Timeline for Why doesn't the frequency of light change during refraction?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2022 at 3:19 | comment | added | nasu | @user4552 These conditions as stated are enough for static fields. For a wave you need phase matching too, to ensure that hese condtions hold at all times. This requires same frequency. | |
| Mar 16, 2022 at 15:35 | review | Suggested edits | |||
| Mar 16, 2022 at 16:02 | |||||
| Mar 22, 2018 at 23:41 | history | edited | stafusa | CC BY-SA 3.0 |
Fixed a small typo.
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| Jan 24, 2014 at 6:15 | review | Suggested edits | |||
| Jan 24, 2014 at 6:24 | |||||
| Sep 15, 2013 at 20:24 | comment | added | user4552 | I'm not sure I quite buy this answer. The things that have to be continuous at the boundary are $D_\perp$, $E_\parallel$, $B_\perp$, and $H_\parallel$. On the other hand, there can be discontinuities in $D_\parallel$, $E_\perp$, $B_\parallel$, and $H_\perp$. So I think there is really more that needs to be filled in to make this a valid argument. | |
| Apr 14, 2013 at 14:51 | vote | accept | Self-Made Man | ||
| Mar 30, 2013 at 17:56 | comment | added | Self-Made Man | I think it's the simplest explanation... | |
| Mar 30, 2013 at 8:11 | history | answered | John Rennie | CC BY-SA 3.0 |