Total internal reflection — find critical angle for glass (n=1.5) to water (n=1.33)

Total Internal Reflection (TIR) occurs when:

1. Light travels from a denser medium to a less dense medium (higher $n$ to lower $n$)
2. The angle of incidence exceeds the critical angle $\theta_c$

Here, glass ($n = 1.5$) is denser than water ($n = 1.33$), so TIR is possible at this interface when light travels from glass into water.

At the critical angle, the refracted ray just grazes the interface (angle of refraction = 90°).

Applying Snell's law:

$n_1 \sin\theta_c = n_2 \sin 90°$

$n_1 \sin\theta_c = n_2 \times 1$

$\sin\theta_c = \frac{n_2}{n_1}$

Here, $n_1 = 1.5$ (glass, denser), $n_2 = 1.33$ (water, less dense):

$\sin\theta_c = \frac{1.33}{1.5} = 0.8867$

$\theta_c = \sin^{-1}(0.8867)$

$\theta_c \approx 62.5°$

So the critical angle at a glass-water interface is approximately 62.5°.

Any ray hitting the glass-water boundary from inside the glass at an angle greater than 62.5° will undergo total internal reflection. At angles less than 62.5°, some light is refracted into water (and some is reflected).

For comparison, the critical angle for glass-to-air (n_air = 1) is: $\sin\theta_c = 1/1.5 = 0.667$ → $\theta_c \approx 41.8°$ — much smaller.