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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
The slit widths are \(a_a, a_b, a_c\) for the respective wavelengths.

Step by step solution

01

Formula for the first minimum in a diffraction pattern

Use the formula for the first minimum in a diffraction pattern, which is given as \(a \sin \theta = m \lambda\). We know that for the first minimum \(m = 1\). So, for our given situation the formula becomes \(a \sin \theta = \lambda\).
02

Calculate the angle using known values

Since the width of the central maximum is given as \(6.00 \mathrm{mm}\) and the distance to the screen is \(2.50 \mathrm{~m}\), we can derive the angle \(\theta\) as \( \theta = \tan^{-1} \frac{y}{L}\), where \(y = 3.00 \mathrm{mm}\) (half of the maximum width) and \(L = 2.50 \mathrm{m}\). As the angle will be small, we can approximate \( \sin \theta \approx \tan \theta\). So, not to carry out these calculations for every part of the problem, it is feasible to use \(\sin \theta\) directly in our main formula.
03

Finding the slit width

After substituting the values of \(\sin \theta\) and \(\lambda\) into the equation from step 1, we can solve for \(a\). Doing so for all three cases gives:\(a) for \(\lambda = 500 \mathrm{nm}\), \(a = \frac{\lambda}{\sin \theta}\) gives \(a_a\);\(b) for \(\lambda = 50.0 \mu \mathrm{m}\), substituting these values in the same equation gives \(a_b\);\(c) for \(\lambda = 0.500 \mathrm{nm}\), REPLACEMENT into our formula gives \(a_c\). It may be necessary to convert all measurements to the same units (meters) for coherent results.

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Key Concepts

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

Wavelength
When we talk about wavelength, we are referring to the distance between successive crests of a wave. This is a critical concept in understanding wave behavior and properties, especially in the context of light and electromagnetic radiation. Wavelength is usually denoted by the Greek letter \(\lambda\) and is measured in meters or its subunits like nanometers (nm) or micrometers (µm).
  • In the context of diffraction, the wavelength of light determines how waves interact with obstacles like slits. Longer wavelengths tend to bend more around obstacles, which affects the overall diffraction pattern.
  • For example, visible light has different wavelengths than infrared or x-rays, which directly influences how each type of light diffracts through a slit.
By understanding wavelength, we can predict and calculate how different types of light will behave in diffraction scenarios. In the exercise, using different wavelengths helps illustrate how the slit width is affected by different types of electromagnetic radiation.
Diffraction Pattern
Diffraction patterns are fascinating and essential to the field of optics. A diffraction pattern is a series of dark and bright regions created when a wave encounters an obstacle or aperture. This occurs due to the bending of waves around the edges of the obstacle.
  • The central maximum is typically the brightest and widest part of a diffraction pattern, caused by most of the waves arriving in phase.
  • Subsequent dark and bright bands are called minima and maxima, and they gradually decrease in brightness with distance from the center.
The shape and size of the diffraction pattern depend on the width of the obstacle (or slit) and the wavelength of the wave passing through. In the given problem, understanding the diffraction pattern was key to calculating the slit width by applying principles like the angle approximation and positions of the minima and maxima. The detailed analysis of these patterns helps scientists and engineers in fields like optics, physics, and material science.
Slit Width
Slit width is a crucial factor in the analysis of diffraction patterns. It refers to the physical measurement of the opening through which waves pass, usually denoted by \(a\) in mathematical equations.
  • The relationship between the slit width and wavelength significantly affects the resulting diffraction pattern, determining the spread and appearance of resulting maxima and minima.
  • Narrower slits cause more significant diffraction, meaning the waves spread over a larger angle, resulting in a wider diffraction pattern.
In the exercise, knowing the slit width was crucial for determining the angle and calculating how it aligns with the wavelength to form a diffraction pattern. By calculating the slit width for different wavelengths, students can gain a practical understanding of how physical dimensions interact with wave properties.
Electromagnetic Radiation
Electromagnetic radiation encompasses various types of waves, including visible light, infrared, and x-rays among others. Each type of electromagnetic wave carries energy through space and exhibits wave-like behavior, including diffraction.
  • These waves have different frequencies and wavelengths, affecting how they interact with matter like slits or obstacles.
  • Higher frequency waves, such as x-rays, have shorter wavelengths which affect their diffraction differently compared to lower-frequency waves like infrared.
The exercise challenges students to work with different types of electromagnetic waves, understanding how each type's wavelength influences the diffraction through a slit. By analyzing diffraction patterns of various electromagnetic types, students can appreciate the vast spectrum and complexity of light and how it applies across scientific and technological fields. These insights are not only foundational for theoretical physics but also practical applications such as imaging, communications, and environmental sensing.

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Most popular questions from this chapter

Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

Light of wavelength \(585 \mathrm{nm}\) falls on a slit \(0.0666 \mathrm{~mm}\) wide. (a) On a very large and 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 calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest that \(m\) can be? \((\mathrm{b})\) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Coherent electromagnetic waves with wavelength \(\lambda=500 \mathrm{nm}\) pass through two identical slits. The width of each slit is \(a,\) and the distance between the centers of the slits is \(d=9.00 \mathrm{~mm}\). (a) What is the smallest possible width \(a\) of the slits if the \(m=3 \max -\) imum in the interference pattern is not present? (b) What is the next larger value of the slit width for which the \(m=3\) maximum is absent?

Quasars, an abbreviation for quasi-stellar radio sources, are distant objects that look like stars through a telescope but that emit far more electromagnetic radiation than an entire normal galaxy of stars. An example is the bright object below and to the left of center in Fig. \(\mathrm{P} 36.58 ;\) the other elongated objects in this image are normal galaxies. The leading model for the structure of a quasar is a galaxy with a supermassive black hole at its center. In this model, the radiation is emitted by interstellar gas and dust within the galaxy as this material falls toward the black hole. The radiation is thought to emanate from a region just a few light-years in diameter. (The diffuse glow surrounding the bright quasar shown in Fig. \(\mathrm{P} 36.58\) is thought to be this quasar's host galaxy.) To investigate this model of quasars and to study other exotic astronomical objects, the Russian Space Agency has placed a radio telescope in a large orbit around the earth. When this telescope is \(77,000 \mathrm{~km}\) from earth and the signals it receives are combined with signals from the ground-based telescopes of the VLBA, the resolution is that of a single radio telescope \(77,000 \mathrm{~km}\) in diameter. What is the size of the smallest detail that this arrangement can resolve in quasar \(3 \mathrm{C} 405,\) which is \(7.2 \times 10^{8}\) light-years from earth, using radio waves at a frequency of \(1665 \mathrm{MHz}\) ? (Hint: Use Rayleigh's criterion.) Give your answer in lightyears and in kilometers.

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

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