Finding the wave length of glass, plug and chug but messing up somewhere

[mr_coffee]

Hello everyone, i don't think I'm understanidn ghits formula correctly.

THe problem says:
The wavelength of yellow sodium light in air is 589 nm.

(a) What is its frequency?
I found, 5.09E14 Hz

(b) What is its wavelength in glass whose index of refraction is 1.74?
? nm

OKay I'm using the equation;
Wave Length of a Light = wave length of medium/index of medium
But I'm not sure if I'm reading it right, it says:
WaveLength1 = WaveLength2/n
where n is the index of Refraction
This equation relatse the wave length of light in any medium to its wavelength in vaccum. It tells us that the greater the index of refraction of a medium, the smaller the wavelength of light in that medium.

So i said, WaveLength1 is the wave length of the medium, which is glass in my case. I said WaveLength2 is the wavelength of yellow sodium light in air is 589 nm. And n is of course 1.74;

WaveLength = (589E-9)(1.74) = .000001, he wanted it in nm, so i said it was: 1000nm, which was wrong. ANy ideas on what I f'ed up on ?
THanks@!

 

[Doc Al]

You multiplied where you should have divided. (And, since they wanted the answer in nm, why did you change units?)

That equation should read:
\lambda_n = \lambda_1/n

 

[Galileo]

You said it yourself. The wavelength in a denser medium (index n) is shorter. In particular equal to \lambda/n where \lambda is the wavelength in vacuum. So why do you multiply the wavelength by n?

 

[mr_coffee]

Thanks Doc,
Is \lambda_n
the Wave Length of Light, in this case 598nm
and
\lambda_1
is the wave length of the medium I'm wanting to find, in this case, glass?

 

[mr_coffee]

Ooo n/m i had those 2 mixed up, thanks guys! i got it right now! wee!

 

[Doc Al]

Thanks Doc,
Is \lambda_n
the Wave Length of Light, in this case 598nm
and
\lambda_1
is the wave length of the medium I'm wanting to find, in this case, glass?

No, just the opposite. (As you've figured out already.) \lambda_n is the wavelength in a medium with index of refraction = n; \lambda_1 is the wavelength in a medium with index of refraction = 1 (vacuum or air). (You are wanting to find the wavelength where n = 1.74.)

 

[mr_coffee]

I was going to start a new thread but you guys already have a background on what I'm doing. There is a part c to this question and it says:
(c) From the results of (a) and (b) find its speed in this glass.
Well here is my work, there is a lot of jibberish on it, but I put a box around part c, and I'm confused on why I'm messing this one up. If your confused on anything I did, i'll explain!
Part A is right, so is part B now:
(a) What is its frequency?
I found, 5.09E14 Hz
(b) What is its wavelength in glass whose index of refraction is 1.74?
338.5 nm


http://img86.imageshack.us/img86/5695/lastscan2sd.jpg


My bad! I figured it out while explaining to you... I used 338.5nm, i thought i was m in my calculation the right answer is:

1.723E8

 

[Galileo]

The speed in the medium is c/n. Actually, it's that result that leads to \lambda/n.