# OPTICS: linewidth

1. May 1, 2008

### 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 im trying to work out a question. but i think the formula i got for it isnt 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 dont get one of the answers listed. i know theres 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.

2. May 1, 2008

### 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!

3. May 1, 2008

### Fat Ryan

i did... (3 x 10^8)\(500 x 10^-9) * (2.5 x 10^-12) = 1500

4. May 2, 2008

### 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

5. May 2, 2008

### Fat Ryan

where did this come from?

6. May 2, 2008

### 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

7. May 2, 2008

### Fat Ryan

i dont understand why youre using these fractoins though. weve never used anything like this. we just used the formula i have listed above. i dont understand how these fractions relate to the formula.

8. May 2, 2008

### Redbelly98

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

$$\frac{\Delta \lambda}{ \lambda} = \frac{\Delta \nu}{ \nu}$$

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

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

$$\Delta \nu = \frac{c \cdot \Delta \lambda }{ \lambda^2}$$

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

9. May 2, 2008

### Fat Ryan

oh ok, so it looks like i was just missing the lambda^2. doh! thanx guys :)