## PHYS102 Study Guide

### 7a. determine the size, location, and nature of images by using the mirror and lens equations

1. How are images formed by mirrors?
2. What is the primary difference between concave and convex mirrors?
3. What are the two possible types of images? How is the type of image formed by a mirror determined by its location relative to the mirror?
4. How are the focal length and curvature of a mirror related?
5. How are the locations of an object and its image related by the mirror equation?
6. How are images formed by lenses?
7. What is the primary difference between converging and diverging lenses?
8. How is the type of an image formed by a lens determined by its location relative to the lens?
9. How are the locations of an object and its image related by the lens equation?
10. What is magnification, and what does its sign (positive/negative) indicate about the image?

Mirrors form images of objects by reflection. As the rays of light generated by an object bounce off the reflective surface of the mirror, they intersect and form an image. The angle of reflection between the reflected ray and the normal of the surface is equal to the angle of incidence between the incident ray and the normal of the surface. A parabolic surface has the property of reflecting all rays parallel to its axis of symmetry in such a way that they all intersect at one point: the focal point of the parabola. Spherical surfaces serve as a good replacement for parabolic ones; as long as the incident rays fall near the axis of symmetry, their reflections also intersect at one focal point that is located on the axis of symmetry halfway between the center of curvature and the mirror. Thus, the focal length $f$ of the spherical mirror and its radius of curvature $R$ are related as $f=\frac{R}{2}$.

Images formed by the intersection of rays reflected back towards the object are real images. If the reflected rays do not intersect, then their extensions intersect behind the mirror, forming a virtual image. A virtual image can be seen, but cannot be captured with the camera film.

When describing the locations of objects and images formed by mirrors, the following conventions are used:

Positions in front of the mirror are considered positive. The distance between an object and the mirror is denoted by $p$ and is always positive, as the object is always located in front of the mirror. The location of an image is denoted by $q$, and is positive when the image is in front of the mirror and negative when the image is behind the mirror. Thus, $q$ is positive for real images and negative for virtual images.

Magnification is defined as the ratio of the sizes of the image and the object: $M=\frac{h_i}{h_o}$. Here, $h_i$ is the height of the image, and $h_o$ is the height of the object. If the image is upside down (or inverted) then it is considered negative, which means $M$ is also negative. From geometric considerations, it can be shown that $M=-\frac{q}{p}$. From here, it follows that $M$ is positive when $q$ is negative, so a virtual image is always upright; and that $M$ is negative when $q$ is positive, so a real image is always inverted.

There are three rays that can be used to draw the images formed by mirrors:

• The reflection of the rays parallel to the axis of symmetry pass through the focal point
• The reflection of the rays passing through the focal point are parallel to the axis of symmetry
• The rays perpendicular to the surface (or passing through the center of curvature, for spherical surfaces) reflected along the same line

Plane mirrors have an infinite radius of curvature, and thus no focal point. The image of an object located in front of a mirror is located at the same distance behind the mirror: $p = -q$. This image is virtual, upright, and has the same size as the object: $h_i = h_o$.

Concave mirrors have a reflective surface on the inside of a spherical surface, so their focal point is in front of the mirror and on the same side as an object. Their focal length is positive: $f > 0$. Concave mirrors can form either real or virtual images, depending on the location of the object.

Convex mirrors have a reflective surface on the outside of a spherical surface, so their focal point is behind the mirror. Their focal length is negative: $f < 0$. Concave mirrors always form virtual images.

If the location $p$ and size $h_o$ of the object are known, the location of the image can be found from the mirror equation: $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$. The size of the image can be found using the magnification formula: $M=-\frac{q}{p}=\frac{h_i}{h_o}$.

To review how to draw images formed by mirrors and determine the location, size, and nature of an image, review Spherical Mirrors and Image Formation by Concave Mirrors, and explore this solved example. Also, watch Image Formation by Convex Mirrors and see this solved example, as well as Image Formation by Plane Mirrors. Finally, see each video in the Mirrors and Lenses series.

Lenses form images of objects by refraction. As rays of light change direction when passing through the material of the lens (typically glass or plastic), they intersect and form an image. The lens has to be very thin compared to the distance between the lens and the object. Lenses are made in a variety of shapes, and they have focal points where the refracted rays converge if the incident rays are parallel to the lens' axis of symmetry. Lenses can be made to be either converging or diverging.

