April 25, 2024, 05:39:30 PM
Forum Rules: Read This Before Posting


Topic: optically active  (Read 4253 times)

0 Members and 1 Guest are viewing this topic.

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
optically active
« on: August 13, 2010, 11:39:19 AM »
how do a plain of polarized light rotated by chiral molecules?

why can this happen?!

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: optically active
« Reply #1 on: August 14, 2010, 07:03:21 AM »
My research: Google Scholar and Researchgate

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: optically active
« Reply #2 on: August 14, 2010, 10:08:05 AM »
wow wow wow ...thank you so much.. i finally understand !!!!

thanks!!!

Offline McCoy

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +4/-2
Re: optically active
« Reply #3 on: August 14, 2010, 01:47:27 PM »
http://www.physicsforums.com/showthread.php?t=52941
wow! physicists explaining chemistry better! hey pls, there is no ill intention ( i just mean they have done a good job). cheers.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27663
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: optically active
« Reply #4 on: August 14, 2010, 04:05:51 PM »
Actually movies is a chemist, few years ago he was quite active here :)
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: optically active
« Reply #5 on: August 15, 2010, 12:45:44 AM »
most of the idea i understand...just a little problem:

as movies said: "intersects where the vertical wave is at 0 and the horizontal wave is at 1"
does it mean that the amplitude of vertical wave is 0 all the time?

and he also said that: frequency of the light is changed when pass through the isomer.
so, how do the speed of light, and wavelength of light changed? and how about amplitude?

Offline Mesogen

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-2
Re: optically active
« Reply #6 on: August 23, 2010, 02:22:58 PM »
most of the idea i understand...just a little problem:

as movies said: "intersects where the vertical wave is at 0 and the horizontal wave is at 1"
does it mean that the amplitude of vertical wave is 0 all the time?

and he also said that: frequency of the light is changed when pass through the isomer.
so, how do the speed of light, and wavelength of light changed? and how about amplitude?

When you're talking about rotating "Plane polarized light" you're talking about putting linearly polarized light into you're sample and seeing some kind of rotation. The source of this rotation is essentially a phase shift.

You can think of linearly polarized light as light oscillating in one direction, i.e. up and down, however, to understand the source of rotation fully you must go a bit further than this...

Linearly polarized light is essentially a summation of two vectors, see here: http://www.chem.nottingham.ac.uk/Tut_CPL.phtml (you'll want to look at the linearly polarized 45 degree option).

These two vectors represent the two components of light that vibrate perpendicularly to each other (as you can see in the animation).

When linearly polarized light enters a chiral material it "sees" the chiral centre which is locked in position (i.e. it is only one enantiomer). What the light actually "sees" is the distribution of electrons around the chiral centre).

You get optical rotation because one of the two components of light is retarded with respect to the other (the two components have different refractive indices), this creates a phase shift between the two components and therefore the light is rotated (this is essentially a birefringence effect). This is what you measure as a rotation.  

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: optically active
« Reply #7 on: August 25, 2010, 12:13:54 AM »
huh? so do u mean that this is a phase change instead of "one component remain unchange, while another change its wavelength"?

why is "this"?

Sponsored Links