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Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{nm}\). The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

Short Answer

Expert verified
(a) Diameter at 2000 km is approximately 34.16 m. (b) Intensity ratio is approximately 3.42 \times 10^{-9}.

Step by step solution

01

Understand Beam Diffraction

The diffraction of the laser beam leads to the spreading over a distance. This spreading causes the beam to widen from its initial diameter as it travels over a distance of 2000 km. We use the formula for the angular width of the central maximum in radians for a circular aperture: \( \theta = \frac{1.22 \lambda}{D} \), where \( \lambda = 1.40 \times 10^{-9} \) m is the wavelength of the laser light and \( D = 0.200 \times 10^{-3} \) m is the diameter of the aperture.
02

Calculate Angular Width of Beam

Substitute the given values into the formula for angular width: \( \theta = \frac{1.22 \times 1.40 \times 10^{-9}}{0.200 \times 10^{-3}} = 8.54 \times 10^{-6} \) radians.
03

Determine Diameter of Beam at Target

To find the diameter of the central beam at the target, we use the formula \( x = 2 L \tan(\theta/2) \). For small angles, \( \tan(\theta) \approx \theta \), so \( x \approx 2 L \cdot \theta \). The distance \( L = 2000 \times 10^{3} \) m. Thus, \( x = 2 \times 2000 \times 10^{3} \times 8.54 \times 10^{-6} \approx 34.16 \) m.
04

Calculate Beam Intensity Ratio

Intensity is inversely proportional to the area. The initial area \( A_0 = \pi (0.100 \times 10^{-3})^2 \), and the target area \( A_t = \pi (17.08)^2 \). The intensity ratio \( \frac{I_t}{I_0} = \frac{A_0}{A_t} \), so \( \frac{I_t}{I_0} = \frac{\pi (0.100 \times 10^{-3})^2}{\pi (17.08)^2} \approx 3.42 \times 10^{-9} \).

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Key Concepts

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

Angular Width
In the context of wave phenomena, such as light, "angular width" refers to the angle over which the majority of the beam's intensity is spread. When light passes through a small opening, it spreads out and this spread is quantified by the angular width. This can often be seen in phenomena such as diffraction.
Consider a nuclear-pumped x-ray laser, which is designed to destroy missiles over great distances. Due to diffraction, the laser beam spreads out as it travels. For a circular aperture, the angular width \( \theta \) of the diffracted beam can be calculated using the formula:
  • \( \theta = \frac{1.22 \lambda}{D} \)
where \( \theta \) is in radians, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture.
In this exercise, the angular width helps determine how much the beam spreads from its starting point to a target 2000 km away. It is essential in understanding how diffraction affects the beam size over a long distance.
Beam Intensity
The intensity of a beam of light or, in this case, an x-ray laser, is essentially the power per unit area. As the beam travels and spreads due to diffraction, its intensity at a distant point decreases. This happens because the same amount of energy is distributed over a larger surface area.
Initial intensity can be computed near the laser's source, using:
  • The formula for area of a circle \( A_0 = \pi r_0^2 \)
where \( r_0 \) is the initial radius of the beam. At a 2000 km distance, the area's radius will be larger, more than 17 meters in this problem.
To find the beam intensity at such a range, the relationship between initial and target intensities can be expressed as a ratio. This ratio indicates that the intensity at the target \( I_t \) becomes a fraction of the initial intensity \( I_0 \), essentially: \[ \frac{I_t}{I_0} = \frac{A_0}{A_t} \] where \( A_t \) is the area where the beam reaches at the target. As beam spread causes significant reduction in intensity, it poses limitations on the effectiveness of such weaponry for distant targets.
X-ray Lasers
X-ray lasers are unique and advanced forms of lasers that use x-rays to generate beams of photons rather than visible light. They emit light at extremely short wavelengths, measured in nanometers, which is suitable for high-precision and high-energy applications, making them potential tools or weapons in advanced technology.
These lasers rely on a compact and powerful energy source, such as a nuclear reaction, to pump the energy levels required for emission in the x-ray spectrum.
Features and applications include:
  • Highly focused x-ray beams.
  • Potential for cutting-edge medical and technological uses, like imaging and delicate material fabrication.
  • Possible military use in targeting through its high energy output and minimal atmospheric interaction.
Despite their potential, x-ray lasers like those described in the exercise are yet to overcome significant challenges, including controlling beam spread and ensuring sustained beam intensity over long distances.

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Most popular questions from this chapter

In the two-slit interference experiment of Fig. \(35-10\), the slit widths are each \(12.0 \mu \mathrm{m},\) their separation is \(24.0 \mu \mathrm{m},\) the wavelength is \(600 \mathrm{nm},\) and the viewing screen is at a distance of \(4.00 \mathrm{~m} .\) Let \(I_{P}\) represent the intensity at point \(\bar{P}\) on the screen, at height \(y=70.0 \mathrm{~cm}\). (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where \(P\) is in the two-slit interference pattern by giving the maximum or minimum on which it lies or the maximum and minimum between which it lies. (c) In the same way, for the diffraction that occurs, determine where point \(P\) is in the diffraction pattern.

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