/*! 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} Q6P The wavelength of yellow sodium ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The wavelength of yellow sodium light in air is 589 nm. (a) What is its frequency? (b) What is its wavelength in glass whose index of refraction is 1.52? (c) From the results of (a) and (b), find its speed in this glass.

Short Answer

Expert verified
  1. The frequency of yellow sodium light is 5.09×1014Hz.
  2. The wavelength of the light when traveling through the glass is 388nm
  3. The light speed when traveling through the glass is1.97×108m/s

Step by step solution

01

Given information

The wavelength of yellow sodium light in air is λ=589nm.

The index of refraction of glass is,n=1.52.

02

Frequency of a wave

The frequency of a wave can be calculated mathematically by dividing the value of the wavelength from the speed of light.

So, if the wavelength increases then the value of frequency of the wave decreases accordingly.

03

(a) Frequency of yellow sodium light

The formula for the frequency of yellow sodium light is given by,

f=cλ

Here, speed of lightc=3×108m/s

Putting the values,

f=3×108m/s589×10-9mf=5.09×1014Hz

Hence, the frequency of yellow sodium light is5.09×1014Hz

04

(b) Wavelength when traveling through the glass

The formula for the wavelength of the light when traveling through the glass is given by,

λn=λnλn=589nm1.52λn=387.5nm

Hence, the wavelength of the light when traveling through the glass is387.5nm

05

(c) Light speed when traveling through the glass

The formula for the light speed when traveling through the glass is given by,

v=fλn

Putting values of f and λn,

v=5.09×1014Hz×387.5×10-9mv=1.97×108m/s

Hence, the light speed when traveling through the glass is1.97×108m/s

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!

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

The lens in a Newton’s rings experiment (see problem 75) has diameter 20 mm and radius of curvature R=5.0m. For A=589nm in air, how many bright rings are produced with the setup (a) in air and
(b) immersed in water (n=1.33)?

In Fig. 35-32a, a beam of light in material 1 is incident on a boundary at an angle of 30o. The extent to which the light is bent due to refraction depends, in part, on the index of refraction n2of material 2. Figure 35-32b gives the angle of refraction θ2versus n2for a range of possible n2values, from na=1.30to nb=1.90. What is the speed of light in material 1?

Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0°. What is the resultant amplitude?

Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3andr4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

If you move from one bright fringe in a two-slit interference pattern to the next one farther out,

(a) does the path length difference ∆Lincrease or decrease and

(b) by how much does it change, in wavelengths λ ?

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.