/*! 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 1 Monochromatic light from a dista... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 \(\mathrm{m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm}\) . Calculate the wavelength of the light.

Short Answer

Expert verified
The wavelength of light is 506 nm.

Step by step solution

01

Understand the Problem

We need to determine the wavelength of monochromatic light incident on a slit by using the distance from the central maximum to the first diffraction minimum. We know the slit width, the distance from the slit to the screen, and the given distance to the first minimum.
02

Apply the Diffraction Formula for Single Slit

For a single slit diffraction, the position of the first minimum (dark fringe) is given by the formula \( a \cdot \sin(\theta) = m\lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction, \( m \) is the order of the minimum (\( m = 1 \) for the first minimum), and \( \lambda \) is the wavelength. We assume small angles, so \( \sin(\theta) \approx \tan(\theta) = \frac{y}{L} \), where \( y \) is the distance from the central maximum to the first minimum and \( L \) is the distance to the screen.
03

Substitute Known Values and Simplify

Substitute \( a = 0.750 \times 10^{-3} \text{ m} \), \( y = 1.35 \times 10^{-3} \text{ m} \), and \( L = 2.00 \text{ m} \) into the equation. \( a \cdot \frac{y}{L} = \lambda \) simplifies to \( 0.750 \times 10^{-3} \cdot \frac{1.35 \times 10^{-3}}{2.00} = \lambda \).
04

Calculate the Wavelength

Perform the calculation: \( \lambda = 0.750 \times 10^{-3} \cdot \frac{1.35 \times 10^{-3}}{2.00} = 5.06 \times 10^{-7} \text{ m} \), or 506 nm.

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.

Wavelength Calculation
Understanding how to calculate the wavelength of light in a diffraction pattern is crucial to grasping the core principle of wave behavior. The classic setup involves light passing through a single slit, followed by observations of the resulting diffraction pattern on a screen. For the calculation of the wavelength (\( \lambda \)), we rely on an important formula: \[ a \cdot \sin(\theta) = m\lambda \]Where:
  • \(a\) is the slit width.
  • \(\theta\) is the angle of diffraction.
  • \(m\) is the order number (first order minimum when \( m = 1 \)).
For small angles, it becomes convenient to use the approximation \( \sin(\theta) \approx \tan(\theta) = \frac{y}{L} \). Here, \(y\) is the observed distance from the central maximum to the first minimum, and \(L\) is the distance from the slit to the observing screen. Thus, the formula modifies to \[ a \cdot \frac{y}{L} = \lambda \] By substituting known values, the wavelength can be determined, demonstrating how simple geometry aids in unraveling the properties of wave nature.
Monochromatic Light
Monochromatic light holds a special role in optics due to its singular, unchanging wavelength. This characteristic makes it ideal for experiments involving diffraction, as it ensures clarity and precision in the resulting diffraction pattern. Unlike white light, which comprises multiple colors and wavelengths, monochromatic light provides a single color with one constant wavelength. Its consistent nature is paramount as it eliminates the complexity of overlapping diffraction patterns produced by multiple wavelengths. Thus, when monochromatic light passes through a slit, the pattern observed is straightforward, typically featuring equally spaced bright and dark fringes corresponding to maxima and minima. In the context of wavelength calculation problems, using monochromatic light simplifies the task, allowing students to directly apply formulas without worrying about variability in light properties.
Diffraction Pattern Analysis
Analyzing diffraction patterns allows us to observe the wave nature of light in action. When monochromatic light passes through a narrow slit, it bends or diffracts, creating a pattern of alternating bright and dark fringes on a screen. These patterns are significant because they provide insights into the properties of the light and the geometry of the experimental setup.
  • Central Maximum: This is the brightest and widest part of the pattern, located directly opposite the slit. It indicates the most intense constructive interference of light waves.
  • Minima: Represented as dark fringes, these occur due to destructive interference, where waves cancel each other out.
  • Secondary Maxima: These are less bright than the central maximum and occur between minima.
The positions of maxima and minima can be predicted using mathematical equations, such as the one used in the wavelength calculation. By measuring these positions, especially the first minimum, we can extract critical information about the wavelength and physical dimensions of the setup. Thus, the meticulous study of diffraction patterns not only confirms the wave nature of light but also serves fundamental roles in experimental physics and technology development.

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

Measuring Refractive Index. A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm 38.2^{\circ}\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm 17.4^{\circ} .\) Find the refractive index of the liquid.

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm} )\) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

Hubble Versus Arecibo. The Hubble Space Telescope has an aperture of 2.4 \(\mathrm{m}\) and focuses visible light \((400-700 \mathrm{nm}) .\) The Arecibo radio telescope in Puerto Rico is 305 \(\mathrm{m}(1000 \mathrm{ft})\) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength \(75 \mathrm{cm} .\) (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

Light of wavelength 585 \(\mathrm{nm}\) falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calclating all the anglest (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be 7 (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute). (a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 550 -nm light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10- diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope ( 5 -m diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

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.