In microwave oven, we use electromagnetic oscillators which produce electromagnetic waves in the wavelength range (A) \(1 \mathrm{~mm}\) to \(10 \mathrm{~m}\) (B) \(0.7 \mu \mathrm{m}\) to \(1 \mathrm{~mm}\) (C) \(0.1 \mathrm{~m}\) to \(1 \mathrm{~mm}\) (D) \(0.1 \mu \mathrm{m}\) to \(0.7 \mu \mathrm{m}\)

Short Answer

Expert verified
The correct range of wavelengths for the electromagnetic oscillations in a microwave oven is (C) \(0.1 \mathrm{~m}\) to \(1 \mathrm{~mm}\), as it covers a significant portion of the microwave range while other options cover either a different range or only a small portion of the microwave range.

Step by step solution

01

Identify the microwave range

First, let's identify the range of wavelengths that belongs to the microwave category of the electromagnetic spectrum. Microwaves fall within the range of \(1 \mathrm{~mm}\) to \(100 \mathrm{~cm}\) wavelengths.
02

Compare the given options to the microwave range

Now, let's analyze the given options and compare them to the identified microwave range we found in Step 1. (A) \(1 \mathrm{~mm}\) to \(10 \mathrm{~m}\) - This option starts in the microwave range, but it extends further out into the radio wave spectrum. (B) \(0.7 \mu \mathrm{m}\) to \(1 \mathrm{~mm}\) - This option starts in the infrared range and ends with the starting point of the microwave range. So, it doesn't cover the entire microwave range. (C) \(0.1 \mathrm{~m}\) to \(1 \mathrm{~mm}\) - This option covers the range from \(10 \mathrm{~cm}\) to \(1 \mathrm{~mm}\), which is part of the microwave range. (D) \(0.1 \mu \mathrm{m}\) to \(0.7 \mu \mathrm{m}\) - This option corresponds to the ultraviolet part of the electromagnetic spectrum, not the microwave range.
03

Choose the correct option

Based on our analysis in Step 2, the best answer is option (C) \(0.1 \mathrm{~m}\) to \(1 \mathrm{~mm}\), as it covers a significant portion of the microwave range, while the other options cover either a different range or only a small portion of the microwave range.

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 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

Electromagnetic wave is produced by oscillating electric and magnetic fields \(E^{-}\) and \(B^{-}\). Choose only the incorrect statement from the following (A) \(\mathrm{E}^{-}\) is perpendicular to \(\mathrm{B}^{-}\). (B) \(E^{-}\) is perpendicular to the direction of propagation of the wave (C) \(\mathrm{B}^{-}\) is perpendicular to the direction of propagation of the wave (D) \(E^{-}\) is parallel to \(\mathrm{B}^{-}\)

If the electric field associated with a radiation of frequency $10 \mathrm{MH} z\( is \)\mathrm{E}=10 \sin (\mathrm{kx}-\omega \mathrm{t}) \mathrm{mV} / \mathrm{m}$ then its energy density is $\mathrm{Jm}^{-3}\left(\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)$ (A) \(4.425 \times 10^{-10}\) (B) \(6.26 \times 10^{-14}\) (C) \(8.85 \times 10^{-16}\) (D) \(8.85 \times 10^{-14}\)

When an electromagnetic wave encounters a dielectric medium, the transmitted wave has (A) same frequency but different amplitude (B) same amplitude but different frequency (C) same frequency and amplitude (D) different frequency and amplitude

The dimensional formula of energy density is (A) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-3}\) (D) \(\mathrm{M}^{\mathrm{l}} \mathrm{L}^{0} \mathrm{~T}^{-3}\)

If the direction of magnetic field \(\mathrm{B}^{\rightarrow}\) at some instant is along + ve \(Z\) direction and the electromagnetic wave is propagating along + ve \(\mathrm{X}\) direction, then the direction of electric field \(\mathrm{E}^{\rightarrow}\) at that instant is (A) along - ve Y direction (B) along + ve Y direction (C) along + ve \(\mathrm{X}\) direction (B) along - ve \(\mathrm{X}\) direction

See all solutions

Recommended explanations on English 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.

Sign-up for free