# Total Internal Reflection of Light Simulation

#### Quiz

Q1) Total internal reflection occurs when the angle of incidence is

A. greater than the angle of refraction
B. equal to the critical angle
C. greater than the critical angle
D. greater than 45°

Q2) Total internal reflection

A. refers to light being reflected from a plane mirror
B. may occur when a fisherman looks at a fish in a lake
C. may occur when a fish looks at a fisherman on a lake

#### Total Internal Reflection of Light Simulation

When the next simulation is not visible, please refer to the following link.

Total Internal Reflection
When you’re in a physics mood and you’re going to take a bath, fill the tub extra deep and bring a waterproof flashlight into the tub with you. Turn the bathroom light off. Shine the submerged light straight up and then slowly tip it and note how the intensity of the emerging beam diminishes and how more light is reflected from the water surface to the bottom of the tub.

The Critical Angle
At a certain angle, called the critical angle, you’ll notice that the beam no longer emerges into the air above the surface. The critical angle is the angle of incidence that results in the light being refracted at an angle of 90° with respect to the normal. As a result, the intensity of the emerging beam reduces to zero. When the flashlight is tipped beyond the critical angle (48° from the normal in water), the beam cannot enter the air; it is only reflected. The beam is experiencing total internal reflection, which is the complete reflection of light back into its original medium.
FIGURE 1. You can observe total internal reflection in your bathtub. a–d. Light emitted in the water at angles below the critical angle is partly refracted and partly reflected at the surface. e. At the critical angle, the emerging beam skims the surface. f. Past the critical angle, there is total internal reflection.

Total internal reflection occurs when the angle of incidence is larger than the critical angle. The only light emerging from the water surface is that which is diffusely reflected from the bottom of the bathtub. This procedure is shown in Figure 1. The proportions of light refracted and reflected are indicated by the relative lengths of the solid arrows. The light reflected beneath the surface obeys the law of reflection: The angle of incidence is equal to the angle of reflection. The critical angle for glass is about 43°, depending on the type of glass. This means that within the glass, rays of light that are more than 43° from the normal to a surface will be totally internally reflected at that surface.

Figure 2. (a) This photo demonstrates several different paths of light radiated from the bottom of an aquarium. (b) At the critical angle, $\theta_{c}$ , a light ray will travel parallel to the boundary. Any rays with an angle of incidence greater than $\theta_{c}$  will be totally internally reflected at the boundary.

Snell’s law can be used to find the critical angle. When the angle of incidence, $\theta_{i}$ , equals the critical angle, $\theta_{c}$ , then the angle of refraction, $\theta_{r}$ , equals 90°. Substituting these values into Snell’s law gives the following relation.

$\ n_{i} sin\theta_{c} = n_{r} sin90°$

Because the sine of 90° equals 1, the following relationship results.

 CRITICAL ANGLE $\sin\theta_{c} = \frac{n_{r}}{n_{i}}~~~~~for~n_{i}>n_{r}$

Note that this equation can be used only when $\ n_{i}$ is greater than $\ n_{r}$. In other words, total internal reflection occurs only when light moves along a path from a medium of higher index of refraction to a medium of lower index of refraction. If ni were less than $\ n_{r}$, this equation would give $\ sin\theta_{c} > 1$, which is an impossible result because by definition the sine of an angle can never be greater than 1. When the second substance is air, the critical angle is small for substances with large indices of refraction.Diamonds, which have an index of refraction of 2.419, have a critical angle of 24.4°. By comparison, the critical angle for crown glass, a very clear optical glass, where $\ n$ = 1.52, is 41.0°. Because diamonds have such a small critical angle, most of the light that enters a cut diamond is totally internally reflected. The reflected light eventually exits the diamond from the most visible faces of the diamond. Jewelers cut diamonds so that the maximum light entering the upper surface is reflected back to these faces.