Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency \(1.20 \times {10^{15}}{\rm{ }}Hz\)?

Short Answer

Expert verified

The smallest detail observable is \(250\;nm\).

Step by step solution

01

Definition of frequency

Frequency is the number of waves that pass through a specific spot in a specific length of time.

02

Given information and Formula to be used

The frequency of ultraviolet light is: \(f = 1.20 \times {10^{15}}\;Hz\)

The speed of light value is: \(c = 3 \times {10^8}\;m/s\)

For an electromagnetic wave, the relationship between propagation speed, wavelength, and frequency is given by,

\(c = f\lambda \)

03

Calculate the wavelength 

Consider the given values and simplify,

We have

\(\begin{array}{c}c = f\lambda \\3 \times {10^8} = 1.20 \times {10^{15}} \times \lambda \\\lambda = 250\;nm\end{array}\)

Therefore, the wavelength is \(250\;nm\).

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 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

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.

Sign-up for free