optical problem with critical angle

[picclock]

(Mods, I know this is way off topic, but I swear the forum members here are the smartest people I know, so please move to the break room if you feel it is too inappropriate)

So I'm building this sight glass that will double as a level sensor and I figure I can use the critical angle of a light source at the water air boundary to prevent light from reaching the photosensor. As you know the critical angle is where the light reflects off of the boundary to internally reflect, thus depriving the sensor of light. So just before I drill the holes for mounting the sensors I'm thinking that I had better check that this works the way its supposed to.

So I set up a laser pointer and a piece of glass with a piece of paper behind it to show the spot which I'm expecting to disappear after the critical angle is exceeded - problem is that it doesn't. I can rotate the glass or change the angle of the pointer and the dot passes through the glass with very little change at all.

Any help much appreciated as I'm way out of my depth here.

Can post pictures if it helps.

Many Thanks

picclock

 

[Henk]

Just a quick guess:

I assume you have a plate of glass and shine the laser pointer through it under varisou angles.

If so, what happens is that inside the glass the laser pointer angle is below the ciritical angle, and therefor nicely goes though the glass-air interface.

If what you did is something different, then a grpah would really help.

Regards,
Henk

 

[picclock]

Hi Henk

Spot on, but the light seems to straight through regardless of the angle. Pictures now.

implication is I'm doing something wrong or things aren't the way they are stated -

http://en.wikipedia.org/wiki/Total_internal_reflection

Many thanks for your interest.

Best Regards

picclock

 

[Ken I]

Picclock,
I just tried it with the same results no critical angle.

Heres a quote from this site

http://www.jiskha.com/display.cgi?id=1209738725

"The critical angle is an angle in which light going from a optically dense substance to a lessor dense substance will undergo total internal reflection. If light is going from air, the optically less dense (lower index of refraction), there will be no total reflection at the boundry."

But to my mind that still does not explain the lack of critical angle of the beam exiting the glass into air - so its got me knackered.

I know optic fibres are actually two different index materials co-axially to get total internal reflection.
So I think you need some sort of coating on your "lens"

Ken

 

[Henk]

Things happen exactly the way they are described in the wikipedia link.

You have to keep in mind that if it happens the total reflectin happens on the glas to air interface. If the incident light has a large angle to the normal total internal reflection happens.

However, the way you have gotten the beam inside the glass plate is via a first air to glass interface. The angle inside the glas will always have a smaller angle (i.e. more perpendicular to the interface) then the angle outside of the glass. Now the funny thing is that as both interfaces are parallel, the ray inside the glas can never have an angle sufficiently large as to see TIR (total internal reflection, marginal theoretical exception : get the light before the plate parallel to the interface...).
That is why in the example on wikipedia, they use a cilindrical front surface to get the internal angle sufficiently high to get TIR.

Now it depends on the actual layout of whatever you intended to do to say something sensible about whether it works yes or no.

Hope it helps.

 

[Lew Hartswick]

What Henk said: The thing to remember the critical angle only applies if the
beam STARTS OUT inside the medium. You are not doing that.
...Lew...

 

[picclock]

Hi Henk

I think you have resolved the issue, in that the light has to be travelling at that angle inside the glass (or denser medium) before the effect takes place.

I did a test with a glass of water shining the light up from the bottom, and tilted the whole thing and the beam of light does disappear at about the correct angle. Fortunately, further tests with a test tube and a small beam angle show that when the liquid rises to the beam level the angular deflection is large enough to move the beam away from the photosensor, thus allowing the level to be sensed by deflection rather than critical angle..

Many thanks for your assistance

Best Regards

picclock