/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Quality control systems have bee... [FREE SOLUTION] | 91Ó°ÊÓ

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Quality control systems have been developed to remotely measure the diameter of wires using diffraction. A wire with a stated diameter of \(170 \mu \mathrm{m}\) blocks the beam of a \(633 \mathrm{nm}\) laser, producing a diffraction pattern on a screen \(50.0 \mathrm{cm}\) distant. The width of the central maximum is measured to be \(3.77 \mathrm{mm}\). The wire should have a diameter within \(1 \%\) of the stated value. Does this wire pass the test?

Short Answer

Expert verified
After performing the steps, one should be able to evaluate whether the wire passes the test. Remember, the wire passes the test if the calculated value from the steps lies within \(1 \%\) of the stated value \(170 \mu \mathrm{m}\).

Step by step solution

01

Understand the problem and formula

The problem is about the diffraction of light by a wire. The light's wavelength (\(633 \mathrm{nm}\)), the distance to the screen (\(50.0 \mathrm{cm}\)), and the width of the central maximum (\(3.77 \mathrm{mm}\)) are given. With these values, it's possible to determine the diameter of the wire using the formula of the angular width of the central maximum in the diffraction pattern by a single slit (or wire), which is \(\Theta = \lambda / d\), where \(\Theta\) is the angular width, \(\lambda\) is the wavelength of the light, and \(d\) is the diameter of the wire.
02

Calculate The Angular Width

To determine the angular width, we can use the formula \(y=L \cdot \tan(\Theta)\), where \(y\) represents the width of the central maximum on the screen, and \(L\) is the distance from the wire to the screen. Hence we can solve for \(\Theta\): \( \Theta = \arctan(y/L)\) Substituting the given values \( y=3.77 \mathrm{mm}=0.00377 \mathrm{m}\) and \(L=50.0 \mathrm{cm}=0.5 \mathrm{m}\), we have \( \Theta = \arctan(0.00377/0.5)\).
03

Calculate the Diameter of the Wire

Using the calculated \(\Theta\), we can find the diameter of the wire using the formula of the angular width \(\Theta = \lambda / d\). Solving for \(d\), we get \(d = \lambda / \Theta\). The wavelength \(\lambda=633 \mathrm{nm}=633 \times 10^{-9} \mathrm{m}\). After calculating \(\Theta\) from the previous step, plug its value and the given \(\lambda\) into the formula, then solve for \(d\).
04

Check the condition

Now compare the calculated diameter with the stated diameter (\(170 \mu \mathrm{m}=170 \times 10^{-6} \mathrm{m}\)). If the calculated diameter lies within \(1 \%\) of the stated value, then the wire passes the test.

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Key Concepts

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

Angular Width
In diffraction patterns, the concept of angular width is crucial to understanding the spread of light when it encounters an obstacle.
When a laser beam hits a wire, it bends around the edges and creates a diffraction pattern that can be captured on a screen. The part of this pattern that primarily draws our attention is the central maximum.
Angular width refers to the angle subtended by the width of this central maximum at the point where the light is obstructed.
To determine the angular width \( \Theta \) in the context of this exercise, we use the formula: \[ \Theta = \arctan\left(\frac{y}{L}\right) \] where:
  • \( y \) is the width of the central maximum,
  • \( L \) is the distance from the wire to the screen.
Knowing the angular width helps in calculating critical information about the obstacle, like its diameter. With an accurate measurement, quality control in manufacturing can be ensured.
Laser Wavelength
The wavelength of a laser is a defining feature of the light it produces, determining its color and how it interacts with objects. In this exercise, the laser has a wavelength \( \lambda = 633 \text{ nm} \), corresponding to red light in the visible spectrum.
The wavelength is vital when examining diffraction, as the dimensions of the diffraction pattern are directly dependent on it.
For diffraction by a wire, the relationship often used is \[ \Theta = \frac{\lambda}{d} \] where:
  • \( \Theta \) is the angular width,
  • \( d \) is the diameter of the wire creating the obstruction.
Having a precise measure of the laser wavelength allows for accurate determination of the geometrical properties of objects through diffraction patterns.
Central Maximum
The central maximum is the brightest and most prominent part of the diffraction pattern formed when light encounters an obstacle. This region is bounded by two dark fringes where light intensity drops sharply.
The width of this central maximum, denoted by \( y \) in the problem, is one of the critical measurements needed to determine the angular width and further calculate the diameter of the wire.
In practical applications, assessing the central maximum's width allows technicians to evaluate whether wire diameters meet specified standards through diffraction testing.
This method is both efficient and accurate, leveraging the predictable nature of light behavior in diffraction.
Diameter Measurement
The measurement of a wire's diameter using diffraction can be precise and non-intrusive.
By calculating the angular width produced by the diffracted laser light and knowing the wavelength, the diameter \( d \) of the wire can be derived from the formula: \[ d = \frac{\lambda}{\Theta} \] Here,
  • \( \lambda \) is the wavelength of the laser,
  • \( \Theta \) is the previously calculated angular width.
Importantly, this technique is useful in quality control to ensure that a wire's actual diameter is within acceptable limits, such as the \(1\%\) range of the stated value in production specifications.

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Most popular questions from this chapter

A physics instructor wants to project a spectrum of visiblelight colors from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 500 lines per \(\mathrm{mm},\) projecting a pattern on a screen \(2.4 \mathrm{m}\) behind the grating. a. How wide is the spectrum that corresponds to \(m=1 ?\) b. How much distance separates the end of the \(m=1\) spectrum and the start of the \(m=2\) spectrum?

Infrared light of wavelength \(2.5 \mu \mathrm{m}\) illuminates a \(0.20-\mathrm{mm}-\) diameter hole. What is the angle of the first dark fringe in radians? In degrees?

Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen \(75.0 \mathrm{cm}\) behind the grating. First-order maxima are observed at distances of \(56.2 \mathrm{cm}, 65.9 \mathrm{cm},\) and \(93.5 \mathrm{cm}\) from the central maximum. What are the wavelengths of light emitted by element X?

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about \(3 \mathrm{cm}\) across, and you estimate that the distance from the window shade to the wall is about \(3 \mathrm{m} .\) Knowing that the average wavelength of sunlight is about \(500 \mathrm{nm},\) estimate the diameter of the pinhole.

Diffraction can be used to provide a quick test of the size of red blood cells. Blood is smeared onto a slide, and a laser shines through the slide. The size of the cells is very consistent, so the multiple diffraction patterns overlap and produce an overall pattern that is similar to what a single cell would produce. Ideally, the diameter of a red blood cell should be between 7.5 and \(8.0 \mu \mathrm{m} .\) If a \(633 \mathrm{nm}\) laser shines through a slide and produces a pattern on a screen \(24.0 \mathrm{cm}\) distant, what range of sizes of the central maximum should be expected? Values outside this range might indicate a health concern and warrant further study.

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