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Using Lloyd's mirror, X-ray fringes were observed, the spacing of which was found to be \(0.0025 \mathrm{cm}\). The wavelength used was \(8.33 \AA .\) If the source-screen distance was \(3 \mathrm{m}\), how high above the mirror plane was the point source of X-rays placed?

Short Answer

Expert verified
The height of the point source above the mirror is approximately 9.00 脳 10鈦 cm.

Step by step solution

01

Convert Units

First, convert the given wavelength from angstroms to centimeters. Since 1 \, AA = 10^{-8} \, cm, the wavelength in centimeters is \( 8.33 \, \text{AA} = 8.33 \times 10^{-8} \, \text{cm} \).
02

Identify Relationship

Using the relationship for fringe spacing in Lloyd's mirror, which is given by \( d = \frac{\lambda \times H}{L} \), where \( d \) is the fringe spacing, \( \lambda \) is the wavelength, \( H \) is the height of the point source above the mirror, and \( L \) is the source-screen distance.
03

Substitute Values

Substitute the known values into the equation: \[ 0.0025 = \frac{8.33 \times 10^{-8} \times H}{3} \].
04

Solve for Height

Rearrange the equation to solve for \( H \):\[ H = \frac{0.0025 \times 3}{8.33 \times 10^{-8}} \].Perform the calculation to find \( H \).
05

Calculate Result

Carry out the calculations:\[ H = \frac{0.0075}{8.33 \times 10^{-8}} \approx 9.00 \times 10^{4} \text{ cm} \].

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Key Concepts

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

X-ray fringes
X-ray fringes are a fascinating display in the study of light and optics. They occur when X-rays interfere after reflecting from a surface, such as in Lloyd's mirror setup. In this setup, X-rays from a narrow point source strike a mirror at a shallow angle. The direct rays and the reflected rays converge and interfere with each other. The outcome of this interference is a pattern of bright and dark fringes, known as interference fringes.

This phenomena is invaluable for studying the properties of light, particularly X-ray light, which has extremely short wavelengths. Understanding these fringe patterns can help us determine various features like wavelength, the nature of the medium through which they pass, and the distances involved in optical experiments. Lloyd鈥檚 mirror, specifically, provides a simple yet powerful way to observe such interference, making it a staple in optical studies.
wavelength conversion
Wavelength conversion is an essential step in the field of optics, especially when dealing with calculations in interference patterns. In many scenarios, wavelengths are given in units such as angstroms (脜), which need to be converted to more convenient units like centimeters, particularly when other distances in the problem are measured in those units. One angstrom is equal to \(10^{-8}\) centimeters.

In the given problem, a wavelength of 8.33 脜 was provided. To work effectively with other components of the problem鈥攍ike fringe spacing in centimeters or source-screen distances in meters鈥攊t was crucial to convert the wavelength: \8.33 \, \text{脜} = 8.33 \, \times 10^{-8} \, \text{cm}\. This conversion allows for a coherent and consistent application of mathematical formulas related to interference and diffraction.
fringe spacing calculation
Fringe spacing calculation is a key step in solving problems related to interference patterns produced by X-rays. The specific formula used in Lloyd's mirror for fringe spacing is \(d = \frac{\lambda \times H}{L}\). In this formula, \(d\) represents the fringe spacing, \(\lambda\) the wavelength, \(H\) the height of the point source above the mirror, and \(L\) the source-screen distance.

This formula highlights how these variables are interrelated and how changes in one can affect the overall pattern. By substituting the known values from a problem鈥攍ike the already converted wavelength, fringe spacing, and source-screen distance鈥攜ou can rearrange the formula to solve for unknowns, such as the height of the point source above the mirror plane.
optics problem solving
Optics problem solving involves using foundational understanding of light behavior, mathematical equations, and unit conversions to arrive at a solution. With Lloyd's mirror and interference fringes, the process often involves several steps:

  • Understanding the Setup: Recognize the relationship between the X-rays, the mirror, and the distance measurements involved.
  • Correct Unit Conversion: Convert all relevant quantities to consistent units, such as angstroms to centimeters.
  • Apply Mathematical Relationships: Use formulas like \(d = \frac{\lambda \times H}{L}\) to connect physical quantities.
  • Reuse and Rearrange: Adjust the formulas to solve for unknown variables, in this instance, calculating the height \(H\) from given measurements.
  • Verify Results: Ensure the result is logical and matches the physical context of the problem, such as checking that distances make sense physically.
This structure of problem solving aids in breaking down complex scenarios into manageable, step-by-step processes, making optics problems more approachable and understandable.

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Most popular questions from this chapter

Newton rings are observed on a film with quasimonochromatic light that has a wavelength of \(500 \mathrm{nm}\). If the 20 th bright ring has a radius of \(1 \mathrm{cm}\), what is the radius of curvature of the lens forming one part of the interfering system?

Imagine that you have an opaque screen with three horizontal very narrow parallel slits in it. The second slit is a center-to-center distance \(a\) beneath the first, and the third is a distance \(5 a / 2\) beneath the first. (a) Write a complex exponential expression in terms of \(\delta\) for the amplitude of the electric field at some point \(P\) at an elevation \(\theta\) on a distant screen where \(\delta=k a \sin \theta .\) Prove that $$I(\theta)=\frac{I(0)}{3}+\frac{2 I(0)}{9}(\cos \delta+\cos 3 \delta / 2+\cos 5 \delta / 2)$$ Verify that at \(\theta=0, I(\theta)=I(O)\)

An expanded beam of red light from a He-Ne laser \(\left(\lambda_{0}=632.8\right.\) nm) is incident on a screen containing two very narrow horizontal slits separated by \(0.200 \mathrm{mm}\). A fringe pattern appears on a white screen held \(1.00 \mathrm{m}\) away. (a) How far (in radians and millimeters) above and below the central axis are the first zeros of irradiance? (b) How far (in \(\mathrm{mm}\) ) from the axis is the fifth bright band? (c) Compare these two results.

Red plane waves from a ruby laser \(\left(\lambda_{0}=694.3 \mathrm{nm}\right)\) in air impinge on two parallel slits in an opaque screen. A fringe pattern forms on a distant wall, and we see the fourth bright band \(1.0^{\circ}\) above the central axis. Kindly calculate the separation between the slits.

A \(3 \times 5\) card containing two pinholes, \(0.08 \mathrm{mm}\) in diameter and separated center to center by \(0.10 \mathrm{mm}\), is illuminated by parallel rays of blue light from an argon ion laser \(\left(\lambda_{0}=487.99 \mathrm{nm}\right) .\) If the fringes on an observing screen are to be \(10 \mathrm{mm}\) apart, how far away should the screen be?

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