/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Monochromatic electromagnetic ra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen \(2.50 \mathrm{~m}\) from the slit. If the width of the central maximum is \(6.00 \mathrm{~mm}\), what is the slit width \(a\) if the wavelength is (a) \(500 \mathrm{~nm}\) (visible light); (b) \(50.0 \mu \mathrm{m}\) (infrared radiation); (c) \(0.500 \mathrm{~nm}\) (x rays)?

Short Answer

Expert verified
(a) 0.417 mm, (b) 41.7 mm, (c) 0.417 µm.

Step by step solution

01

Understanding the Diffraction Pattern

When monochromatic light passes through a slit, it creates a diffraction pattern. The width of the central maximum in this pattern is related to the slit width, the wavelength of light, and the distance from the slit to the screen.
02

Formula for Central Maximum Width

The width of the central maximum (central fringe) in single-slit diffraction can be determined by the formula:\[ W = \frac{2 \lambda D}{a} \]where \(W\) is the width of the central maximum, \(\lambda\) is the wavelength of light, \(D\) is the distance from the slit to the screen, and \(a\) is the slit width we want to find.
03

Re-arranging the Formula to Solve for Slit Width

To find the slit width \(a\), we rearrange the formula:\[ a = \frac{2 \lambda D}{W} \]
04

Substituting Values for Part (a)

For visible light with wavelength \(500 \text{ nm} = 500 \times 10^{-9} \text{ m}\), substitute \(\lambda = 500 \times 10^{-9} \text{ m}\), \(D = 2.50 \text{ m}\), and \(W = 6.00 \text{ mm} = 6.00 \times 10^{-3} \text{ m}\) into the formula:\[a = \frac{2 \times 500 \times 10^{-9} \times 2.5}{6.00 \times 10^{-3}} = \frac{2500 \times 10^{-9}}{6 \times 10^{-3}} = 416.7 \times 10^{-6} \text{ m} = 0.417 \text{ mm}\]
05

Substituting Values for Part (b)

For infrared radiation with wavelength \(50.0 \text{ µm} = 50.0 \times 10^{-6} \text{ m}\), substitute \(\lambda = 50.0 \times 10^{-6} \text{ m}\) into the formula:\[a = \frac{2 \times 50.0 \times 10^{-6} \times 2.5}{6.00 \times 10^{-3}} = \frac{250.0 \times 10^{-6}}{6 \times 10^{-3}} = 41.7 \times 10^{-3} \text{ m} = 41.7 \text{ mm}\]
06

Substituting Values for Part (c)

For x-rays with wavelength \(0.500 \text{ nm} = 0.500 \times 10^{-9} \text{ m}\), substitute \(\lambda = 0.500 \times 10^{-9} \text{ m}\) into the formula:\[a = \frac{2 \times 0.500 \times 10^{-9} \times 2.5}{6.00 \times 10^{-3}} = \frac{2.5 \times 10^{-9}}{6 \times 10^{-3}} = 0.417 \times 10^{-6} \text{ m} = 0.417 \text{ µm}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Diffraction Pattern
When light, or any wave, passes through a narrow opening or slit, the wave bends around the edges of the slit. This phenomenon is known as diffraction. As a result, a diffraction pattern is formed on a screen behind the slit. This pattern typically consists of a bright central area known as the central maximum, flanked by a series of alternating dark and bright bands. The appearance of this pattern is due to the interference of light waves spreading out after passing through the slit.

Understanding the diffraction pattern helps in predicting how light will distribute upon passing through the slit. The formation of these patterns is governed by the principles of wave interference and can be mathematically described by diffraction equations. These patterns are not only fascinating to observe but are also crucial in various scientific and engineering applications, such as the analysis of optical instruments, and even in understanding the wave nature of light itself.
Central Maximum
The central maximum is the brightest and widest band in the diffraction pattern. It is located directly in line with the slit and the light source. Essentially, it is the primary area where most of the light focuses due to the constructive interference of waves.

For single-slit diffraction, the width of the central maximum is an important characteristic. It is related to the wavelength, the distance to the screen, and the slit width. The formula \[ W = \frac{2 \lambda D}{a} \] is used to calculate the width of the central maximum, where:
  • \( W \) is the width of the central maximum,
  • \( \lambda \) is the wavelength of the light,
  • \( D \) is the distance from the slit to the screen,
  • \( a \) is the width of the slit.

This formula shows that wider slits result in narrower central maxima since the waves have less of an angle to spread. Conversely, as the slit narrows, diffraction becomes more pronounced, and the central maximum becomes wider.
Wavelength
The wavelength of light is a crucial factor in determining the characteristics of a diffraction pattern. It refers to the distance between successive peaks of a wave and is typically measured in units like meters, nanometers (nm), or micrometers (µm).

Different types of electromagnetic waves (such as visible light, infrared, or x-rays) have different wavelengths. This affects how they diffract through a slit. Longer wavelengths lead to wider diffraction patterns because the waves spread out more as they pass through the slit. In the original exercise, you encounter wavelengths from visible light at 500 nm, infrared at 50 µm, to x-rays at 0.5 nm.

Understanding wavelength is crucial for various applications, including telecommunications, microscopy, and the study of crystal structures, where the diffraction of x-rays is employed to understand atomic arrangements.
Slit Width
The slit width, representing the opening through which the light or wave passes, is another critical factor influencing the diffraction pattern. It is typically represented by the variable \( a \) in equations. The slit width determines how much the incoming wave is allowed to spread out after passing through.

Manipulating the slit width affects the central maximum directly. A narrow slit causes a wider central maximum due to greater diffraction, and a wider slit results in a narrower central maximum. This idea is encapsulated by the formula:\[ a = \frac{2 \lambda D}{W} \] This equation helps in calculating the slit width when you know the other variables, like the wavelength and the width of the central maximum. Changes in slit width are utilized in different scientific instruments to analyze light and other waves. Understanding how slit width interacts with other factors allows us to harness diffraction for a variety of technological and scientific applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A converging lens \(7.20 \mathrm{~cm}\) in diameter has a focal length of \(300 \mathrm{~mm}\). If the resolution is diffraction limited, how far away can an object be if points on it \(4.00 \mathrm{~mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{~nm}\).

Parallel rays of green mercury light with a wavelength of \(545 \mathrm{~nm}\) 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 \(10.9 \mathrm{~mm}\). What is the width of the slit?

In a YDSE apparatus, two identical slits are separated by \(1 \mathrm{~mm}\) and distance between slits and screen is \(1 \mathrm{~m}\). The wavelength of light used is \(6000 \AA\). Find the minimum distance between two points on the screen having \(75 \%\) intensity of the maximum intensity.

Photography. A wildlife photographer uses a moderate telephoto lens of focal length \(135 \mathrm{~mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is \(11.5 \mathrm{~m}\) away. Assume the wavelength is \(550 \mathrm{~nm}\). (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 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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.