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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 for visible light, 41.7 mm for infrared, 0.417 µm for x-rays.

Step by step solution

01

Understand the Problem

We need to find the width of a slit, \(a\), that causes a central maximum of \(6.00 \mathrm{mm}\) in a diffraction pattern observed \(2.50 \mathrm{m}\) away. We'll solve this for different wavelengths: 500 nm, 50.0 µm, and 0.500 nm.
02

Use the Diffraction Formula

The formula for the width of the central maximum for a single-slit diffraction pattern is \(w = \frac{2 \lambda L}{a}\), where \(w\) is the width of the central maximum, \(\lambda\) is the wavelength, \(L\) is the distance to the screen, and \(a\) is the slit width.
03

Rearrange the Formula to Solve for Slit Width

Rearrange the formula to solve for \(a\):\[a = \frac{2 \lambda L}{w}\]
04

Calculate for \(\lambda = 500\) nm (visible light)

Substitute \(\lambda = 500 \times 10^{-9} \mathrm{m}\), \(L = 2.50 \mathrm{m}\), and \(w = 6.00 \times 10^{-3} \mathrm{m}\) into the formula:\[a = \frac{2 \times 500 \times 10^{-9} \times 2.50}{6.00 \times 10^{-3}}\]Calculate \(a\):\[a = \frac{2500 \times 10^{-9}}{6 \times 10^{-3}} = 4.17 \times 10^{-4} \mathrm{m} = 0.417 \mathrm{mm}\]
05

Calculate for \(\lambda = 50.0\) µm (infrared radiation)

Substitute \(\lambda = 50.0 \times 10^{-6} \mathrm{m}\), \(L = 2.50 \mathrm{m}\), and \(w = 6.00 \times 10^{-3} \mathrm{m}\):\[a = \frac{2 \times 50.0 \times 10^{-6} \times 2.50}{6.00 \times 10^{-3}}\]Calculate \(a\):\[a = \frac{250 \times 10^{-6}}{6 \times 10^{-3}} = 0.0417 \mathrm{m} = 41.7 \mathrm{mm}\]
06

Calculate for \(\lambda = 0.500\) nm (x-rays)

Substitute \(\lambda = 0.500 \times 10^{-9} \mathrm{m}\), \(L = 2.50 \mathrm{m}\), and \(w = 6.00 \times 10^{-3} \mathrm{m}\):\[a = \frac{2 \times 0.500 \times 10^{-9} \times 2.50}{6.00 \times 10^{-3}}\]Calculate \(a\):\[a = \frac{2.5 \times 10^{-9}}{6 \times 10^{-3}} = 4.17 \times 10^{-7} \mathrm{m} = 0.417 \mathrm{\mu m}\]

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation already plays a pivotal role in our day-to-day life. It refers to a variety of energies that travel through space in the form of waves. These energies can range from visible light that we see with our eyes, to infrared that warms us up, and even to x-rays that help us see inside our bodies. Each type of radiation is characterized by its wavelength, which is the distance between two consecutive peaks of the wave. Wavelengths can vary significantly, from the tiny x-rays with wavelengths measured in nanometers, to long radio waves that span several kilometers.

Understanding electromagnetic radiation helps us work with technologies like smartphones, wireless internet, and medical imaging. Moreover, it's also central to the study of optics, the branch of physics that deals with light and its properties.
Diffraction Pattern
Diffraction is a phenomenon that occurs when a wave encounters an obstacle or opening. Imagine the wave as water passing through a narrow hole. Instead of continuing straight on, the wave spreads out, creating a pattern. This is what we call a diffraction pattern. In the context of light, the diffraction pattern results from light waves bending as they pass through small slits.

When light waves encounter a slit in an obstacle, they generate a series of dark and bright regions on a screen placed behind the slit. This happens because the waves interact with each other, sometimes amplifying the brightness (constructive interference) and sometimes canceling it out (destructive interference). The central bright spot is usually wider and more pronounced.

The study of diffraction patterns is vital in understanding wave behaviors. It allows scientists to determine properties of light, such as its wavelength, and is used in various technologies, including optical instruments and even examining the structure of molecules in crystallography.
Wavelength Calculation
Calculating the wavelength of electromagnetic radiation is crucial for understanding the ways it interacts with matter. In the single-slit diffraction method, we use the wavelength to determine the properties of the generated patterns. The characteristic formula used in such calculations is derived from geometry and wave interference principles.

An essential formula in single-slit diffraction is:
  • \[w = \frac{2 \lambda L}{a}\]
Where:
  • \( w \) is the width of the central diffraction maximum,
  • \( \lambda \) is the wavelength of the incoming wave,
  • \( L \) is the distance from the slit to the screen,
  • \( a \) is the width of the slit.
By rearranging this formula, we can solve for the slit width \( a \) when given the wavelength. This process highlights how each wavelength uniquely determines the diffraction pattern, underscoring the significance of wavelength analysis in various scientific and industrial applications.
Central Maximum
In the study of single-slit diffraction, the central maximum is the central bright band of the diffraction pattern. It is the most visible part of the sequence produced when light passes through a slit and spreads out onto a screen. This pattern arises due to the intervention of waves spreading from the slit, causing overlapping zones of light and dark.

The central maximum is particularly significant because it is wider and more intense than the other maxima, which gives it the name 'central.' The size of this central maximum can be used to deduce properties of the slit and the light passing through it. By knowing the screen distance and measuring the central maximum's width, we can determine the slit width using the wavelength of the light.

This is pivotal for understanding many optical phenomena and is applied in creating precise instruments that rely on diffraction patterns for accurate measurements, such as spectrometers and microscopes.

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Most popular questions from this chapter

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength 681 \(\mathrm{nm}\) , find (a) the number of slits per centimeter for the grating and \((b)\) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

Red light of wavelength 633 \(\mathrm{nm}\) from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

Monochromatic light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is \(\frac{1}{7}\) the center-to-center distance between the slits. (a) Which interference maxima are missing in the pattem on the screen? (b) Does the answer to part (a) depend on the wavelength of the light used? Does the location of the missing maxima depend on the wavelength?

The Very Long Baseline Array can resolve (by Rayleigh's criterion) signals from sources separated by \(1.0 \times 10^{-8} \mathrm{rad} .\) If the effective diameter of the receiver is 8000 \(\mathrm{km}\) , what is the wavelength of these signals?

Light of wavelength 585 \(\mathrm{nm}\) falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calclating all the anglest (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be 7 (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

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