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What is the wavelength of light having a frequency of 600 THz

 Mar 28, 2019
 #1
avatar+4779 
+3

\(c = \lambda \nu\\ 3\times 10^8 m/s = \lambda m\cdot 600\times 10^{12}s^{-1}\\ \lambda = \dfrac{3\times 10^8 m/s}{6\times 10^{14}s^{-1}}= \dfrac 1 2 \times 10^{-6}m=500nm\)

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 Mar 28, 2019
 #2
avatar+983 
+2

Where did you get all the numbers? I asked the question so I could understand it, not just get the answer.

SmartMathMan  Mar 28, 2019
edited by SmartMathMan  Mar 28, 2019
 #3
avatar+4779 
+3

The speed of light is about 3x108 m/s, that's a standard physical constant you will eventually have memorized.

 

600 THz is 600 x1012 s-1 , that came with the problem.

Rom  Mar 28, 2019
edited by Rom  Mar 28, 2019
 #4
avatar+983 
+1

Thx for explaining it. I'm not trying to call you out, I was just confused.

SmartMathMan  Mar 28, 2019
 #5
avatar+100042 
+3

Hi Rom, I am not trying to hijack your question, I started this ages ago.

I will post it, even though you have already given a good answer.

---------------

 

Hi SmartMathMan :)

 

I think you are expected to memorize the speed of light. 

You can google it of course, which is what i just did.

 

https://goo.gl/en2ZTc

 

So we have the speed of light = 3*10^8   m/s

 

1 Herz means that the wave passes a point in seocnd.  That is  1 Herz=1 wavelength/sec.

Your wave has  600THz which is     600*10^12 wavelengths/sec

 

You want to know wavelegth

 

For any wave

speed of wave (m/s)= wavelength (m)* number of waves/sec

 

In this case the speed is the speed of light which is 3*10^8 m/s

the wavelength is unknown

The number of waves per sec is the number of Herz.

     600THz = 600*10^12 Herz = 6*10^14 Herz       (6*10^14 waves pass a certain point each second)

 

speed of wave (m/s)= wavelength (m)* number of waves/sec

\(c=\lambda * v\\ \lambda=c\div v\\ \lambda=\frac{3*10^8\;m}{sec}\div\frac{ 6*10^{14}waves}{sec}\\ \lambda=\frac{3*10^8\;m}{sec}\div\frac{ 6*10^{14}waves}{sec}\\ \lambda=\frac{3*10^8\;m}{sec}\times\frac{sec}{ 6*10^{14}waves}\\ \text{the secs cancel out leaving}\\ \lambda=\frac{3*10^8}{ 6*10^{14}}\frac{metres}{wave}\\~\\ \lambda=\frac{3*10^8}{ 6*10^{14}}metres\\~\\ \text{the word 'wave' is just understood} \)

.
 Mar 28, 2019
 #6
avatar+983 
+1

sorry, what is the right answer? I did the math myself. but now that you posted this i'm not sure if i'm right or wrong.

SmartMathMan  Mar 28, 2019
 #7
avatar+4779 
+2

1/2 x 10-6m = 500 x 10-9m = 500 nm is the correct answer

Rom  Mar 28, 2019
 #8
avatar+100042 
+2

\(\lambda = \frac{3*10^8}{6*10^{14}}metres\\ \lambda = 0.5*10^{-6}metres\\ \text{there are 10^9 nm in a metre}\\ \lambda = 0.5*10^{-6}*10^9 nanometres\\ \lambda = 0.5*10^{3}nanometres\\ \lambda =500 \;nm\\ \)

 

 

Just like Rom has said     cool

Melody  Mar 28, 2019
edited by Melody  Mar 28, 2019
 #9
avatar+1406 
+2

Note that Rom used engineering notation to depict the numerical values.   Engineering notation is a specialized subset of scientific notation, and its use is common (and preferred) in all branches of physical sciences. 

 

Reviewing this will help you understand the (different) notations presented by Rom and Melody

https://en.wikipedia.org/wiki/Engineering_notation

 

Start by familiarizing yourself with the eight prefixes that modify the unit value:

Observe that engineering notation uses base 1000; here it’s contrasted to base 10 used in standard scientific notation.   

 

tera     T           10004     1012    

giga     G         10003     109     

mega   M         10002     106     

kilo      k          10001     103     

UNIT                10000     100     

milli     m         1000−1   10−3   

micro  μ          1000−2   10−6   

nano    n         1000−3   10−9   

pico     p         1000−4   10−12

 

Having an instant working knowledge of these prefixes will help you understand these questions.

 

Another note:

Also, in the Wiki article pay attention to the overview, starting with:

“... Compared to normalized scientific notation, one disadvantage of using SI prefixes and engineering notation is that significant figures are not always readily apparent. ...”   Note in the text, following this extract, that the speed of light is used specifically as an example for significant figures. 

 

The speed of light constant is used in your question, and though significant figures is not an explicit concern for your presented question or its solution, it is always an implied concern. An understanding of significant figures for constants and measurements will quickly become significant as you pursue your studies in physics.

 

Other general helps for learning physics:

Learn the Greek alphabet and the context in which the symbols are used (in mathematics and physics). In your question, the Greek letter \(\lambda\) (lambda) is used and its context means Wave Length. The Latin letter (c) is used to denote the constant for the speed of light in a vacuum.  

 

Remember this constant (c) is speed of light in a vacuum; when light passes through other media, it slows down, and further, it slows down in proportion to its frequency (wave length). This is why you can see “rainbow” colors when composite (white) light passes through the atmosphere or a prism.     

 

Another thing to remember is “light” in this context refers to the full electromagnetic radiation spectrum, not just the small portion that is visible.    

 

 

 

GA    

 Mar 28, 2019

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