MCAT Textbooks / MCAT Physics and Math Review 2026-2027 / Ch 8 of 12

📐 Light and Optics

12,648 words · 28 figures · ≈55 min read · MCAT Physics and Math Review 2026-2027

Chapter 8: Light and Optics

Chapter 8: Light and Optics with fiber optic lighting in the background

Chapter 8: Light and Optics

Science Mastery Assessment

Every pre-med knows this feeling: there is so much content I have to know for the MCAT! How do I know what to do first or what’s important?

While the high-yield badges throughout this book will help you identify the most important topics, this Science Mastery Assessment is another tool in your MCAT prep arsenal. This quiz (which can also be taken in your online resources) and the guidance below will help ensure that you are spending the appropriate amount of time on this chapter based on your personal strengths and weaknesses. Don’t worry though— skipping something now does not mean you’ll never study it. Later on in your prep, as you complete full-length tests, you’ll uncover specific pieces of content that you need to review and can come back to these chapters as appropriate.

How to Use This Assessment

If you answer 0–7 questions correctly:

Spend about 1 hour to read this chapter in full and take limited notes throughout. Follow up by reviewing all quiz questions to ensure that you now understand how to solve each one.

If you answer 8–11 questions correctly:

Spend 20–40 minutes reviewing the quiz questions. Beginning with the questions you missed, read and take notes on the corresponding subchapters. For questions you answered correctly, ensure your thinking matches that of the explanation and you understand why each choice was correct or incorrect.

If you answer 12–15 questions correctly:

Spend less than 20 minutes reviewing all questions from the quiz. If you missed any, then include a quick read-through of the corresponding subchapters, or even just the relevant content within a subchapter, as part of your question review. For questions you got correct, ensure your thinking matches that of the explanation and review the Concept Summary at the end of the chapter.

Answer Key

Chapter 8: Light and Optics

CHAPTER 8

LIGHT AND OPTICS

In This Chapter

8.1 Electromagnetic Spectrum

Electromagnetic Waves

Color and the Visible Spectrum

8.2 Geometrical Optics

Reflection

Refraction

Lenses

Dispersion

8.3 Diffraction

Single Slit

Slit–Lens System

Multiple Slits

X-Ray Diffraction

8.4 Polarization

Plane-Polarized Light

Circular Polarization

Concept Summary

CHAPTER PROFILE

The content in this chapter should be relevant to about 14% of all questions about physics on the MCAT.

This chapter covers material from the following AAMC content category:

4D: How light and sound interact with matter

Introduction

The next time you’re browsing your local convenience store, take a look at the security mirrors—the ones that bulge out from the wall, usually above eye level. Looking into one of these mirrors, notice not just that the image you see of the world is distorted but how it is distorted: the image is still right-side up, but everything is much smaller than you’d expect, and the curve of the mirror introduces some slopes that are not present in reality. Additionally, you see a much wider field of vision than you would if the mirror were a simple plane mirror. This is why security mirrors are useful: they are a convenient, low-tech solution that allows the cashier to survey the entire store in one glance. All these features result from the fact that the security mirror is a convex, diverging optical system. Parallel light rays that hit the mirror are reflected in multiple directions, which allows observers to see a large field of vision, even if the image is somewhat distorted and the objects in the image are closer than they appear. Indeed, the passenger-side mirror of a car that bears that same message is also a convex mirror, allowing the driver to see a wider view of the other vehicles behind the car.

This chapter will first complete a topic from Chapter 7 of MCAT Physics and Math Review by analyzing the transverse waveform of visible light and other electromagnetic (EM) waves. We will then consider in detail the rules of optics, which describe the behavior of electromagnetic waves as they bounce off of and travel through various shapes and compositions of matter. The optical systems covered are those tested on the MCAT: concave and convex mirrors, which produce images by reflection, and concave and convex lenses, which produce images by refraction. To finish, we will discuss the phenomena of thin-slit experiments (diffraction) and light polarization.

8.1 Electromagnetic Spectrum

LEARNING OBJECTIVES

After Chapter 8.1, you will be able to:

The full electromagnetic spectrum includes radio waves on one end (long wavelength, low frequency, low energy) and gamma rays on the other (short wavelength, high frequency, high energy). Between the two extremes, we find, in order from lowest energy to highest energy, microwaves, infrared, visible light, ultraviolet, and x-rays. This chapter will focus primarily on the range of wavelengths corresponding to the visible spectrum of light (400 nm to 700 nm).

Electromagnetic Waves

A changing magnetic field can cause a change in an electric field, and a changing electric field can cause a change in a magnetic field. Because of the reciprocating nature of these two fields, we can see how electromagnetic waves occur in nature. Each oscillating field causes oscillations in the other field completely independent of matter, so electromagnetic waves can even travel through a vacuum.

Electromagnetic waves are transverse waves because the oscillating electric and magnetic field vectors are perpendicular to the direction of propagation. The electric field and the magnetic field are also perpendicular to each other. This is illustrated in Figure 8.1.

two waves perpendicular to each other reaching amplitude and equilibrium position at the same time along propagation path

Figure 8.1. Electromagnetic Wave The electric field (E) oscillates up and down the page; the magnetic field (B) oscillates into and out of the page.

