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Chapter 36: Problem 2

Parallel rays of green mercury light with a wavelength of 546 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 60.0 \(\mathrm{cm} .\) In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 \(\mathrm{mm} .\) What is the width of the slit?

Short Answer

Expert verified
The slit width is approximately 32.1 碌m.

Step by step solution

01

Understanding the Problem

We need to find the width of the slit through which parallel rays of green mercury light pass. We are given the wavelength of the light (546 nm), the focal length of the lens (60 cm), and the distance from the central maximum to the first minimum in the diffraction pattern in the focal plane (10.2 mm).
02

Identifying the Formula

For a single slit diffraction pattern, the position of the first minimum is given by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( m \) is the order of the minimum (\( m = 1 \) for the first minimum), and \( \lambda \) is the wavelength of the light.
03

Relating Angle and Distance

Since the angle \( \theta \) for small angles can be approximated using \( \theta \approx \tan \theta \approx \frac{y}{f} \), where \( y \) is the distance from the central maximum to the first minimum (10.2 mm or 0.0102 m) and \( f \) is the focal length (60 cm or 0.60 m).
04

Calculating the Angle \( \theta \)

Using the small angle approximation, \( \theta \approx \frac{0.0102}{0.60} \).
05

Substituting Values into the Equation

The sine of the angle can be written as \( \sin \theta = \frac{0.0102}{0.60} \). Substituting \( m=1 \) and \( \lambda = 546 \times 10^{-9} \ m \) into \( a \sin \theta = m \lambda \), we get \( a \cdot \frac{0.0102}{0.60} = 1 \cdot 546 \times 10^{-9} \).
06

Solving for Slit Width \( a \)

Rearranging \( a \cdot \frac{0.0102}{0.60} = 546 \times 10^{-9} \) gives \( a = \frac{546 \times 10^{-9} \times 0.60}{0.0102} \). Calculating this gives \( a \approx 3.21 \times 10^{-5} \ m \) or \( 32.1 \ \mu m \).

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Key Concepts

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

Single Slit Diffraction
Single slit diffraction is an intriguing phenomenon that occurs when waves pass through a narrow slit and spread out or diffract. This effect is most noticeable when the slit width is comparable to the wavelength of the light. Instead of traveling in a straight path, light waves bend around the edges of the slit, creating a diffraction pattern of light and dark bands.
This phenomenon is explained by the principle of superposition, where the waves interfere with each other, creating constructive and destructive interference. Constructive interference occurs at points where waves combine to increase amplitude, resulting in bright bands, while destructive interference causes dark bands. The diffraction pattern's characteristics can be used to analyze and calculate the properties of the slit or the wave, such as in our exercise to find the slit width.
Wavelength
Wavelength plays a vital role in diffraction patterns. It is the distance between consecutive crests or any identical points of a light wave, usually measured in nanometers (nm) for light. In the exercise, the green mercury light has a wavelength of 546 nm.
Wavelength determines how much the light will diffract when it encounters a slit or obstacle. Shorter wavelengths bend less and longer wavelengths bend more. This relationship signifies that the diffraction pattern is sensitive to changes in wavelength. Knowing the wavelength is essential when calculating the diffraction pattern's features, such as the position of minima and maxima in the pattern.
Focal Length
Focal length is a crucial concept in optics. It represents the distance between the lens and the point where it converges light rays to a focus. In the given exercise, the focal length is 60 cm. This measurement is necessary for understanding how the diffraction pattern is projected onto the focal plane of the lens.
By using the lens, we can project the diffraction pattern onto a screen or plane, allowing us to measure distances, like the one from the central maximum to the first minimum. The focal length helps relate physical distances in the pattern to angles using trigonometric approximations, which are then crucial for calculating the characteristics of the slit.
Diffraction Pattern
A diffraction pattern emerges when a wave, such as light, encounters an obstruction or opening. This pattern consists of alternating light and dark fringes. When dealing with single slit diffraction, the pattern typically includes a central bright fringe or maximum flanked by multiple dark and bright fringes.
The central maximum is usually the brightest and widest, surrounded by decreasing brightness as you move away from the center. Minima occur at points of destructive interference where the wave peaks and troughs cancel out. These patterns are an essential tool for understanding wave behavior, especially in fine measurements, such as determining slit widths from light interference experiments.

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Most popular questions from this chapter

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

On December \(26,2004,\) a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than \(150,000\) people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 \(\mathrm{h}\) and the speed of the wave to be 800 \(\mathrm{km} / \mathrm{h}\) . Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about \(4500 \mathrm{km},\) while the distance between the southern end of Australia and Antarctica is about 3700 \(\mathrm{km} .\) As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

\(\cdot\) Plane monochromatic waves with wavelength 520 \(\mathrm{nm}\) are incident normally on a plane transmission grating having 350 slits/mm. Find the angles of deviation in the first, second, and third orders.

The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm} .\) White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example \(36.4 . )\)

A glass sheet is covered by a very thin opaque coating. In the middle of this sheet there is a thin scratch 0.00125 \(\mathrm{mm}\) thick. The sheet is totally immersed beneath the surface of a liquid. Parallel rays of monochromatic coherent light with wavelength 612 \(\mathrm{nm}\) in air strike the sheet perpendicular to its surface and pass through the scratch. A screen is placed in the liquid a distance of 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on the screen are 22.4 \(\mathrm{cm}\) apart. What is the refractive index of the liquid?
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