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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: θ ≈ 23.3° for m=1, θ ≈ 52.0° for m=2; (b) DVD: θ ≈ 58.7° 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

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

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 is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of 11.3\(^\circ\). What is the angular position of the fourth-order maximum?

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about 8000 km. When this radio telescope is focusing radio waves of wavelength 2.0 cm, what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength 550 nm so that the visible-light telescope has the same resolution as the radio telescope?

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum \((\theta = 0^\circ)\) is \(I_0\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_0\)/2?

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