Timeline for Trapping light by total internal reflection
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8 at 13:35 | vote | accept | Nightwing | ||
| Mar 7 at 19:03 | comment | added | SAKhan | Light has to be incident from denser to rarer for this to happen. | |
| Mar 6 at 20:36 | answer | added | Kellan Heerdegen | timeline score: 0 | |
| Mar 6 at 19:51 | answer | added | HerrAlvé | timeline score: 7 | |
| Mar 6 at 15:52 | answer | added | The Fright Of The Night | timeline score: 1 | |
| Mar 6 at 14:21 | comment | added | Nightwing | Yeah thats totally true. But can it be mathematically contradicted that since all conditions for total internal reflection are satisfied within the prism hence the light will not get refracted. That is what I need. | |
| Mar 6 at 12:07 | comment | added | MichaelK | Optics are entirely reversible. The "one-way mirror" does not exist. If a ray can enter at some angle, then by simply reversing the direciton, is can also exit by that same angle. | |
| Mar 6 at 12:01 | answer | added | Jagerber48 | timeline score: 1 | |
| Mar 6 at 11:41 | comment | added | naturallyInconsistent | Yes, if you considered Snell's Law in enough detail, you will realise that either the light can enter and leave at those angles, or it will neither enter nor leave at those angles. You cannot create a scenario whereby the light gets trapped like how you think it can. | |
| Mar 6 at 11:36 | comment | added | Nightwing | It can get refracted by the glass because it enters from a rarer medium from where total internal reflection is not possible.However it attempts to leave at an angle of 45 degrees which is greater than the critical angle of the glass air interface. I know that it would leave if TIR was not possible from the inside. Do you possibly mean that the light can't enter in the manner shown because if it does it must be able to leave too? | |
| Mar 6 at 9:23 | comment | added | naturallyInconsistent | Look at your Snell's Law. If it can refract into the thing and move at 45 degrees, then it must also be able to leave at 45 degrees. | |
| Mar 6 at 9:20 | comment | added | Nightwing | No sir, it can enter by refraction but if it tries to leave from the inside at the given angle of 45 degrees it can only get reflected and not refracted. | |
| Mar 6 at 8:54 | comment | added | naturallyInconsistent | I'm trying to tell you that if you have a pretty symmetric situation like this ellipse, then if it can enter by refraction, then it can also exit by refraction. | |
| Mar 6 at 7:35 | history | edited | Nightwing | CC BY-SA 4.0 |
added 211 characters in body
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| Mar 6 at 6:02 | history | edited | Nightwing | CC BY-SA 4.0 |
added 17 characters in body
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| Mar 6 at 6:00 | comment | added | Nightwing | It enters the prism by refraction and not through a hole. | |
| Mar 6 at 5:56 | comment | added | naturallyInconsistent | If it can enter, then it will be able to exit when it gets to a similar point the other way. | |
| Mar 6 at 5:31 | history | asked | Nightwing | CC BY-SA 4.0 |