Geometrical Optics Study Guide

Example

Two planes are set such that they form an angle of 100° with each other. A light is incident on the first surface at an angle of 60°. Find the angle of the reflected light from the second plane with respect to the normal to that plane.

Solution

First, we need to set our angle diagram and use the laws of reflection to determine the angle of the outgoing ray (see Figure 18.4).

In the ABC triangle, the angle at A can be found from the first law of reflection:

θ_i = θ_r = 60°

And since the triangle totals 180° with A at 30°, C is 100° and B is 50°.

θ_i = 90° – B = 90° – 50° = 40°

θ* _i = 40°

Again, we use the law of reflection on the second surface:

θ* _i = θ* _r = 40°

Refraction of Light

The examples we studied previously are probably easy to understand if you have performed experiments such as shining a light on a mirror or a shiny metal surface. But what happens when light meets a transparent medium and not an opaque one? We all know that light is transmitted beyond the surface of separation. Light goes through the water in your glass and even through the glass itself. This phenomenon of light propagation from a transparent medium to another transparent medium is called refraction. But refraction is affected by the properties of the medium itself, and the direction of propagation is changed by it. The corresponding rays are called incident and refracted rays, as are the angles. Figure 18.6 shows an incident wave and the refraction on the separation surfaces.

On the far left, we have the incident ray refracting on the surface and propagating inside Medium 2 at an angle smaller than the incident angle:

θ_i < θ_r

This behavior happens when light comes from a medium where the speed of light is larger and propagates into a medium with reduced speed. The light bends toward the normal when v_l > v₂ where v_l and v₂ are the speed of light in Medium 1 and Medium 2.

On the right side of the picture, we see the light reflecting from the bottom of Medium 2 and refracting at the surface of separation a second time. This time, the angle of refraction is larger than the angle of incidence because the light returns to Medium 1 where the speed of propagation is larger.

θ* _i > θ* _r

The light bends away from the normal when v_l < v₂

Analyzing the geometry of the figure and relating to the speed in respective media, one can determine after some calculations a law of refraction relating the angles of incidence and refraction.

The laws of refraction are as follows:

• The incident, refracted ray, and normal are in the same plane.
• The angle of incidence and the angle of refraction are related by

In optics, we also work with a coefficient that relates the speed of light in the specific medium with the speed in vacuum, and it is called the index of refraction:

The index of refraction is a unitless quantity and it is material dependent. A few values for the most common substances are given in Table 18.l.

Because the index of refraction is so close to 1 for air, we will consider n = 1. One note on propagation in a different medium is that the frequency of the wave will remain constant while the wavelength changes with speed.

Returning to the laws of refraction, we will find a second version of the law of the angles by multiplying the right-hand-side fraction with the speed oflight in a vacuum:

and with the definition of the index of refraction:

or separating on the same side factors related to the incident and refracted rays, respectively:

n₁ · sin θ_i = n₂ · sin θ_r

This is also known as Snell's law of refraction (Willebroad van Roijen Snell, 1580-1626).

For the purpose of this lesson, we will consider the index of refraction constant, although in reality, n is wavelength dependent. This dependence is the explanation for phenomena such as the dispersion of light through a prism.

Example

Light travels from a vacuum through the medium of an index of refraction of 1.55. Find the speed of light in the second medium and the wavelength in each medium if the frequency is 6 · 10¹⁴ Hz. If the light comes at an incident angle of 30°, what is the refraction angle?

Solution

We want to set all known data and then determine what expressions are useful to determine the unknowns. We know that the first medium is a vacuum and the second is a material that has an index of refraction of 1.55.

n₁ = 1

n₁ = 1.55

θ₁ = 30°

f = 6 · 10¹⁴ Hz

v = ?

λ₁ = ?

λ₂ = ?

θ₂ = ?

From the definition of the index of refraction, we can determine v₂:

We remember that v = λ · f, and that frequency stays the same with refraction. Then:

v₁ = λ · f

And since the first medium is a vacuum, the speed v₁ equals c:

λ₁ = 600 nm

And in the second medium,

λ₂ = 323 nm

Then to determine the refraction angle, we apply Snell's law:

n₁ · sin θ_i = n₂ · sin θ_r

1 · sin 30° = 1.55 · n₁ · sin θ_i

sin θ_r = 0.32

Internal Reflection

Let's summarize with a specific case: light rays going from a medium with a large optical density (large n) to a smaller n. In this case, the angle of refraction is larger than the angle of incidence, and so the light will move away from the normal. What happens when the light is at 90° from the normal? In this case, the light is no longer transmitted into the second medium but is returned to the first medium. This is the case of total internal reflection, and the limit angle of internal reflection can be determined by applying Snell's law (see Figure 18.12).

n₁ · sin θ_i = n₂ · sin θ_r

n₁ · sin θ_i = n₂ · sin 90°

n₁ · sin θ_i = n₂

We will call the angle of incidence for which total internal reflection happens a critical angle θ_c.

n₁ · sin θ_i = n₂

Example

Can a light ray propagating from air to water suffer total internal reflection at any angle?

Solution

This is a general question that addresses the expression of the critical angle of total internal reflection:

and n > 1 for any of the two media in the experiment.

If we have Medium 1 to be air, then n_l = 1 and the equation becomes

But n₂ > 1 and because the limits for the sine function are between –1 and + 1, the above equation has no solution. Hence, no critical angle exists for total internal reflection to happen.

Practice problems of this concept can be found at: Geometrical Optics Practice Questions