Petrine shines a laser inside a glass block toward its top surface, first steeply, then at a larger and larger angle from the normal. At a small angle, most of the light leaves the glass. Past one particular angle, none of it leaves: the whole beam reflects back down into the glass.
Light trying to LEAVE a dense medium can be trapped. This lesson describes that trapping, and names the angle where it begins.
Set the scene: the light travels in the optically denser medium, the glass, and strikes the boundary with the less dense air.
Leaving a denser medium, the refracted ray bends AWAY from the normal: the angle outside is larger than the angle inside.
At a small angle of incidence, most of the light refracts out into the air, bent away from the normal. A little reflects back inside.
As the angle of incidence grows, the refracted ray leans further and further toward the surface.
The reflected beam obeys the reflection law you already use: it leaves at the same angle from the normal as it arrived, on the other side.
For Petrine's glass and air, the critical angle is about 42°. Strike the surface at 50° from the normal, and the beam has no way out.
Light inside a diamond strikes the boundary with air at 40° from the normal. The critical angle for diamond and air is about 24°. What happens to the light?
The light is in the denser medium, heading for a less dense one: the trapping setup.
Compare the angle of incidence with the critical angle: 40° is past 24°.
Past the critical angle, all of the light reflects back inside and none refracts out.
The whole beam reflects back into the diamond: total internal reflection.
Light inside the same diamond strikes the boundary at 15° from the normal. What happens now?
Compare again: 15° is below the 24° critical angle.
Below the critical angle, the light can still get out.
Most of it refracts out into the air, bent away from the normal; a little reflects back inside.
Most of the light refracts out of the diamond; only a little reflects back.
Light inside a diamond strikes the boundary with air at an angle past the critical angle. What happens to the light?
Light inside a glass paperweight strikes the boundary with air at an angle smaller than the critical angle. What happens to the light?
Maud says: 'Past the critical angle, none of the light leaves the glass: the boundary reflects all of it back inside.' Is Maud correct?
Which statement describes total internal reflection?
A swimmer underwater at night shines a flashlight up toward the surface at 60° from the normal. The critical angle for water and air is about 49°. What happens to the beam at the surface?
Light inside a glass block strikes the top surface at 55° from the normal, past the glass-air critical angle, and totally internally reflects. At what angle from the normal does the reflected beam travel back into the glass?