Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 2183

In electromagnetic spectrum, the visible light lie between (A) radiowaves and microwaves (B) ultraviolet rays and infrared rays (C) ultraviolet rays and \(\mathrm{x}\) rays (D) infrared rays and microwaves

Short Answer

Expert verified
In the electromagnetic spectrum, visible light lies between infrared rays (longer wavelengths) and ultraviolet rays (shorter wavelengths). Therefore, the correct answer is (B) Ultraviolet rays and infrared rays.

Step by step solution

01

List the components of the electromagnetic spectrum in order

Here's the general order of the electromagnetic spectrum, from the longest wavelength to the shortest wavelength: 1. Radiowaves 2. Microwaves 3. Infrared rays 4. Visible light 5. Ultraviolet rays 6. X-rays 7. Gamma rays
02

Examine each choice

Now, let’s examine each of the choices and see which one correctly places visible light between two other types of electromagnetic radiation: (A) Radiowaves and microwaves: Visible light lies between infrared rays and ultraviolet rays, not between radiowaves and microwaves. (B) Ultraviolet rays and infrared rays: This choice correctly places visible light between infrared rays and ultraviolet rays. (C) Ultraviolet rays and X-rays: Visible light lies between infrared rays and ultraviolet rays, not between ultraviolet rays and X-rays. (D) Infrared rays and microwaves: Visible light lies between infrared rays and ultraviolet rays, not between infrared rays and microwaves.
03

Choose the correct answer

Based on our analysis, the correct answer is: (B) Ultraviolet rays and infrared rays. Visible light lies between infrared rays (longer wavelengths) and ultraviolet rays (shorter wavelengths) in the electromagnetic spectrum.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 an electromagnetic wave, if the amplitude of magnetic field is $3 \times 10^{-10} \mathrm{~T}$, the amplitude of the associated electric field will be (A) \(9 \times 10^{-2} \overline{\mathrm{Vm}^{-1}}\) (B) \(3 \times 10^{-10} \mathrm{Vm}^{-1}\) (C) \(3 \times 10^{-2} \mathrm{Vm}^{-1}\) (D) \(1 \times 10^{-18} \mathrm{Vm}^{-1}\)

What is the name associated with the equation \(E^{\longrightarrow} \cdot d t^{-}=-(d \Phi \beta / d t)\) (A) Gauss law for electricity (B) Gauss law for magnetism (C) ampere's law (D) faraday's law

An electric charge oscillating with a frequency of 1 kilo cycles/s can radiates electromagnetic waves of wavelength (A) \(100 \mathrm{~km}\) (B) \(200 \mathrm{~km}\) (C) \(300 \mathrm{~km}\) (D) \(400 \mathrm{~km}\)

A plane electromagnetic wave of wave intensity \(10 \mathrm{~cm}^{-2}\) strikes a small mirror of area \(20 \mathrm{~cm}^{2}\), held perpendicular to the approaching wave. The radiation force on the mirror will be (A) \(6.6 \times 10^{-11} \mathrm{~N}\) (B) \(1.33 \times 10^{-11} \mathrm{~N}\) (C) \(1.33 \times 10^{-10} \mathrm{~N}\) (D) \(6.6 \times 10^{-10} \mathrm{~N}\)

Bolometer is used to detect (A) infrared rays (B) ultraviolet rays (C) x rays (D) \(\gamma\) rays
See all solutions

Recommended explanations on English Textbooks

Discourse

Read Explanation

Language and Social Groups

Read Explanation

Essay Prompts

Read Explanation

Phonology

Read Explanation

Cues and Conventions

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation
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.

Sign-up for free