# Waves/Reflection and Refraction

Waves : Geometrical Optics
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## Reflection and RefractionEdit

Most of what we need to know about geometrical optics can be summarized in two rules, the laws of reflection and refraction. These rules may both be inferred by considering what happens when a plane wave segment impinges on a flat surface. If the surface is polished metal, the wave is reflected, whereas if the surface is an interface between two transparent media with differing indices of refraction, the wave is partially reflected and partially refracted. Reflection means that the wave is turned back into the half-space from which it came, while refraction means that it passes through the interface, acquiring a different direction of motion from that which it had before reaching the interface.

Figure 3.1: Sketch showing the reflection of a wave from a plane mirror. The law of reflection states that $\theta _I = \theta _R$.

Figure 3.1 shows the wave vector and wave front of a wave being reflected from a plane mirror. The angles of incidence, $\theta _I$, and reflection, $\theta _R$, are defined to be the angles between the incoming and outgoing wave vectors respectively and the line normal to the mirror. The law of reflection states that $\theta_R = \theta_I$. This is a consequence of the need for the incoming and outgoing wave fronts to be in phase with each other all along the mirror surface. This plus the equality of the incoming and outgoing wavelengths is sufficient to insure the above result.

Figure 3.2: Sketch showing the refraction of a wave from an interface between two dielectric media with $n_2 > n_1$.

Refraction, as illustrated in figure 3.2, is slightly more complicated. Since $n_R > n_I$, the speed of light in the right-hand medium is less than in the left-hand medium. (Recall that the speed of light in a medium with refractive index $n$ is $c_{medium} = c_{vac}/n$.) The frequency of the wave packet doesn't change as it passes through the interface, so the wavelength of the light on the right side is less than the wavelength on the left side.

Let us examine the triangle ABC in figure 3.2. The side AC is equal to the side BC times $\sin ( \theta_I )$. However, AC is also equal to $2 \lambda_I$, or twice the wavelength of the wave to the left of the interface. Similar reasoning shows that $2 \lambda_R$, twice the wavelength to the right of the interface, equals BC times $\sin ( \theta_R )$. Since the interval BC is common to both triangles, we easily see that

$\frac{\lambda_I}{\lambda_R} = \frac{\sin ( \theta_I )}{\sin ( \theta_R )} .$ (4.1)

Since $\lambda_I = c_I T = c_{vac} T/n_I$ and $\lambda_R = c_R T = c_{vac} T/n_R$ where $c_I$ and $c_R$ are the wave speeds to the left and right of the interface, $c_{vac}$ is the speed of light in a vacuum, and $T$ is the (common) period, we can easily recast the above equation in the form

$n_I \sin ( \theta_I ) = n_R \sin ( \theta_R ) .$ (4.2)

This is called Snell's law, and it governs how a ray of light bends as it passes through a discontinuity in the index of refraction. The angle $\theta _I$ is called the incident angle and $\theta _R$ is called the refracted angle. Notice that these angles are measured from the normal to the surface, not the tangent.

## Derivation for Law of ReflectionEdit

The derivation of Law of Reflection using Fermat's principle is straightforward. The Law of Reflection can be derived using elementary Calculus and Trigonometry. The generalization of the Law of Reflection is Snell's law, which is derived bellow using the same principle.

The medium that light travels through doesn't change. In order to minimize the time for light travel between to points, we should minimize the path taken.

θi = θr.
the angle of incidence equals the angle of reflection

1. Total path length of the light is given by

$L=d_1+d_2\,$

2. Using Pythagorean theorem from Euclidean Geometry we see that

$d_1=\sqrt{x^2 + a^2}\,$ and $d_2=\sqrt{(l-x)^2 + b^2}\,$

3. When we substitute both values of d1 and d2 for above, we get

$L=\sqrt{x^2 + a^2} + \sqrt{(l-x)^2 + b^2}$

4. In order to minimize the path traveled by light, we take the first derivative of L with respect to x.

$\frac{dL}{dx}=\frac{x}{\sqrt{x^2 + a^2}} + \frac{-(l-x)}{\sqrt{(l-x)^2 + b^2}}=0$

5. Set both sides equal to each other.

$\frac{x}{\sqrt{x^2 + a^2}} = \frac{(l-x)}{\sqrt{(l-x)^2 + b^2}}$

6. We can now tell that the left side is nothing but $\sin\theta_i$ and the right side $\sin\theta_r$ means

$\sin\theta_i=\sin\theta_r \!\$

7. Taking the inverse sine of both sides we see that the angle of incidence equals the angle of reflection

$\theta_i=\theta_r \!\$

## Derivation for Snell's LawEdit

The derivation of Snell's Law using Fermat's Priciple is straightforward. Snell's Law can be derived using elementary calculus and trigonometry. Snell's Law is the generalization of the above in that it does not require the medium to be the same everywhere.

To mark the speed of light in different media refractive indices named n1 and n2 are used.

$v_1=\frac{c}{n_1} \!\$
$v_2=\frac{c}{n_2}$
Snell's law relates n1, θ1 and n2, θ2.

Here $c$ is the speed of light in the vacuum and $n_1,n_2 \ge 1 \,$ because all materials slow down light as it travels through them.

1. Time for the trip equals distance traveled divided by the speed.

$t=\frac{d_1}{v_1}+\frac{d_2}{v_2}$

2. Using the Pythagorean theorem from Euclidean Geometry we see that

$\frac{d_1}{v_1}=\frac{\sqrt{x^2 + a^2}}{v_1}\,$ and $\frac{d_2}{v_2}=\frac{\sqrt{b^2 + (l-x)^2}}{v_2}\,$

3. Substituting this result into equation (1) we get

$t=\frac{\sqrt{x^2 + a^2}}{v_1} + \frac{\sqrt{b^2 + (l-x)^2}}{v_2}$

4. Differentiating and setting the derivative equal to zero gives

$\frac{dt}{dx}=\frac{x}{v_1\sqrt{x^2 + a^2}} + \frac{-(l-x)}{v_2\sqrt{(l-x)^2 + b^2}}=0$

5. After careful examination the above equation we see that it is nothing but

$\frac{dt}{dx}=\frac{\sin\theta_1}{v_1} - \frac{\sin\theta_2}{v_2}=0$

6. Thus

$\frac{v_1}{\sin\theta_1}=\frac{v_2}{\sin\theta_2}$

7. Multiplying both sides by $\sin\theta_1\sin\theta_2$ we get

$v_1\sin\theta_2=v_2\sin\theta_1 \!\$

8. Substituting $\frac{c}{n_1}$ for v1 and $\frac{c}{n_2}$ for $v_2$ we get

$\frac{c}{n_1}\sin\theta_2=\frac{c}{n_2}\sin\theta_1$

9. Simplifying both sides we get our final result

$n_1\sin\theta_1=n_2\sin\theta_2 \!\$

Waves : Geometrical Optics
1 - 2 - 3 - 4 - 5 - 6
Problems