The INbetween part

[mendhak]

It took me twenty minutes on Google before I realized I had lambda on MS-word.

anyways, here's my question:

ëƒ = c

where lambda is the wavelength, f is the frequency and c is the speed of light in a vacuum.

Now, in the book, it said:

If we solve the equation for f and differentiate with respect to ë we get

df/dë = -c/ë^2 ------------------------ (A)


If we now go to finite differences instead of differentials and only look at absolute values we get

Äf = cÄë/ë^2 -------------------------(B)


Thus given the width of a wavelength band, Äë we can compute the corresponding frequency band.

Ok, the part I'm lost at is how did they get (A) and how the **** did they get (B).

Someone please explain it!!

[bugzpodder]

not sure about be, the for a,

treat lambda as x, and f as y (well usually you take the dervitive to by dy/dx -- with respect to x, now its df/d(lambda) or with respect to lambda)

so you have xy=c

rearrange says y=c/x, where c can be treated as a constant.

so dy/dx=-c/x^2

[bugzpodder]

a reasonable explaination for b, it would be the same as a, execept i guess they introduced Ae. btw it says absolute value so i guess you take out the - sign.

[mendhak]

****, that was simple!!
Thanks.