/*! 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 5 A single slit is illuminated by ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A single slit is illuminated by light of wavelengths \(\lambda_{a}\) and \(\lambda_{b}\), chosen so that the first diffraction minimum of the \(\lambda_{a}\) component coincides with the second minimum of the \(\lambda_{b}\) component. (a) If \(\lambda_{b}=350 \mathrm{~nm}\), what is \(\lambda_{a} ?\) For what order number \(m_{b}\) (if any) does a minimum of the \(\lambda_{b}\) component coincide with the minimum of the \(\lambda_{a}\) component in the order number (b) \(m_{a}=2\) and (c) \(m_{a}=3 ?\)

Short Answer

Expert verified
(a) \(\lambda_{a} = 700 \text{ nm}\); (b) \(m_{b}=4\); (c) \( m_{b}=6 \).

Step by step solution

01

Understanding the Diffraction Condition

The condition for a diffraction minimum for a single slit is given by the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction, \( m \) is the order of the minimum, and \( \lambda \) is the wavelength. For minima to coincide, their conditions must be equal.
02

Forming the Equation for Coinciding Minima

We are given that the first diffraction minimum for \( \lambda_{a} \) coincides with the second minimum for \( \lambda_{b} \). Therefore, we can write: \( m_a \lambda_{a} = m_b \lambda_{b} \). Substituting the values \( m_a = 1 \) and \( m_b = 2 \): \( \lambda_{a} = \frac{2 \cdot 350 \text{ nm}}{1} = 700 \text{ nm} \).
03

Checking Coincidence for New Orders

Let's now check for any coincidence in minima for the cases given by \(m_a = 2\) and \(m_a = 3\).
04

Coinciding Minima for m_a=2

If \( m_a = 2 \), we equate it to \( m_b \lambda_{b} = 2 \lambda_{a} \). Substituting \( \lambda_{a} = 700 \text{ nm} \) and \( \lambda_{b} = 350 \text{ nm} \): \( 2 \times 700 \text{ nm} = m_b \cdot 350 \text{ nm} \). Solving gives \( m_b = 4 \).
05

Coinciding Minima for m_a=3

Similarly, for \( m_a = 3 \), we have \( m_b \lambda_{b} = 3 \lambda_{a} \). Therefore, \( 3 \times 700 \text{ nm} = m_b \cdot 350 \text{ nm} \). Solving gives \( m_b = 6 \).

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.

Diffraction Minimum
In the study of light and optics, the concept of diffraction is essential to understanding how light interacts with obstacles like slits. A single slit diffraction pattern features regions of minimum light intensity, called diffraction minima, which occur at specific angles. These minima are defined by the condition \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction, \( m \) is the order number (an integer), and \( \lambda \) is the wavelength of light. The order of the minimum, represented by \( m \), indicates which minimum it is: the first, second, third, and so on.
This calculation is crucial, as it determines the angle at which light will constructively or destructively interfere after passing through the slit. In simpler terms, diffraction minima are spots where the light waves cancel out, resulting in darkness in the pattern projected on a screen.
Wavelength Coincidence
Wavelength coincidence occurs when diffraction minima for light of different wavelengths align at the same angular position. In the case study, the first minimum of wavelength \( \lambda_a \) exactly coincides with the second minimum of wavelength \( \lambda_b \). This is expressed mathematically by the equation \( m_a \lambda_a = m_b \lambda_b \).
This phenomenon is important as it allows for a comparison between different wavelengths under similar conditions. By adjusting the orders \( m_a \) and \( m_b \), you can find specific wavelengths that align in terms of their diffraction patterns. This alignment is not arbitrary; it occurs due to the physical properties of the light and the geometry of the setup, allowing for precise calculations in experiments.
Order of Minimum
The order of a diffraction minimum, symbolized by \( m \), represents the series number of a particular minimum. For example, \( m = 1 \) indicates the first minimum, \( m = 2 \) is the second, and so forth. This concept is crucial when comparing different wavelengths of light, as each will have its own set of minima based on its wavelength.
Understanding this order is key in solving exercises involving wavelength coincidence. In the given problem, the coincidence occurs when the first order of \( \lambda_a \) matches the second order of \( \lambda_b \). This relationship helps calculate unknown values, such as another wavelength or the order at other minima coincidences.
Diffraction Equation
The diffraction equation serves as the foundation for calculating the behavior of light through a single slit. It is given by \( a \sin \theta = m \lambda \), where all terms are already familiar from previous sections. This equation allows for predicting exactly where minima and maxima will occur on a pattern.
In the given problem, it is used to compute the wavelength \( \lambda_a \) when the first diffraction minimum of \( \lambda_a \) coincides with the second of \( \lambda_b \). By manipulating this equation, observations can be analyzed to determine key qualities like wavelength, the angle of diffraction, and the order number of the minima. The versatility of the diffraction equation makes it an invaluable tool in optics, especially when dealing with problems involving multiple wavelengths and conditions of coincidence.

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

In the single-slit diffraction experiment of Fig. \(36-4\), let the wavelength of the light be \(500 \mathrm{~nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=3.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{P}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm} .\) (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

When monochromatic light is incident on a slit \(22.0 \mu \mathrm{m}\) wide, the first diffraction minimum lies at \(1.80^{\circ}\) from the direction of the incident light. What is the wavelength?

Suppose that the central diffraction envelope of a double-slit diffraction pattern contains 11 bright fringes and the first diffraction minima eliminate (are coincident with) bright fringes. How many bright fringes lie between the first and second minima of the diffraction envelope?

If you look at something \(40 \mathrm{~m}\) from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh's criterion? Assume the pupil of your eye has a diameter of \(4.00 \mathrm{~mm}\), and use \(500 \mathrm{~nm}\) as the wavelength of the light reaching you.

Suppose that two points are separated by \(2.0 \mathrm{~cm} .\) If they are viewed by an eye with a pupil opening of \(5.0 \mathrm{~mm}\), what distance from the viewer puts them at the Rayleigh limit of resolution? Assume a light wavelength of \(500 \mathrm{~nm}\).

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.