Optics: Solving for Linewidth Frequency Bandwidth

  • Thread starter Thread starter Fat Ryan
  • Start date Start date
  • Tags Tags
    Optics
Click For Summary

Homework Help Overview

The discussion revolves around calculating the frequency bandwidth associated with the linewidth of green light, specifically using the wavelength of 500 nm and a linewidth of 2.5 x 10-3 nm. Participants are exploring the relationship between wavelength and frequency in the context of optics.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for converting linewidth in wavelength to frequency bandwidth, with some questioning the correctness of the original formula provided by the original poster. Others suggest alternative methods for calculating the frequency based on the wavelength and the given linewidth.

Discussion Status

The discussion is active, with various interpretations of the formula being explored. Some participants have provided alternative approaches and clarifications regarding the relationship between wavelength and frequency. There is acknowledgment of potential errors in the original formula, and participants are engaging in a productive exchange of ideas.

Contextual Notes

Participants note the importance of unit conversion and the implications of using fractional bandwidth in calculations. There is also mention of the original poster's confusion regarding the formula and its application, highlighting the need for clarity in understanding the relationships involved.

Fat Ryan
Messages
48
Reaction score
0
i wasnt sure where to put this so feel free to move it, mods, if its in the wrong place. just reviewing for my optics final and I am trying to work out a question. but i think the formula i got for it isn't right.

QUESTION:
The linewidth of a green light (lambda= 500nm) is DELTA-lambda=2.5 X 10^(-3) nm, what is the frequency bandwidth DELTA-nu?

a) 3X10^8 Hz
b) 3X10^9 Hz
c) 3X10^6 Hz
d) 3X10^-8 Hz
e) none o the above

SOLUTION:
i have the formula...DELTA-nu= (-c / lambda)*DELTA-lambda...my prof gave us this forumula to use on the last day of class when we were going over these problems. but when i use it i don't get one of the answers listed. i know there's an option E, but he virtually never uses that. and also my answer is 1500 which is off by a factor of 2 X 10^x...i found that odd. and lastly when you compute the units of the above mentioned formula you end up with m/s...shouldnt the units be in 1/s? i must have this formula wrong.
 
Physics news on Phys.org
It's just the same formula as c = wavelength * frequency.
You can confirm it by checking the dimensions.

You probably just made a mistake in the units - convert everything to metres/seconds first!
 
mgb_phys said:
It's just the same formula as c = wavelength * frequency.
You can confirm it by checking the dimensions.

You probably just made a mistake in the units - convert everything to metres/seconds first!

i did... (3 x 10^8)\(500 x 10^-9) * (2.5 x 10^-12) = 1500
 
A simpler way to think of it.
500nm light has a frequency of ( 3E8 ms-1 / 500E-9 m ) = 6*10^14Hz
2.5E-3 is 0.0025/500 of this
so bandwidth = 6*10^14 Hz * (0.0025/500) = 3 * 10^9
 
mgb_phys said:
A simpler way to think of it.
500nm light has a frequency of ( 3E8 ms-1 / 500E-9 m ) = 6*10^14Hz
2.5E-3 is 0.0025/500 of this
so bandwidth = 6*10^14 Hz * (0.0025/500) = 3 * 10^9

where did this come from?
 
Think of bandwidth as the difference between two signals.
So an FM station braodcasts on 100Mhz with a bandwidth of 10KHz means it uses between 100,000,000 and 100,010,000 Hz.
In other words it uses 10,000/100,000,000 = 0.0001 fractional bandwidth or 0.01%

The linewidth/bandwidth you quoted was 2.5E-3nm ie 0.0025nm and a wavelength of 500nm.
So the fractional bandwidth is 0.0025/500 = 5E-6 or 0.0005%
then just take the frequency of the light 6E14Hz and apply this fraction.


ps '6E14' is just the computer shorthand for 6*10^14
 
mgb_phys said:
Think of bandwidth as the difference between two signals.
So an FM station braodcasts on 100Mhz with a bandwidth of 10KHz means it uses between 100,000,000 and 100,010,000 Hz.
In other words it uses 10,000/100,000,000 = 0.0001 fractional bandwidth or 0.01%

The linewidth/bandwidth you quoted was 2.5E-3nm ie 0.0025nm and a wavelength of 500nm.
So the fractional bandwidth is 0.0025/500 = 5E-6 or 0.0005%
then just take the frequency of the light 6E14Hz and apply this fraction.


ps '6E14' is just the computer shorthand for 6*10^14
i don't understand why youre using these fractoins though. we've never used anything like this. we just used the formula i have listed above. i don't understand how these fractions relate to the formula.
 
As long as the linewidth is a small fraction of the wavelength (as it is in this case),

[tex] \frac{\Delta \lambda}{ \lambda} = \frac{\Delta \nu}{ \nu}[/tex]

I.e., they have the same fractional width whether it's wavelength or frequency.

Using [tex]\nu = c/\lambda[/tex],
the above equation becomes

[tex] \Delta \nu = \frac{c \cdot \Delta \lambda }{ \lambda^2}[/tex]

Which is different that the formula quoted in Ryan's original post.
 
Redbelly98 said:
As long as the linewidth is a small fraction of the wavelength (as it is in this case),

[tex] \frac{\Delta \lambda}{ \lambda} = \frac{\Delta \nu}{ \nu}[/tex]

I.e., they have the same fractional width whether it's wavelength or frequency.

Using [tex]\nu = c/\lambda[/tex],
the above equation becomes

[tex] \Delta \nu = \frac{c \cdot \Delta \lambda }{ \lambda^2}[/tex]

Which is different that the formula quoted in Ryan's original post.
oh ok, so it looks like i was just missing the lambda^2. doh! thanks guys :)
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
3K
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K