For lenses, the object and image locations are similar to that of mirrors. The distance between an object and the lens is denoted by $p$, which is always positive. However, real images are located on the opposite side (behind the lens), so the distance between the image and the lens, $q$, is positive is when the image is behind the lens (a real image) and negative when the image is in front of the lens (a virtual image)

For lenses, magnification is defined as with mirrors, as the ratio of the sizes of the image and the object: $M=\frac{h_i}{h_o}$. Here, $h_i$ is the height of the image, and $h_o$ is the height of the object. If the image is upside down (or inverted) then it is considered negative, which means $M$ is also negative. From geometric considerations, it can be shown that $M=-\frac{q}{p}$. From here, it follows that $M$ is positive when $q$ is negative, so the virtual image is always upright, and that $M$ is negative when $q$ is positive, so the real image is always inverted.

Converging lenses refract incident rays so that they bend toward the axis of symmetry, while diverging lenses bend incident rays away from the axis of symmetry. There are three rays that can be used to draw the images formed by the lenses:

• The incident rays parallel to the axis of symmetry refract so that the refracted rays pass through the focal point behind the converging lens, or so that their extensions pass through the focal point in front of the diverging lens
• The incident rays passing through the focal point refract so that they are parallel to the axis of symmetry
• The rays passing through the center of the lens are not refracted

Converging lenses have a positive focal length and can form real and virtual images, depending on the location of the object. Diverging lenses have a negative focal length and always form virtual images.

If the location p and size ho of the object are known, the location of the image can be found from the lens equation: $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$. Then, the size of the image can be found using the magnification formula: $M=-\frac{q}{p}=\frac{h_i}{h_o}$.

To review how to draw images formed by lenses and determine the location, size, and nature of an image, read Thin Lenses and Image Formation by Thin Lenses and see these examples of converging and diverging lenses. Also, watch the Mirrors and Lenses series.

### 7b. solve problems using the law of refraction

1. How are the angle of incidence and angle of refraction related?
2. When does total internal reflection occur? What is a critical angle?

Refraction is the change of direction, or bending, of a light ray when it crosses a boundary between two media. It occurs because the light propagates at different speeds in media with different optical properties, which are determined by the electric permittivity and the magnetic permeability of the media. The speed of light in vacuum is $c=3\times 10^8\ \mathrm{m/s}$. In other media, it is decreased by a factor of $n$, the index of refraction of the medium: $v_\mathrm{light} = \frac{c}{n}$. In air, the index of refraction is very close to 1, so the speed of light in air is considered to be equal to that of vacuum. Other transparent media have indices of refraction greater than 1.

As a ray of light crosses the boundary between two media, its frequency remains unchanged (since frequency depends only on the source of the light), but its speed of propagation, and therefore wavelength, changes. This results in a change of direction of the ray. The direction is determined by the angle the ray makes with the normal of the boundary between the media. The law relating the angle of incidence $\theta_i$ to the angle of refraction $\theta_r$ is called Snell's Law: $\frac{\sin{\theta_i}}{\sin{\theta_r}}=\frac{n_r}{n_i}$ . Alternatively, it can be written as $n_i\sin{\theta_i}=n_r\sin{\theta_r}$.

Notice that if $\theta_i = 0$, that is, the incident light is perpendicular to the boundary, then $\theta_r = 0$ as well, so the light will not be refracted.

From Snell's Law, it follows that when light crosses the boundary to the medium with the greater index of refraction (for example, from air to water), the refracted ray will be closer to the normal than the incident ray. Also, the expression for the angle of refraction $\sin{\theta_r}=\frac{n_i\sin{\theta_i}}{n_r}$ will always have solutions, since the right-hand side of the equation will be less than 1 for any incident angle. However, when light goes from a medium with a greater index of refraction to a medium with a smaller one (as with from water to air), the refracted ray will be further away from the normal than the incident ray. It can make a 90° angle with the normal, but it cannot go further, since it would then longer be in the refractive medium and would reflect back to the incident medium. This phenomenon is known as total internal reflection. When the incident angle is larger than a certain value called the critical angle, light will not refract, just reflect. The value of critical angle can be found by setting the angle of refraction to 90°, which means $\sin{\theta_r}=1$. Then, $\sin{\theta_c}=\frac{n_r}{n_i}$.

