PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
bartleby

Videos

Question
Book Icon
Chapter 31, Problem 29P

(a)

To determine

The speed of the helium neon laser light in air, water and glass.

(a)

Expert Solution
Check Mark

Answer to Problem 29P

The speed of the helium neon laser light in air is 3×108m/s , in water is 2.25×108m/s and in glass is 2×108m/s .

Explanation of Solution

Given:

The wavelength of the helium-neon laser is λ=632.8nm .

Formula used:

The expression for the speed is given as,

  v=cn

Here, c is the speed of light and n is the refractive index of that particular material.

Calculation:

The value of speed of light is c=3×108m/s .

The refractive index of air is na=1 .

The value of speed of helium neon laser light in air can be calculated as,

  va=cna=3× 108m/s1=3×108m/s

The refractive index in water is nw=1.33 .

The value of speed of helium neon laser light in water can be calculated as,

  vw=cnw=3× 108m/s1.33=2.25×108m/s

The refractive index in glass is ng=1.50 .

The value of speed of helium neon laser light in glass can be calculated as,

  vg=cng=3× 108m/s1.5=2×108m/s

Conclusion:

Therefore, the speed of the helium neon laser light in air is 3×108m/s , in water is 2.25×108m/s and in glass is 2×108m/s .

(b)

To determine

The wavelength of the helium neon laser light in air, water and glass.

(b)

Expert Solution
Check Mark

Answer to Problem 29P

The wavelength of the helium neon laser light in air is 6.329×107m , in water is 4.57×107m and in glass is 4.21×107m .

Explanation of Solution

Formula used:

The expression for the frequency is given as,

  f=cλ

The expression for the wavelength is given as,

  λ=cnf

Calculation:

The value of frequency can be calculated as,

  f=cλ=3× 108m/s632.8nm( 1m 10 9 nm )=4.74×1014Hz

The value of wavelength of the helium neon laser light in air can be calculated as,

  λa=cnaf=3× 108m/s1×4.74× 10 14Hz=6.329×107m

The value of wavelength of the helium neon laser light in water can be calculated as,

  λw=cnwf=3× 108m/s1.33×4.74× 10 14Hz=4.57×107m

The value of wavelength of the helium neon laser light in glass can be calculated as,

  λg=cngf=3× 108m/s1.5×4.74× 10 14Hz=4.21×107m

Conclusion:

Therefore, the wavelength of the helium neon laser light in air is 6.329×107m , in water is 4.57×107m and in glass is 4.21×107m .

(c)

To determine

The frequency of the helium neon laser light in air, water and glass.

(c)

Expert Solution
Check Mark

Answer to Problem 29P

The frequency of the helium neon laser light in air is 4.74×1014Hz , in water is 3.56×1014Hz and in glass is 3.16×1014Hz .

Explanation of Solution

Formula used:

The expression for the frequency can be given as,

  f=cnλ

Calculation:

The value of frequency of the helium neon laser light in air can be calculated as,

  fa=cnaλ=3× 108m/s1×632.8nm( 1m 10 9 nm )=4.74×1014Hz

The value of frequency of the helium neon laser light in water can be calculated as,

  fw=cnwλ=3× 108m/s1.33×632.8nm( 1m 10 9 nm )=3.56×1014Hz

The value of frequency of the helium neon laser light in glass can be calculated as

  fg=cngλ=3× 108m/s1.50×632.8nm( 1m 10 9 nm )=3.16×1014Hz

Conclusion:

Therefore, the frequency of the helium neon laser light in air is 4.74×1014Hz , in water is 3.56×1014Hz and in glass is 3.16×1014Hz .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Hi! I need help with these calculations for part i and part k for a physics Diffraction Lab. We used a slit width 0.4 mm to measure our pattern.
Examine the data and % error values in Data Table 3 where the angular displacement of the simple pendulum decreased but the mass of the pendulum bob and the length of the pendulum remained constant. Describe whether or not your data shows that the period of the pendulum depends on the angular displacement of the pendulum bob, to within a reasonable percent error.
In addition to the anyalysis of the graph, show mathematically that the slope of that line is 2π/√g . Using the slope of your line calculate the value of g and compare it to 9.8.

Chapter 31 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Polarization of Light: circularly polarized, linearly polarized, unpolarized light.; Author: Physics Videos by Eugene Khutoryansky;https://www.youtube.com/watch?v=8YkfEft4p-w;License: Standard YouTube License, CC-BY