/*! 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 23 A set of parallel slits for opti... [FREE SOLUTION] | 91影视

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Chapter 28: Problem 23

A set of parallel slits for optical interference can be made by holding two razor blades together (carefully!) and scratching a pair of lines on a glass microscope slide that has been painted black. When monochromatic light strikes these slits at normal incidence, an interference pattern is formed on a distant screen. The thickness of each razor blade used to make the slits is \(0.230 \mathrm{mm}\), and the screen is \(2.50 \mathrm{m}\) from the slits. If the centerto-center separation of the fringes is \(7.15 \mathrm{mm},\) what is the wavelength of the light?

Short Answer

Expert verified
Wavelength of the light is 657.8 nm.

Step by step solution

01

Identify the Given Information

We are given the following values: thickness of each razor blade (which is the width between the two slits) \(d = 0.230\text{ mm} = 0.230 \times 10^{-3}\text{ m}\), distance from slit to screen \(L = 2.50\text{ m}\), and center-to-center fringe separation \(\Delta y = 7.15\text{ mm} = 7.15 \times 10^{-3}\text{ m}\). We need to find the wavelength of the light \(\lambda\).
02

Understand the Relationship

In a double-slit interference pattern, the distance between adjacent fringes \(\Delta y\) is related to the wavelength \(\lambda\), the slit separation \(d\), and the distance to the screen \(L\) by the equation:\[\Delta y = \frac{\lambda L}{d}\]This equation allows us to solve for \(\lambda\) given \(\Delta y\), \(L\), and \(d\).
03

Rearrange the Equation

To find the wavelength \(\lambda\), rearrange the equation from Step 2:\[\lambda = \frac{\Delta y \cdot d}{L}\]This rearranged equation allows us to solve directly for the wavelength.
04

Substitute the Known Values

Substitute the given values into the rearranged equation:\[\lambda = \frac{(7.15 \times 10^{-3}) \cdot (0.230 \times 10^{-3})}{2.50}\]We use the values of \(\Delta y = 7.15 \times 10^{-3}\text{ m}\), \(d = 0.230 \times 10^{-3}\text{ m}\), and \(L = 2.50\text{ m}\).
05

Calculate the Wavelength

Perform the multiplication and division:\[\lambda = \frac{(7.15 \times 0.230) \times 10^{-6}}{2.50} = \frac{1.6445 \times 10^{-6}}{2.50} = 0.6578 \times 10^{-6} \text{ meters} = 657.8 \text{ nm}\]Thus, the wavelength of the light is 657.8 nm.

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Key Concepts

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

Monochromatic Light
Monochromatic light refers to light that consists of a single wavelength. This is important in the context of optical interference because it produces clear and consistent interference patterns. When a beam of monochromatic light passes through a double-slit apparatus, it spreads out and overlaps on the other side of the slits. This overlap leads to an interference pattern. Such a pattern is characterized by alternating bright and dark fringes. The brightness results from constructive interference, where the waves align perfectly to enhance each other. The dark fringes are due to destructive interference, where the waves cancel each other out. Monochromatic light is commonly used in experiments involving interference and diffraction because it simplifies the analysis of the resulting patterns. For example, lasers are often used in these experiments because they emit highly coherent and monochromatic light. Understanding this concept is vital as it forms the foundation for examining phenomena like fringe separation and wavelength calculation.
Fringe Separation
Fringe separation, also known as fringe spacing, refers to the distance between adjacent bright or dark fringes in an interference pattern. It is an essential aspect of studying interference patterns because it helps quantify how the light interferes after passing through the slits. In the context of double-slit interference, fringe separation is directly related to the wavelength of the light, the distance between the slits (known as slit separation), and the distance from the slits to the screen.The formula that describes fringe separation is given by:\[\Delta y = \frac{\lambda L}{d}\]where:
  • \(\Delta y\) is the fringe separation
  • \(\lambda\) is the wavelength of the light
  • \(L\) is the distance from the slits to the screen
  • \(d\) is the slit separation
By understanding fringe separation, we can determine how varying factors like wavelength or slit separation affect the interference pattern. Calculating the fringe separation is crucial when attempting to determine the wavelength of unknown light, as seen in experiments and many practical applications.
Wavelength Calculation
Calculating the wavelength in a double-slit experiment involves using known measurements of the interference pattern. The formula used to find the wavelength \(\lambda\) is derived from the fringe separation formula:\[\lambda = \frac{\Delta y \cdot d}{L}\]To calculate the wavelength:1. Gather all necessary measurements: - Fringe separation \(\Delta y\) - Slit separation \(d\) - Distance from the slits to the screen \(L\) 2. Substitute these values into the wavelength formula. For example, if \(\Delta y\) is measured to be 7.15 mm, \(d\) is 0.230 mm, and \(L\) is 2.50 m, convert these to meters by multiplying mm values by \(10^{-3}\).3. Perform the calculation: \[\lambda = \frac{(7.15 \times 10^{-3}) \times (0.230 \times 10^{-3})}{2.50} = 657.8\,\text{nm}\]This process yields the wavelength of the monochromatic light, essential for understanding its properties and applications. Wavelength calculation is a critical component in optics, helping to identify materials and analyze light sources in scientific and industrial contexts.

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Most popular questions from this chapter

The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter is \(4.25 \mathrm{mm}\) ? Assume light with a wavelength of \(\lambda=550 \mathrm{nm} .\) (Note: The distance from the pupil to the retina is \(25.4 \mathrm{mm} .\) In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is \(n=1.36 .\) )

A thin soap film \((n=1.33)\) suspended in air has a uniform thickness. When white light strikes the film at normal incidence, violet light \(\left(\lambda_{\mathrm{V}}=420 \mathrm{nm}\right)\) is constructively reflected. (a) If we would like green light \(\left(\lambda_{G}=560 \mathrm{nm}\right)\) to be constructively reflected, instead, should the film's thickness be increased or decreased? (b) Find the new thickness of the film. (Assume the film has the minimum thickness that can produce these reflections.)

Splitting Binary Stars As seen from Earth, the red dwarfs Kruger \(60 \mathrm{A}\) and \(\mathrm{Kruger} 60 \mathrm{B}\) form a binary star system with an angular separation of 2.5 arc seconds. What is the smallest diameter telescope that could theoretically resolve these stars using \(550-\mathrm{nm}\) light? \(\left(\right.\) Note: 1 arc \(\left.\sec =1 / 3600^{\circ}\right)\)

Two plates of glass are separated on both ends by small wires of diameter d. Derive an expression for the condition for constructive interference when light of wavelength \(\lambda\) is incident normally on the plates. Consider only interference between waves reflected from the bottom of the top plate and the top of the bottom plate.

Suppose you want to produce a diffraction pattern with X-rays whose wavelength is \(0.030 \mathrm{nm}\). If you use a diffraction grating, what separation between lines is needed to generate a pattern with the first maximum at an angle of \(14^{\circ}\) ? (For comparison, a typical atom is a few tenths of a nanometer in diameter.)
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