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Chapter 27: Problem 47

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

Short Answer

Expert verified
The estimated slit spacing is approximately 1.95 碌m.

Step by step solution

01

Understand the Problem

We are given a diffraction grating problem where we need to estimate the spacing between the slits of the grating, known as the slit separation or grating spacing, denoted by \( d \). We have a wavelength \( \lambda = 780 \text{ nm} \), a distance \( L = 3.0 \text{ mm} \) from the grating to the point where the beams are measured, and the distance between first-order beams \( x = 1.2 \text{ mm} \).
02

Apply the Diffraction Grating Formula

The formula for the position \( x \) of the first-order maximum in a diffraction grating pattern for small angles is \( x = \frac{m\lambda L}{d} \), where \( m \) is the order of the beam, \( \lambda \) is the wavelength, \( L \) is the distance from the grating, and \( d \) is the slit spacing. Given that \( m = 1 \), rearrange the formula to solve for \( d \): \( d = \frac{m\lambda L}{x} \).
03

Convert Units

Convert all units to consistent units. Since wavelength is given in nanometers, convert \( \lambda \) to millimeters: \( \lambda = 780 \text{ nm} = 0.780 \times 10^{-3} \text{ mm} \). Use the given \( L = 3.0 \text{ mm} \) and \( x = 1.2 \text{ mm} \).
04

Calculate Slit Spacing

Substitute the converted values into the formula: \[d = \frac{1 \times 0.780 \times 10^{-3} \times 3.0}{1.2} \]Perform the calculation: \[d = \frac{0.00234}{1.2} = 0.00195 \, \text{mm} \].
05

Conclusion

The slit spacing \( d \) is approximately \( 1.95 \times 10^{-3} \text{ mm} \) or \( 1.95 \text{ 碌m} \). This is the estimated distance between the slits of the diffraction grating.

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Key Concepts

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

Understanding Wavelength
Wavelength is a fundamental concept in the study of waves, including light waves. It represents the distance between two consecutive peaks (or troughs) of a wave. The symbol for wavelength is \( \lambda \), and it is typically measured in nanometers (nm) for light waves. One nanometer is one-billionth of a meter. Wavelength is crucial in defining the color of light. For instance, visible light ranges from about 400 nm (violet) to 700 nm (red). With shorter wavelengths like UV light, the beam is more energetic, while longer wavelengths correspond to less energetic infrared light. In our problem about the compact disc player's laser, the specified wavelength is 780 nm. This wavelength puts it in the infrared spectrum, which is invisible to the human eye, but often used in devices such as remote controls and CD players. Knowing the wavelength is critical because it directly influences diffraction patterns produced when light passes through a grating.
  • Wavelength (\( \lambda \)) determines the interference pattern.
  • Measured in nanometers (1 nm = \( 10^{-9} \) meters).
  • Impacts the resultant color or whether it is visible or invisible light.
Slit Spacing in Diffraction Gratings
Slit spacing, also known as grating spacing, is the distance between adjacent slits in a diffraction grating. It is denoted by the symbol \( d \). The slit spacing is a critical factor in determining the diffraction pattern produced when waves pass through the grating.When light waves encounter a grating, they diffract, or spread out, and interfere with each other to form a pattern of light and dark bands. The separation of these bands depends on the wavelength of the light and the slit spacing. Specifically, the formula that guides this relationship is \( d = \frac{m \lambda L}{x} \).
  • \( d \) stands for slit spacing and is usually measured in micrometers (碌m).
  • The separation \( x \) can be calculated when slit spacing is known.
  • A smaller slit spacing results in more spread out diffraction patterns.
In our example, we used this understanding and formula to calculate the slit spacing in a CD player's diffraction grating. This calculation relies on precise measurements and conversion of units to maintain consistency.
First-Order Maximum
A first-order maximum in a diffraction pattern refers to the first bright spot or fringe that appears on either side of the central bright spot (zero-order maximum). These maxima occur due to constructive interference where the path difference between light from adjacent slits is an integer multiple of the wavelength.In our scenario, we're looking at the first maximum or fringe, denoted by the order \( m = 1 \). The formula used to determine this position is derived to include the integer order of the maximum, the wavelength, the distance from the grating, and the slit spacing.
  • First-order maxima (\( m = 1 \)) are crucial for determining slit spacing and beam separation.
  • Correct interpretation is crucial for deriving formula outcomes like \( d \).
  • First-order maxima provide insights into structural properties of materials.
In the provided solution, knowing the distance between first-order maxima helps to reverse engineer the grating's slit spacing, which is vital for applications such as assessing the quality of materials in CDs or even conducting precise scientific measurements.

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Most popular questions from this chapter

A soap film \((n=1.33)\) is \(375 \mathrm{nm}\) thick and coats a flat piece of glass \((n=1.52) .\) Thus, air is on one side of the film and glass is on the other side, as the figure illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through air and strikes the film nearly perpendicularly. Concepts: (i) What, if any, phase change occurs when light, traveling in air, reflects from the air-film interface? (ii) What, if any, phase change occurs when light, traveling in the film, reflects from the film-glass interface? (iii) Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum? Calculations: For what wavelengths in the range of 380 to 750 nm does constructive interference cause the film to look bright in reflected light?

How many dark fringes will be produced on either side of the central maximum if light \((\lambda=651 \mathrm{nm})\) is incident on a single slit that is \(5.47 \times 10^{-6} \mathrm{m}\) wide?

The light shining on a diffraction grating has a wavelength of \(495 \mathrm{nm}\) (in vacuum). The grating produces a second-order bright iringe whose position is defined by an angle of \(9.34^{\circ} .\) How many lines per centimeter does the grating have?

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure \(27.6 a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the monochromatic light?

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm}),\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths. The ability to exhibit interference effects is a fundamental characteristic of any kind of wave. Our understanding of these effects depends on the principle of linear superposition, which we first encountered in Chapter 17\. Only by means of this principle can we understand the constructive and destructive interference of light waves that lie at the heart of every topic in this chapter. Problem 67 serves as a review of the essence of this principle. Problem 68 deals with thin-film interference and reviews the factors that must be considered in such cases.
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