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

In summary, the conversation is about understanding the formula for finding the wavelength of light in a medium with a certain index of refraction. The formula is \lambda_n = \lambda_1/n, where \lambda_n is the wavelength in the medium and \lambda_1 is the wavelength in vacuum or air. The conversation also discusses how to find the frequency and speed of the light in the medium.
  • #1
mr_coffee
1,629
1
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@!
 
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  • #2
You multiplied where you should have divided. (And, since they wanted the answer in nm, why did you change units?)

That equation should read:
[tex]\lambda_n = \lambda_1/n[/tex]
 
  • #3
You said it yourself. The wavelength in a denser medium (index n) is shorter. In particular equal to [itex]\lambda/n[/itex] where [itex]\lambda[/itex] is the wavelength in vacuum. So why do you multiply the wavelength by n?
 
  • #4
Thanks Doc,
Is [tex]\lambda_n[/tex]
the Wave Length of Light, in this case 598nm
and
[tex]\lambda_1[/tex]
is the wave length of the medium I'm wanting to find, in this case, glass?
 
  • #5
Ooo n/m i had those 2 mixed up, thanks guys! i got it right now! wee!
 
  • #6
mr_coffee said:
Thanks Doc,
Is [tex]\lambda_n[/tex]
the Wave Length of Light, in this case 598nm
and
[tex]\lambda_1[/tex]
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.) [tex]\lambda_n[/tex] is the wavelength in a medium with index of refraction = n; [tex]\lambda_1[/tex] 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.)
 
  • #7
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 [Broken]


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

1.723E8
 
Last edited by a moderator:
  • #8
The speed in the medium is c/n. Actually, it's that result that leads to [itex]\lambda/n[/itex].
 

1. What is the "plug and chug" method for finding the wave length of glass?

The "plug and chug" method is a common term used in science and math to describe a step-by-step approach to solving a problem. In the context of finding the wavelength of glass, it involves plugging in values for known variables (such as the speed of light and the refractive index of glass) into an equation and solving for the unknown variable (wavelength).

2. What are common mistakes made when using the "plug and chug" method to find the wave length of glass?

One common mistake is not using the correct units for the variables in the equation. For example, the speed of light is usually given in meters per second, so it is important to convert this to the correct unit (such as nanometers) before plugging it into the equation. Another mistake is not using the correct refractive index for the specific type of glass being studied.

3. How can I avoid making mistakes when using the "plug and chug" method to find the wave length of glass?

To avoid mistakes, it is important to double check all units and variables before plugging them into the equation. It can also be helpful to have a colleague review your work to catch any errors. Additionally, using a calculator or computer program can reduce the chances of making a calculation mistake.

4. What is the significance of finding the wave length of glass?

Finding the wavelength of glass is important in understanding the behavior of light as it passes through different materials. This information is used in various fields such as optics, materials science, and engineering. It can also help in the design and development of optical devices and instruments.

5. Are there other methods for finding the wave length of glass besides the "plug and chug" approach?

Yes, there are other methods such as using spectroscopy techniques or mathematical modeling. These methods may be more complex and require more advanced knowledge and equipment, but they can provide more accurate results in certain situations.

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