Optics: Solving for Linewidth Frequency Bandwidth

[Fat Ryan]

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 forumla...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 forumla you end up with m/s...shouldnt the units be in 1/s? i must have this formula wrong.

 

[mgb_phys]

It's just the same formula as c = wavelenght * frequency.
You can confirm it by checking the dimensions.

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

 

[Fat Ryan]

It's just the same formula as c = wavelenght * 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

 

[mgb_phys]

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

 

[Fat Ryan]

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?

 

[mgb_phys]

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

 

[Fat Ryan]

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.

 

[Redbelly98]

As long as the linewidth is a small fraction of the wavelength (as it is in this case),

<br /> \frac{\Delta \lambda}{ \lambda} = \frac{\Delta \nu}{ \nu}<br />

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

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

<br /> \Delta \nu = \frac{c \cdot \Delta \lambda }{ \lambda^2}<br />

Which is different that the formula quoted in Ryan's original post.

 

[Fat Ryan]

As long as the linewidth is a small fraction of the wavelength (as it is in this case),

<br /> \frac{\Delta \lambda}{ \lambda} = \frac{\Delta \nu}{ \nu}<br />

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

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

<br /> \Delta \nu = \frac{c \cdot \Delta \lambda }{ \lambda^2}<br />

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 :)