/*! 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 83 A microwave of an unknown wavele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A microwave of an unknown wavelength is incident on a single slit of width \(6 \mathrm{cm} .\) The angular width of the central peak is found to be \(25^{\circ} .\) Find the wavelength.

Short Answer

Expert verified
The wavelength of the microwave is approximately \(2.62\times10^{-2} \text{m}\).

Step by step solution

01

1. Setup the single-slit diffraction formula

To find the wavelength of the microwave, we will be using the formula for the angular width of the central peak in single-slit diffraction: \( \theta \approx \dfrac{\lambda}{a} \) Where: - \( \theta \) is the angular width of the central peak (in radians). - \( \lambda \) is the wavelength of the incident microwave. - \(a\) is the width of the single slit.
02

2. Convert the angular width to radians

Given, the angular width in degrees is \(25^{\circ}\), we need to convert it into radians for the formula. We know that \(1^{\circ} = \pi/180\) radians. Therefore, \( \theta = 25^{\circ} \times \dfrac{\pi}{180} \text{ radians} \) Calculating the value, we get: \( \theta = 0.4363 \text{ radians} \)
03

3. Arrange the formula to find the wavelength

We know the relationship \( \theta \approx \dfrac{\lambda}{a} \) for single-slit diffraction. Our goal is to find the wavelength \( \lambda \). To isolate \( \lambda \) in the equation, we can multiply both sides of the equation by \( a \): \( \theta \times a \approx \lambda \)
04

4. Plug in the given values to the formula

Now, let's plug in the given values for \( \theta \) and \( a \) into the formula: \( 0.4363 \times 6\times10^{-2} \text{m} \approx \lambda \) Calculating the value, we get: \( \lambda \approx 2.62\times10^{-2} \text{m} \) So, the wavelength of the microwave is approximately \(2.62\times10^{-2} \text{m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Wavelength Calculation
Wavelength refers to the distance between two consecutive peaks or troughs in a wave, a vital concept in physics that characterizes waves of all types, including sound, light, and microwaves. The calculation of a wave's wavelength becomes particularly important in scenarios such as single-slit diffraction experiments. In such experiments, a wave encounters an obstacle with an opening (the slit) comparable in size to its wavelength, leading to a pattern of bright and dark regions known as a diffraction pattern.

The width of the central bright peak in this pattern can provide clues about the wave's wavelength. In the given exercise, the key to finding the unknown wavelength of microwave radiation is the known angular width of the central peak and the width of the slit through which the microwave has passed. To find the wavelength, one simply manipulates the diffraction formula which, in its basic form for small angles, is given by \( \theta \approx \dfrac{\lambda}{a} \), where \( \theta \) is the angle of the peak's width in radians, \( \lambda \) is the wavelength, and \( a \) is the slit width.

By isolating \( \lambda \) in the equation and plugging in the known values for \( \theta \) and \( a \) from the experimental setup, the wavelength can be determined. This simple yet powerful calculation not only is essential for understanding wave behavior in a diffraction scenario but also sees application in fields ranging from optics to quantum mechanics where wave properties are fundamental.
Converting Angular Width from Degrees to Radians
In physics, angles can be measured in degrees or radians. While everyday measurements often use degrees, radians are the standard unit of angular measurement in mathematical equations, including the single-slit diffraction formula. Understanding how to convert degrees to radians is crucial for solving physics problems correctly.

The conversion is straightforward: since a full circle is \(360^\circ\) or \(2\pi\) radians, one degree is equivalent to \(\pi/180\) radians. This means that to convert an angular width from degrees to radians, you multiply the degree measurement by \(\pi/180\). In the context of the exercise, the angular width of \(25^\circ\) is converted to radians by multiplying \(25\times\frac{\pi}{180}\), yielding approximately \(0.4363\) radians.

This conversion is essential because diffraction formulas rely on radian measures. Remember that for small angles, which are common in diffraction problems, the approximation that the sine of the angle is roughly equal to the angle itself (in radians) is often acceptable, simplifying calculations significantly. Being comfortable with these conversions will enhance a student's ability to tackle a wide range of physics problems beyond just wave diffraction scenarios.
Applying the Diffraction Formula
The diffraction formula, central to understanding wave behavior as they encounter obstacles, mathematically describes how the waves spread out. For single-slit diffraction, the formula is remarkably simple for small angles, expressed as \( \theta \approx \frac{\lambda}{a} \), where \( \theta \) is the angular width of the central diffraction peak in radians, \( \lambda \) is the wavelength of the wave, and \( a \) is the width of the slit.

By rearranging the formula to solve for the wavelength, \( \lambda \approx \theta \times a \), we set the stage for plugging in the known quantities. In our exercise, after converting the angular width to radians, we multiply it by the known slit width to solve for the wavelength of the microwave. This formula, while deceptively simple, elegantly encapsulates the physical principles at play and illustrates how the structure of the slit affects the wave. The quantitative relationship predicted by the diffraction formula allows students to connect theoretical concepts with observable physical phenomena, thereby solidifying their understanding of wave optics. Furthermore, knowledge of the diffraction formula is not just limited to academic exercises; it's used in designing optical instruments and understanding the limitations imposed by the wave nature of light and other electromagnetic radiation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The analysis shown below also applies to diffraction gratings with lines separated by a distance \(d .\) What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 600 -nm light, if the screen is \(1.50 \mathrm{m}\) away? (Hint: The distance between adjacent fringes is \(\Delta y=x \lambda / d, \quad\) assuming the slit separation \(d\) is comparable to \(\lambda_{-}\) )

A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be projected onto a satellite or the moon. (a) If this is done with the Mount Wilson telescope, producing a 2.54 -m-diameter beam of 633 -nm light, what is the minimum angular spread of the beam? (b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the moon, assuming a lunar distance of \(3.84 \times 10^{8} \mathrm{m} ?\)

Consider a single-slit diffraction pattem for \(\lambda=589 \mathrm{nm},\) projected on a screen that is \(1.00 \mathrm{m}\) from a slit of width \(0.25 \mathrm{mm}\). How far from the center of the pattern are the centers of the first and second dark fringes?

Crystal lattices can be examined with X-rays but not UV. Why?

(a) What visible wavelength has its fourth-order maximum at an angle of \(25.0^{\circ}\) when projected on a 25,000-line per centimeter diffraction grating? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.