Question

explain why total internal reflection does not occur when light goes from air into glass

Answer 1

"Why" is a very challenging word in science/humanity.
When light goes from air into glass, it's going into a medium of a HIGHER REFRACTIVE INDEX. The surface of the glass acts as a sort of "translucent" mirror reflecting some light, passing more light, so that the energy of the incoming light equals that of the transmitted and partially reflected light, and all the other mathematical bits and pieces are balanced.

This doesn't seem to happen when light "tries" to come OUT of the glass into the air. Beyond the CRITICAL angle, ALL the light (at least I think it's all the light) is TOTALLY INTERNALLY REFLECTED. It surprised Snell, I guess, when he and his mates discovered it, and is now much loved in OPTICAL FIBRES and other bits of jazzy super hi tech ...

This prob won't address the word "why". A greater mind than my scruffy thing is needed for that, methinks.

Answer 2

You can take a heuristic approach into understanding why. From Snell's law, we have: \[\frac{Sin\,\theta_i}{Sin\,\theta_r}=\frac{n_r}{n_i}\] or, rearranging,
\[Sin\,\theta_i=\frac{n_r}{n_i}Sin\,\theta_r\]
Note that Sine values go from -1 to 1, so we cannot exceed 1. From 0 to 90 degrees, Sine is an increasing function. In the case when the light is incident from air into glass, we have:
\[Sin\,\theta_i = 1.5\,Sin\,\theta_r\]
which says that the angle of refraction is always going to be smaller than the angle of incidence (since sine only increases between 0 to 90 degrees), and so the law of refraction holds nicely without problem!

In the case when the light is incident from air into glass, we have:
\[Sin\,\theta_i = \frac{1}{1.5}\,Sin\,\theta_r\]
Now we have a problem. This says that the Sine of the angle of incidence is 1.5 times larger than the Sine of the angle of refraction. But refraction, the bending of light, can only go up to 90 degrees. Beyond that, the formula does not describe the correct physics. To see why it would require a trip into boundary conditions of electromagnetism which might be a little daunting. Nevertheless, we can define the critical angle (of incidence as:
\[Sin{\,\theta_c}=\frac{1}{1.5}Sin\,90^o\]
Which tells us the "largest" angle of incidence we can use for refraction to occur when light is exiting from glass into air.

Answer 3

@Festinger nice one.