To review the laws of reflection and refraction, watch Reflection and Refraction and read Law of Geometric Propagation, Law of Reflection, Law of Refraction, and Total Internal Reflection. Also, review this solved example.

### 7c. describe the interference pattern in a double-slit experiment and explain the experiment's results

1. In the double-slit experiment, light is emitted onto a sheet with two small openings, and a pattern is observed on a screen some distance away. If the light is modeled as a beam of particles, what pattern should be observed on the screen?
2. If light is instead modeled as a wave, what pattern should be observed? What wave property explains this pattern?

The double-slit experiment conclusively demonstrated the wave nature of light. If light was a beam of particles, the pattern on the screen in the double-slit experiment would have consisted of two bright spots in front of the slits. Instead, the double-slit experiment yielded a pattern of alternating bright and dark bands. This pattern can be explained, and the location of the bright and dark bands predicted, by explaining light as a wave. Two light waves with the same frequency and same initial phase leave the slits. When the waves reach the screen, they recombine and interfere with one another. By the time they reach the screen, they have traveled different distances and are no longer in phase. The difference between the distance each wave travels is called path difference and is denoted by $\delta$.

If the path difference equals an integer number m of wavelengths: $\delta = m\lambda$, then the waves undergo constructive interference, and their amplitudes add up. In this case, a bright band is observed. However, if the path difference is an odd multiple of half wavelengths: $\delta=\left ( m+\frac{1}{2} \right )\lambda$, the waves will undergo destructive interference and cancel each other out. In this case, a dark band is observed.

From geometric considerations and the assumption that the screen is far away from the sheet with the slits, the locations of bright and dark bands can be found as $y_\mathrm{bright}=\frac{m\lambda L}{a}$ and $y_\mathrm{dark}=\frac{\left (m+\frac{1}{2} \right )\lambda L}{a}$. Here, $L$ is the distance between the sheet and the screen, and $a$ is the distance between the slits. Note that the "zeroth" bright band occurs at $y = 0$, in the center of the screen, opposite the midpoint between the slits. Note also that the dark and bright bands will be equidistant from one another as long as they are close enough to the center of the screen. Further away from the center of the screen, the brightness of the bright bands becomes less intense, and the bright bands spread further away from one another.

Review the details of the double-slit experiment by watching Interference of Light Waves and reading Young's Double-Slit Experiment. Also, review this solved example.

### 7d. explain how rainbows are produced

1. Why do light waves of different frequencies refract at different angles?
2. How do these frequencies and angles correspond to the different colors of visible light?

Snell's Law describes how the angle of refraction depends on the ratio of the indices of refraction of incident and refractive media. For many media (including water and glass), the index of refraction depends slightly on the frequency or wavelength of incident light. This dependency is called dispersion, and a medium for which $n$ depends on $f$ (or $\lambda$) is called dispersive.

The most familiar natural phenomenon that demonstrates dispersion is a rainbow. The light from the sun contains waves with all frequencies of the visible spectrum. The combination of all these waves is perceived as white light. If there are water droplets in the air, the light refracts when it enters the droplets, and then refracts again as it leaves the droplets. Since the angle of refraction is different for light of different frequencies, waves of different colors separate. This is perceived by observers as a rainbow.

Watch the last video in the Reflection and Refraction series. Also, review Unweaving the Rainbow, which illustrates why rainbows appear as arcs.

### Unit 7 Vocabulary

This vocabulary list includes terms that might help you with the review items above and some terms you should be familiar with to be successful in completing the final exam for the course.

Try to think of the reason why each term is included.

• Concave mirror
• Converging lens
• Convex mirror
• Diffraction
• Dispersion
• Diverging lens
• Focal length
• Focal point
• Interference, constructive and destructive
• Magnification
• Mirror equation
• Lens equation
• Path difference
• Plane mirror
• Real Image
• Reflection, specular and diffuse
• Refraction
• Snell's Law
• Total internal reflection
• Virtual Image