference occurs at specific angles. These angles are determined by the equation:
  • \[ m \lambda = d \sin \theta \]
Here, \( m \) refers to the order of maximum, \( \lambda \) is the wavelength of the laser light, \( d \) is the distance between the grating lines, and \( \theta \) is the angle of reflection producing maximum brightness.
Laser Beam
A laser beam is a focused source of light characterized by its coherent and monochromatic properties. This means the laser light waves have a single wavelength and are aligned in phase, making them highly predictable and stable.
This uniformity is crucial in experiments where precise measurements are needed, like diffraction. When a laser beam hits a CD or DVD, its reflection patterns depend heavily on these properties. The beam interacts with the nanostructures on the disc's surface, gaining insights into how the light waves interfere with each other.
This interaction is key in tasks involving wavelength calculation and determining angles where the light intensity becomes maximal.
Wavelength Calculation
Calculating the wavelength in a diffraction grating setup involves using the observed angles of reflection where maximum brightness is noted. The formula is used:
  • \[ m \lambda = d \sin \theta \]
Where \( \lambda \) is the unknown wavelength if the other values are known. The distance \( d \) is the spacing between the reflecting lines on the disc, and \( \theta \) is the angle observed where the light is brightest.
If you know the wavelength, you can rearrange the equation to find the different angles \( \theta \), which tell us where constructive interference occurs. This calculation helps understand how the disc's grating interacts with the particular laser light.
Angle of Reflection
Understanding the angle of reflection in the context of a diffraction grating is important for determining how light behaves when it encounters a surface with closely spaced lines, such as a CD or DVD. The angle of reflection is measured from an imaginary line perpendicular to the disc's surface, known as the normal.
Using the diffraction formula \( m \lambda = d \sin \theta \), we calculate this angle to find points of constructive interference, where the light's intensity is maximized. It's important to experiment with different values of \( m \) to find all possible angles \( \theta \) where brightness peaks.
Understanding these angles enables the design and production of optical storage media and informs the principles of optics technology in devices.

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Most popular questions from this chapter

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

A slit \(0.240 \mathrm{~mm}\) wide is illuminated by parallel light rays of wavelength \(540 \mathrm{nm} .\) The diffraction pattern is observed on a screen that is \(3.00 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda\) = 587.9782 nm and \(\lambda\) = 587.8002 nm. You wish to resolve these spectral lines in first order using a grating 1.20 cm in length. What minimum number of slits per centimeter must the grating have?

Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?
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