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# 9.4: Summary

• Contributed by Wendell Potter and David Webb et al.
• Physics at UC Davis

1. Become familiar with the idea of wavefronts and rays.
2. Geometric optics is the approximation that rays always travel in straight lines. This approximation is good provided that the wavelength is much smaller than anything it encounters (i.e. we are neglecting diffraction). The geometric optics approximation allows us to perform ray-tracings to locate images.

3. When a wave encounters an interface between two media, some of the wave's energy can reflect into the original medium while the rest can be transmitted into the new medium. Because the two media have different allowed wave speeds, the transmitted wave is typically deformed, a phenomenon called refraction.

4. The law of reflection states that $$\theta_{inc} = \theta_{ref}$$, where both angles are measured from the normal of the reflecting surface.

5. Objects with rough surfaces have normals that change over their surface. As a result, light incident on rough surfaces is reflected in all directions. This is calleddiffuse reflection.

6. Each non-absorbing material has a refractive index that describes how quickly light travels through it. The higher the refractive index, the slower light travels in that medium. The refractive index in a medium is defined as $$n_{medium} = c/v_{medium}$$, where $$v_{medium}$$ is the speed of the wave in the medium and $$c$$ is the speed of light in vacuum.

7. To find the direction that light bends when refracted, we use Snell’s law $$n_1 \sin \theta_1 = n_2 \sin \theta_2$$. Both $$\theta_1$$ and $$\theta_2$$ are measured from the normal of the interface between the two media.

8. If Snell’s law cannot be satisfied then none of the wave can be transmitted; instead it is all reflected. This phenomenon is called total internal reflection. Total internal reflection can only occur when the light is coming from a faster medium and reaches the boundary between media.

9. Our eyes can only track back the rays that reach our eyes, and so if rays appear to be coming from somewhere then our brain thinks there is an object there. If there is no object there, the object our brain thinks it sees is called an image.

10. Images come in two types: real and virtual.

• ​​A real image is where the light rays actually come to a point and then spread out again. This sort of image can be placed on a screen.

• A virtual image is an image where the light rays do not cross, but our brain traces back the rays and is tricked into thinking that they cross.

11. For thin lenses or particular curved mirrors there is a focal length $$f$$. The relationship between the object distance $$o$$ and image distance $$i$$ is $\dfrac{1}{o} + \dfrac{1}{i} = \dfrac{1}{f}$which is known as the thin lens equation. If $$i$$ is positive,this is a real image; if $$i$$ is negative this is a virtual image.

12. The image of an object is typically a different size than the object. We use the magnification $$m = −i/o$$ to describe the change in size; for example $$m=2$$ means the image is twice as big as the original object. If the magnification is negative then this means the image is inverted.

# Derivations

In this chapter we simply presented Snell's Law, magnification, and the thin lens equation as true instead of deriving these results from what we know. We present the derivations here for the interested reader. While not necessary to learn and apply these equations, understanding these derivations will deepen your understanding of the concepts they reflect.

## Snell's Law

Let us draw both the peaks (as wavefronts) and the rays of light together for light that is traveling from air into water.

Look at the distance between the wavefronts on the boundary, shown as a bold line between the two indicated normals (dashed lines). Let us call this distance $$h$$ for hypotenuse, because it is the hypotenuse of both right-angled triangles indicated in the water and in the air. We know that the distance between the wavefronts (which makes up the opposite side of these triangles) is given by the wavelength in that medium. Writing this out for the triangle in the water we have

$\sin \theta_w = \dfrac{\lambda_w}{h} \implies h = \dfrac{\lambda_w}{\sin \theta_w}$

For the triangle in the air we have a similar relationship:

$\sin \theta_a = \dfrac{\lambda_a}{h} \implies h = \dfrac{\lambda_a}{\sin \theta_a}$

Because we know that the hypotenuse is the same in both of these equations, we are lead to conclude that

$\dfrac{\lambda_w}{\sin \theta_w} = \dfrac{\lambda_a}{\sin \theta_a}$

By multiplying this equation by the frequency $$f$$ and recalling that $$v_{wave} = f \lambda$$ we have

$\dfrac{f \lambda_w}{\sin \theta_w} = \dfrac{f \lambda_a}{\sin \theta_a}$

$\dfrac{v_w}{\sin \theta_w} = \dfrac{v_a}{\sin \theta_a}$

Finally we recall that $$v_a = c/n_a$$ (and a similar result for water) we have

$\dfrac{c}{n_w \sin \theta_w} = \dfrac{c}{n_a \sin \theta_a}$

This results holds if and only if

$n_w \sin \theta_w = n_a \sin \theta_a$

is true. This has been Snell's Law.

## Magnification

Consider an object with height $$h_o$$ and a converging lens that produces a real image with height $$h_i$$. Examine at the principal ray that goes through the center. Because it is a straight line, the gradient (slope) does not change.

We see that the gradient on the left hand side is

$\text{gradient }= \dfrac{\Delta y}{\Delta x} = \dfrac{-h_o}{o}$

We can use the information on the right-hand side to calculate the gradient we get

$\text{gradient }= \dfrac{\Delta y}{\Delta x} = \dfrac{h_i}{i}$

Because this ray does not bend, we know these gradients are the same. Therefore: $\dfrac{-h_o}{o} = \dfrac{h_i}{i}$ Rearranging this equation we have

$m \equiv \dfrac{h_i}{h_o} = -\dfrac{i}{o}$

## The Thin Lens Equation

We will prove the thin lens equation for a converging lens that produces a real image. The other cases can be shown in a similar manner. Examine the ray that enters the lens parallel to the optical axis and refracts through the focal point.

We know that there are two ways of calculating the gradient of the ray that passes through the focal point. The first manipulates the fact that the incoming ray has the same height as the object, and drops to the optical axis within a focal length

$\text{gradient} = \dfrac{\Delta y}{\Delta x} = -\dfrac{h_o}{f}$

The second way of calculating the gradient uses the fact that the height of the ray drops to the location of the image in the distance $$i$$. Because $$h_i < 0$$ we should be slightly careful with the sign of $$\Delta y$$

$\Delta y = y_f - y_i = h_i - h_o \implies \text{gradient} = \dfrac{h_i-h_o}{i}$

We can get rid of $$h_i$$ by applying our equation for the magnification:

$h_i = m h_o = -\dfrac{i}{o}h_o$

Writing out our gradient again we obtain

$\text{gradient} = \dfrac{-\frac{i}{o} h_o - h_o}{i} = - \left( \dfrac{1}{o} + \dfrac{1}{i} \right) h_o$

As a straight line has a constant gradient, the segment we use should not matter. Therefore these two expressions for the gradient must be equal:

$- \left( \dfrac{1}{o} + \dfrac{1}{i} \right) h_o = - \dfrac{h_o}{f}$

Cancelling the $$-h_o$$ from both sides leaves the thin lens equation

$\dfrac{1}{o} + \dfrac{1}{i} = \dfrac{1}{f}$