From the nature of light unit.

Click For Summary

Homework Help Overview

The discussion revolves around a problem related to electromagnetic radiation, specifically focusing on calculating wavelength in different mediums, identifying the visibility of the radiation, and understanding the index of refraction. The subject area includes concepts from optics and wave physics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculations for wavelength in vacuum and water, express confusion over the visibility of radiation, and question the relevance of color based on frequency versus wavelength. There is also uncertainty regarding the index of refraction and its calculation.

Discussion Status

Some participants have provided feedback on calculations and clarified concepts related to color and frequency. There is an ongoing exploration of the correct answers to the questions posed, with no explicit consensus reached on all points.

Contextual Notes

Participants mention the lack of direct guidance from teachers due to the independent study format, which may contribute to the confusion regarding the questions and assumptions being made.

nblu
Messages
56
Reaction score
0
Hi and thank you for viewing this post.
I'll begin with the question from the textbook;

Q: An electromagnetic radiation has a frequency of 5.00 x 10^14 Hz.
a) Calculate its wavelength in a vacuum, in meters and nanometers.
b) Calculate its wavelength in water.
c) Is this radiation visible? If so, what is its colour?
d) What is the index of refraction of a medium in which the speed of this radiation
is 2.54 x 10^8 m/s?
e) Where would you encounter this radiation in your daily life?

and here are my attempts to the above questions;

A: a) given that the speed of light in "vacuum" is 3.00 x 10^8 m/s,
(not sure how to type-in the wavelength sign)
wavelength = 3.00 x 10^8 m/s / 5.00 x 10^14 Hz
Then it comes out to be, 6.00 x 10^-17 m which also equals to 600nm.
b) given that the speed of light in "water" is 2.2 x 10^8 m/s,
wavelength = 2.2 x 10^8 m/s / 5.00 x 10^14 Hz
which turns out to be 4.4 x 10^-7 m, and 440 nm.
c) I had a confusion because I wasn't really sure whether the question was asking
about the radiation in vacuum or in water (Sorry this is an independent study course
so there aren't any teachers whom I can contact). So, I gave the answer for both
situations. Orange(590-610nm) for a), and Violet(400-450nm) for b).
d) Again, confusion from the question c.. not really sure whether to use
orange(neon gas) or violet(mercury vapour) as the source. According to the textbook
the equation looks like this, n=c/v (n=absolute index of refraction, c=speed of light,
v=speed of light in substance) so am I looking to solve for the variable n?
e) Our ordinary main way to gather information about the world.


So it would be greatly appreciated if anyone could read this over
and correct me any wrongs. Any suggestions or advice are also appreciated,
Thanks again!
 
Physics news on Phys.org
nblu said:
A: a) given that the speed of light in "vacuum" is 3.00 x 10^8 m/s,
(not sure how to type-in the wavelength sign)
wavelength = 3.00 x 10^8 m/s / 5.00 x 10^14 Hz
Then it comes out to be, 6.00 x 10^-17 m which also equals to 600nm.
b) given that the speed of light in "water" is 2.2 x 10^8 m/s,
wavelength = 2.2 x 10^8 m/s / 5.00 x 10^14 Hz
which turns out to be 4.4 x 10^-7 m, and 440 nm.
Good. (Small typo in part a: -7, not -17.) You could also just have used the index of refraction for water.
c) I had a confusion because I wasn't really sure whether the question was asking
about the radiation in vacuum or in water (Sorry this is an independent study course
so there aren't any teachers whom I can contact). So, I gave the answer for both
situations. Orange(590-610nm) for a), and Violet(400-450nm) for b).
Color depends on frequency, not wavelength. (You don't see color until it hits your eye, so it doesn't matter what happened to it along the way as long as the frequency hasn't changed.)
d) Again, confusion from the question c.. not really sure whether to use
orange(neon gas) or violet(mercury vapour) as the source. According to the textbook
the equation looks like this, n=c/v (n=absolute index of refraction, c=speed of light,
v=speed of light in substance) so am I looking to solve for the variable n?
The source doesn't matter as you are given the speed in the medium.
e) Our ordinary main way to gather information about the world.
Sure.
 
Doc Al said:
Good. (Small typo in part a: -7, not -17.) You could also just have used the index of refraction for water.

Color depends on frequency, not wavelength. (You don't see color until it hits your eye, so it doesn't matter what happened to it along the way as long as the frequency hasn't changed.)

The source doesn't matter as you are given the speed in the medium.

Sure.

Thank you Doc Al,
I have read the textbook over again and I've realized that my prediction was incorrect.
I'm aware of the fact that the frequency range of visible light is 10^14 Hz to 10^15 and therefore, saying "Yes" to question c) is correct, however I'm not sure which colour I
should put down as my answer.
I have this chart called "Visible Spectrum" and it has all the colours listed with respect
to their "Range of wavelength in a vacuum".
I was thinking "Orange(590nm - 610nm)" is the only answer to write down because
the chart is discussing only about the wavelength in a vacuum.

Am I correct?

Thanks again Doc!
 
You are correct!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
15
Views
3K
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K