Explanation 1) Concept: By using the formula of Rayleigh’s criterion, we can find the central beam at the target. To find the ratio of beam intensity, we can consider the ratio of diameter of beam at the end of the wire and at the target. 2) Given: i) Nuclear-pumped X-ray lasers are seen as a possible weapon to destroy ICBM booster rockets, at ranges up to 2000 km. ii) One limitation of this device is the spreading of the beam due to diffraction, resulting in dilution of the beam intensity. iii) A laser is operating with a wavelength, λ = 1.40 × 10 - 9 m iv) The element that is emitting this light, is at the end of a wire of diameter d = 2.00 × 10 - 3 m v) The laser is fired from space, so neglect any atmospheric absorption. 3) Formulas θ R = 1.22 λ d = D L 4) Calculation a) We know the Rayleigh’s angle: θ R = 1.22 λ d = D L 1.22 1.40 × 10 - 9 2.00 × 10 - 3 = D 2000 × 10 3 D = 1

10th Edition

ISBN: 9781118230718

Author: David Halliday

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Concept explainers

Compton effect

The incoming photons' energy must be in the range of an X-ray frequency to generate the Compton effect. The electron does not lose enough energy that reduces the wavelength of scattered photons towards the visible spectrum. As a result, with visible ligh…

Recoil Velocity

The amount of backward thrust or force experienced by a person when he/she shoots a gun in the forward direction is called recoil velocity. This phenomenon always follows the law of conservation of linear momentum.

Question

Chapter 36, Problem 33P

To determine

To calculate:

a) the diameter of central beam at a target 2000 km away from the beam source.

b) the ratio of beam intensity at the target to that at the end of the wire.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 36, Problem 32P

Chapter 36, Problem 34P

Students have asked these similar questions

For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =

Chapter 36 Solutions

Fundamentals of Physics

Ch. 36 - You are conducting a single-slit diffraction...Ch. 36 - In a single-slit diffraction experiment, the top...Ch. 36 - For three experiments, Fig. 36-30 gives the...Ch. 36 - For three experiments, Fig. 36-31 gives versus...Ch. 36 - Figure 36-32 shows four choices for the...Ch. 36 - Prob. 6Q Ch. 36 - At night many people see rings called entoptic...Ch. 36 - a For a given diffraction grating, does the...Ch. 36 - Figure 36-33 shows a red line and a green line of...Ch. 36 - For the situation of Question 9 and Fig. 36-33, if...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

a) the diameter of central beam at a target 2000 km away from the beam source. b) the ratio of beam intensity at the target to that at the end of the wire.

Solution Summary: The author calculates the diameter of the central beam at a target 2000 km away from the beam source.

Concept explainers

To calculate: a) the diameter of central beam at a target 2000 km away from the beam source. b) the ratio of beam intensity at the target to that at the end of the wire.

Want to see the full answer?

Chapter 36 Solutions

To calculate:

a) the diameter of central beam at a target 2000 km away from the beam source.

b) the ratio of beam intensity at the target to that at the end of the wire.