In this video, we're going to look at
some applications of Snell's Laws in the laboratory.
In the laboratory, we often use a laser beam,
or a single ray of white light,
to represent the rays that travel through our optical elements.
So, today we've got a white light source,
and I've got just a piece of glass that has parallel sides.
And I can send a beam of light into this glass,
and get a small fraction of that that is reflected off the front surface.
And most of it travels through the glass and comes out the other side.
And we see, the laser is seen with the larger angles,
but we see that the light bends into the material,
and bends when it hits the air again.
The light that exits the material,
is at the same angle as light entering but,
two beams of light are parallel.
But they're shifted transversely a little bit.
And the light that is reflected off the front surface
also is parallel to the light that's reflected off the back surface, and also enters out.
And we can see that the angle at which we entered,
this piece of glass affects this transverse shift.
So, if I enter at a very small angle,
you get a very small shift.
And as I enter at a much larger angle, that shift increases.
We can also look at an optical element that doesn't have parallel sides with this prism.
So, if I send in light,
into the front surface here,
then light exits the back surface, which is not parallel.
Put a 45 degree angle here,
and it exit this at an angle.
So, we only get the instant beam,
and the exiting beam,
to be parallel when we have a surface with parallel sides.
One last structure we can look at,
is something a little bit more complicated with some curvature.
With this lens, we can see the light entering into the center of the lens.
Exits at the same angle.
I'm entering it at zero degrees,
and we have parallel surfaces in the center.
As I send the light though the top of the lens,
we 'll get deflection downward.
And as I send the light through the bottom of the lens,
I get deflection upward.
And you can imagine, if we have rays coming in from
the top and the bottom, we would get focusing.
I would like to know what happens when I send a beam of light,
a ray of light, into the hypotenuse of this prism.
It's a 90 degree angle prism,
45 degree are their internal angles.
I want to send this light in,
the zero degree angle,
incident angle, to this front surface.
So, I don't expect it to bend at the front surface, and it doesn't.
You see the light coming into the prism,
and it doesn't bend at the front surface.
And at the back surface, it reflects.
So, I actually don't get any light coming out here at all.
And if I change the angle of this prism,
and I make that incident angle on the back surface,
glass to air surface,
smaller, there's a point where I get light
that is refracted out back into the air, on this back surface.
And as I make that angle larger,
there's a point where that beam goes away,
and all of my light is reflected from that back surface.
This is called total internal reflection,
which happens if I'm leaving material of higher index to go to a material like this,
like air, of lower index.
And there's a critical angle at which all of the light becomes reflected inside.
I'd like to be able to figure out at
what angle will I get total internal reflection in this prism.
And at one angle will I get light refracted into the air.
So, what is this critical angle, above which,
all the light reflects in the material in this prism, and below it,
some of the light refracts out of the prism?
Start with Snell's Laws, tells us that the index refraction of the first material,
in this case our prism,
times the sin of the incident angle,
is equal to the index refraction of our second material,
air in this case, times the sin of the transmitted angle.
In our case our first material has an index of about 1.5.
And our second material is air,
so it has an index of about 1.0.
Now at the critical angle,
this is the point where the sin of the transmitted angle goes to 1.
It can't get any larger than that.
So, this is the point where the transmitted angle goes to 90 degrees.
So, our transmitted angle is 90 degrees.
But we'd like to find our critical angle.
And we can go ahead and solve this.
We find that our critical angle in this case,
for an index of 1.5 material,
and index 1.0 air,
is about 42 degrees.
So, when we put the prism in the lab in this orientation,
where the incident angle is 45 degrees,
we get total internal reflection.
When I decrease that,
even just a little bit,
we start to get light that's refracted out into the air.
Amy SullivanResearch Associate
Robert McLeodProfessor
