7th edition HW Problem 1.A15
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7th edition HW Problem 1.A15
Hello, can someone please explain how to solve HW Problem 1A.15 in the 7th edition of the textbook?
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Re: 7th edition HW Problem 1.A15
For a line at 102.6nm, we know that n1=1.
Then use the equation c=(wavelength)(frequency), but rearrange it to look like:
v= c/(wavelength)
So:
v=(2.998x10^8)/(102.6x10^-9) =2.922x10^15 s^-1.
We then need to find n2.
Use this equation: v=R((1/n1^2) - (1/n2^2).
Then you will get this if you rearrange:
1/n2^2 =(1/1^2) -(v/R)
Then plug in your givens.
(1/n2^2)=1-((2.922x10^15)/(3.29x10^15))
This is equal to o.112.
So...
(1/n2^2) =0.112.
n2^2=9
Take the square root on each side to solve for n2
n2=3.
n1=1 and n2=3.
Then use the equation c=(wavelength)(frequency), but rearrange it to look like:
v= c/(wavelength)
So:
v=(2.998x10^8)/(102.6x10^-9) =2.922x10^15 s^-1.
We then need to find n2.
Use this equation: v=R((1/n1^2) - (1/n2^2).
Then you will get this if you rearrange:
1/n2^2 =(1/1^2) -(v/R)
Then plug in your givens.
(1/n2^2)=1-((2.922x10^15)/(3.29x10^15))
This is equal to o.112.
So...
(1/n2^2) =0.112.
n2^2=9
Take the square root on each side to solve for n2
n2=3.
n1=1 and n2=3.
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- Posts: 33
- Joined: Fri Sep 28, 2018 12:27 am
- Been upvoted: 1 time
Re: 7th edition HW Problem 1.A15
Wait, actually, sorry. My TA said we can't use the equation from the book. We are supposed to use another one, so the above explanation is wrong. Sorry.
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