/*! 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 62 Figure \(32-57\) shows a liquid-... [FREE SOLUTION] | 91Ó°ÊÓ

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Figure \(32-57\) shows a liquid-detecting prism device that might be used inside a washing machine or other liquid-containing appliance. If no liquid covers the prism's hypotenuse, total internal reflection of the beam from the light source produces a large signal in the light sensor. If liquid covers the hypotenuse, some light escapes from the prism into the liquid and the light sensor's signal decreases. Thus a large signal from the light sensor indicates the absence of liquid in the reservoir. If this device is designed to detect the presence of water, determine the allowable range for the prism's index of refraction \(n\). Will the device work properly if the prism is constructed from (inexpensive) lucite? For lucite, \(n=1.5\)

Short Answer

Expert verified
The prism should have \( n > 1.33 \). Yes, lucite with \( n = 1.5 \) will work properly.

Step by step solution

01

Understanding Total Internal Reflection

Total Internal Reflection (TIR) occurs when the angle of incidence exceeds the critical angle for the interface between two media with different indices of refraction. For TIR at the boundary between the prism and air (when no liquid is present), we need the light to be completely reflected back into the prism.
02

Finding the Critical Angle for Air-Prism Interface

The critical angle \( \theta_c \) at the interface of the prism and air can be calculated using Snell's Law: \( n = \frac{1}{\sin(\theta_c)} \), where the index of refraction of air is approximately 1.
03

Calculating Index of Refraction for Total Internal Reflection

For total internal reflection to occur at the air-prism boundary, the index of refraction \( n \) must be such that the critical angle \( \theta_c \) satisfies \( \sin(\theta_c) = \frac{1}{n} \). Solving for \( n \), we find that \( n > 1 \).
04

Analyzing the Prism in Contact with Water

When water covers the hypotenuse, some light escapes depending on the critical angle for the water-prism interface. Here, \( n_{water} \approx 1.33 \), which determines a new critical angle using \( \sin(\theta'_c) = \frac{n_{water}}{n} \). For some light to escape and register less intensity on the sensor, this must satisfy \( \theta' < \theta_c \).
05

Range for Index of Refraction

Combining conditions for total internal reflection with air and partial transmission when submerged, the index must satisfy \( n > 1.33 \).
06

Conclusion on Lucite

Since lucite has \( n = 1.5 \), and 1.5 is greater than 1.33, the prism will achieve total internal reflection when uncovered by the liquid and allow some light to escape when submerged, thus functioning correctly.

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Key Concepts

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

Snell's Law
Snell's Law is a fundamental principle used to understand how light behaves at the boundary between two different media. It explains how light bends, also called refraction, as it passes through surfaces like glass, water, or air. This is described by the formula:\[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]Where:
  • \( n_1 \) and \( n_2 \) are the indices of refraction for the two media.
  • \( \theta_1 \) is the angle of incidence (angle between incoming light and the normal).
  • \( \theta_2 \) is the angle of refraction (angle between refracted light and the normal).
In the context of the prism device in a washing machine, Snell's Law helps us determine how light will behave as it travels through the prism and is crucial for determining if total internal reflection will occur.
This can affect the functionality of devices like a liquid-detecting sensor, ensuring the correct signaling based on whether the prism is submerged in liquid or not.
Index of Refraction
The index of refraction, or refractive index, is a number that describes how fast light travels through a material. Symbolized by \( n \), it tells us how much light slows down compared to its speed in a vacuum. Every medium has its own index of refraction. For example,
air has an index close to 1, while water’s index is about 1.33.A higher refractive index indicates light travels slower in that medium, resulting in more bending or refraction. In the prism sensor of a washing machine, the refractive index of the prism material (like lucite) is critical. It determines whether total internal reflection can occur.
A key requirement for the sensor to function properly to detect liquid is to select a material with the right index, ensuring the light behavior is as expected. By comparing indices between the prism and surrounding liquids, engineers calculate the optimal index to avoid signal errors.
Critical Angle
The critical angle is the specific angle of incidence above which total internal reflection occurs. It is relevant when light passes from a denser to a less dense medium, such as from glass or water to air.
The critical angle \( \theta_c \) can be calculated using:\[\sin(\theta_c) = \frac{n_2}{n_1}\]Where:
  • \( n_1 \) is the index of refraction of the denser medium.
  • \( n_2 \) is the index of refraction of the less dense medium.
For the liquid-detecting prism device, the critical angle determines the threshold at which the light ray will no longer exit through the hypotenuse when it's uncovered.
When the angle of incidence exceeds this critical angle, the light reflects internally, causing a high sensor signal. Adjustments based on the monitored medium’s properties ensure reliable liquid detection through changes in signal levels.
Washing Machine Sensor
Washing machine sensors that use prisms leverage the principle of total internal reflection to detect if a liquid is covering the sensor's surface. This works by utilizing the prism's specific index of refraction and critical angle. As described, if no liquid is present, the light bounces fully back into the prism and produces a high signal.
Conversely, when liquid covers the prism, some light escapes into the liquid, reducing the signal the sensor detects. Therefore, differing light intensities help indicate the presence or absence of liquid.
For materials like lucite, with a refractive index of 1.5, this device operates efficiently. This technology is essential for functions like monitoring water levels to prevent flooding or ensuring adequate washing cycles in household appliances. By ensuring proper light reflection and refraction, these sensors offer a non-invasive and reliable method for real-time liquid level assessment.

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Most popular questions from this chapter

A small candle is \(35 \mathrm{~cm}\) from a concave mirror having a radius of curvature of \(24 \mathrm{~cm} .\) (a) What is the focal length of the mirror? (b) Where will the image of the candle be located? (c) Will the image be upright or inverted?

Show that if two plane mirrors meet at an angle \(\phi,\) a single ray reflected successively from both mirrors is deflected through an angle of \(2 \phi\) independent of the incident angle. Assume \(\phi<90^{\circ}\) and that only two reflections, one from each mirror, take place.

A solar cooker, really a concave mirror pointed at the Sun, focuses the Sun's rays \(18.8 \mathrm{~cm}\) in front of the mirror. What is the radius of the spherical surface from which the mirror was made?

A light beam strikes a piece of glass at a \(60.00^{\circ}\) incident angle. The beam contains two wavelengths, \(450.0 \mathrm{nm}\) and \(700.0 \mathrm{nm},\) for which the index of refraction of the glass is 1.4831 and \(1.4754,\) respectively. What is the angle between the two refracted beams?

(II) Two rays \(A\) and \(B\) travel down a cylindrical optical fiber of diameter \(d=75.0 \mu \mathrm{m},\) length \(\ell=1.0 \mathrm{km},\) and index of refraction \(n_{1}=1.465 .\) Ray A travels a straight path down the fiber's axis, whereas ray \(\mathrm{B}\) propagates down the fiber by repeated reflections at the critical angle each time it impinges on the fiber's boundary. Determine the extra time \(\Delta t\) it takes for ray \(B\) to travel down the entire fiber in comparison with ray A (Fig. \(58 ),\) assuming \((a)\) the fiber is surrounded by air, (b) the fiber is surrounded by a cylindrical glass "cladding" with index of refraction \(n_{2}=1.460 .\)

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