/*! 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 38 Light passes through a 200 line/... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light passes through a 200 line/mm grating and is observed on a 1.0 -m-wide screen located \(1.0 \mathrm{m}\) behind the grating. Three bright fringes are seen on both sides of the central maximum. What are the minimum and maximum possible values of the wavelength (in mm)?

Short Answer

Expert verified
The minimum possible value for the wavelength is 0.001 m or 1 mm and the maximum possible value is 0.0015 m or 1.5 mm.

Step by step solution

01

Convert units

Before we get to the calculations, we need to make sure that the units on our input variables match. For this, we need to convert length of grating line from per mm to per meter. Thus, \(d = 200 \,lines / mm = 200000 \,lines / m \). Which can be rewritten as \(d = 1 / 200000 \,m\)
02

Calculate minimum possible value of the wavelength

We are given that 3 bright fringes were observed on both sides of the central maximum. These 3 fringes form 4 spaces uniformely divided over half of the screen width (0.5m), so each space has a size of \(0.5 / 4 = 0.125 \, m \). The angle of diffraction for the third order fringe (\(m = 1,2,3\)) is then \(\theta= \arctan(0.125/1) \). The equation of grating \(d\sin(\theta) = m\lambda\) will be used to calculate the minimum possible value for \(\lambda\). If we solve for \(\lambda\), assuming \(m = 1\) (to get the minimum value), it becomes \(\lambda= d\sin(\theta)/m\). Substitute known variables to solve for \(\lambda\).
03

Calculate maximum possible value of the wavelength

For the maximum possible value of the wavelength, \(\theta\) will be assumed to be 90° and \(m = 3\). Substituting these values in the equation \(d\sin(\theta) = m\lambda\), we get the solution for \(\lambda\).

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.

Wave-optics
Wave-optics, also known as physical optics, deals with the study of waves and their interactions. In this realm, light is not considered to travel merely in straight lines, as in geometric optics, but as a wave that can exhibit phenomena like interference, diffraction, and polarization.

When it comes to wave-optics, one key principle is that light waves can interfere with each other constructively and destructively, creating patterns of light and dark fringes. This happens when light waves that are coherent, meaning they have a constant phase difference, meet. Analysis of such patterns can reveal important information about the light source, such as its wavelength, which is a crucial aspect in the study of diffraction gratings. A diffraction grating is a tool used in wave-optics to separate different wavelengths of light, often into a spectrum, through the process of diffraction.
Spectral Lines
Spectral lines are unique frequencies of electromagnetic radiation emitted or absorbed by substances, typically seen as discrete lines on a spectrogram. They result from electrons transitioning between different energy levels within an atom, emitting or absorbing photons of specific wavelengths in the process.

Understanding spectral lines is essential in fields such as astronomy and chemistry, as they are identifiers for chemical composition and physical properties of substances. They can indicate not just what elements are present but also provide clues about temperature, density, and movement of the source material. In the context of our exercise, a spectroscope equipped with a diffraction grating separates the incoming light into its component spectral lines, which can then be analyzed.
Light Diffraction
Light diffraction is a wave phenomenon where light waves bend around corners or spread out as they pass through small openings or across sharp edges. A prime example of this occurrence is when light encounters a diffraction grating, which consists of many closely spaced slits or grooves.

The gratings cause the light waves to spread out and overlap, producing interference patterns that manifest as bright and dark fringes on a screen. The equation for a diffraction grating, which was used in our exercise, is given as \(d \sin(\theta) = m\lambda\), where \(d\) is the separation of the grating lines, \(\theta\) is the angle of diffraction, \(m\) is the order number of the fringe, and \(\lambda\) is the wavelength of light. By analyzing the angle at which the bright fringes appear, we can calculate the wavelength of the light using the diffraction grating equation, providing insight into the physical properties of the light source.

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

Light of wavelength 600 nm passes though two slits separated by \(0.20 \mathrm{mm}\) and is observed on a screen \(1.0 \mathrm{m}\) behind the slits. The location of the central maximum is marked on the screen and labeled \(y=0\) a. At what distance, on either side of \(y=0,\) are the \(m=1\) bright fringes? b. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by \(5.0 \times 10^{-16} \mathrm{s}\) in comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay? c. With the glass in place, what is the phase difference \(\Delta \phi_{0}\) between the two waves as they leave the slits? d. The glass causes the interference fringe pattern on the screen to shift sideways. Which way does the central maximum move (toward or away from the slit with the glass) and by how far?

A helium-neon laser \((\lambda=633 \mathrm{nm})\) illuminates a diffraction grating. The distance between the two \(m=1\) bright fringes is \(32 \mathrm{cm}\) on a screen \(2.0 \mathrm{m}\) behind the grating. What is the spacing between slits of the grating?

You want to photograph a circular diffraction pattern whose central maximum has a diameter of \(1.0 \mathrm{cm} .\) You have a heliumneon laser \((\lambda=633 \mathrm{nm})\) and a 0.12 -mm-diameter pinhole. How far behind the pinhole should you place the viewing screen?

A double-slit interference pattern is created by two narrow slits spaced \(0.20 \mathrm{mm}\) apart. The distance between the first and the fifth minimum on a screen \(60 \mathrm{cm}\) behind the slits is \(6.0 \mathrm{mm} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the light used in this experiment?

Light consisting of two nearly equal wavelengths \(\lambda+\Delta \lambda\) and \(\lambda\) where \(\Delta \lambda \ll \lambda\), is incident on a diffraction grating. The slit separation of the grating is \(d\) a. Show that the angular separation of these two wavelengths in the \(m\)th order is $$\Delta \theta=\frac{\Delta \lambda}{\sqrt{(d / m)^{2}-\lambda^{2}}}$$ b. Sodium atoms emit light at \(589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm} .\) What are the first-order and second-order angular separations (in degrees) of these two wavelengths for a 600 line/mm grating?

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.