The electromagnetic spectrum describes the full range of frequencies and wavelengths of electromagnetic waves. Wavelengths are often given in the following units: mm (10–3 m), µm (10–6 m), nm (10–9 m), and Å (ångström, 10–10 m). The full spectrum is broken up into many regions, which in descending order of wavelength are radio (109–1 m), microwave (1 m–1 mm), infrared (1 mm–700 nm), visible light (700–400 nm), ultraviolet (400–50 nm), X-rays (50–10–2 nm), and γ-rays (less than 10–2 nm). The electromagnetic spectrum is depicted in Figure 8.2.

from highest frequency (lowest wavelength): gamma rays, x rays, UV, visible light (violet, blue, green, yellow, orange, red), infrared, microwave, FM and AM radio, long radio waves

Figure 8.2. The Electromagnetic Spectrum

Electromagnetic waves vary in frequency and wavelength, but in a vacuum, all electromagnetic waves travel at the same speed, called the speed of light. This constant is represented by c and is approximately 3.00×108 ms. To a first approximation—and for the purposes of all MCAT-related equations—electromagnetic waves also travel in air with this speed. In reference to electromagnetic waves, the familiar equation ν = fλ becomes

c = fλ

Equation 8.1

where c is the speed of light in a vacuum and, to a first approximation, also in air, f is the frequency, and λ is the wavelength.

MNEMONIC

To recall the order of the colors in the visible spectrum, remember the grade-school “rainbow” of ROY G. BV (red, orange, yellow, green, blue, violet).

Color and the Visible Spectrum

The only part of the spectrum that is perceived as light by the human eye is the visible region. Within this region, different wavelengths are perceived as different colors, with violet at one end of the visible spectrum (400 nm) and red at the other (700 nm).

MCAT EXPERTISE

Wavelengths in the visible range are common on the MCAT. Remembering the boundaries of the visible spectrum (about 400–700 nm) will save you time and energy on Test Day.

Light that contains all the colors in equal intensity is perceived as white. The color of an object that does not emit its own light is dependent on the color of light that it reflects. Thus, an object that appears red is one that absorbs all colors of light except red. This implies that a red object under green illumination will appear black because it absorbs the green light and has no light to reflect. The term blackbody refers to an ideal absorber of all wavelengths of light, which would appear completely black if it were at a lower temperature than its surroundings.

MCAT CONCEPT CHECK 8.1

Before you move on, assess your understanding of the material with these questions.

_____> ___> ___> ___> ___> ___> _____

_________________________________

8.2 Geometrical Optics

LEARNING OBJECTIVES

After Chapter 8.2, you will be able to:

When light travels through a homogeneous medium, it travels in a straight line. This is known as rectilinear propagation. The behavior of light at the boundary of a medium or interface between two media is described by the theory of geometrical optics. Geometrical optics explains reflection and refraction, as well as the applications of mirrors and lenses.

Reflection

Reflection is the rebounding of incident light waves at the boundary of a medium. Light waves that are reflected are not absorbed into the second medium; rather, they bounce off of the boundary and travel back through the first medium. Figure 8.3 illustrates reflection on a plane mirror.

light bouncing off surface and returning at same angle on the other side of the normal

Figure 8.3. Reflection According to the law of reflection, θ1 = θ2.

The law of reflection is

θ1 = θ2

Equation 8.2

where θ1 is the incident angle and θ2 is the reflected angle, both measured from the normal. The normal is a line drawn perpendicular to the boundary of a medium; all angles in optics are measured from the normal, not the surface of the medium.

Plane Mirrors

In general, images created by a mirror can be either real or virtual. An image is said to be real if the light actually converges at the position of the image. An image is virtual if the light only appears to be coming from the position of the image but does not actually converge there. One of the distinguishing features of real images is the ability of the image to be projected onto a screen.

Parallel incident light rays remain parallel after reflection from a plane mirror; that is, plane mirrors—being flat reflective surfaces—cause neither convergence nor divergence of reflected light rays. Because the light does not converge at all, plane mirrors always create virtual images. In a plane mirror, the image appears to be the same distance behind the mirror as the object is in front of it, as shown in Figure 8.4. In other words, plane mirrors create the appearance of light rays originating behind the mirrored surface. Because the reflected light remains in front of the mirror but the image appears behind the mirror, the image is virtual. Plane mirrors include most of the common mirrors found in our homes. To assist in our discussion of spherical mirrors, plane mirrors can be conceptualized as spherical mirrors with an infinite radius of curvature.

light rays bouncing off plane mirror; point of theoretical convergence shown at

Figure 8.4. Reflection in a Plane Mirror O is the object and I is the (virtual) image; all incident angles (θ) are equal to their respective reflected angles (θ′).

Spherical Mirrors

Spherical mirrors come in two varieties: concave and convex. The word spherical implies that the mirror can be considered a spherical cap or dome taken from a much larger spherically shaped mirror. Spherical mirrors have an associated center of curvature (C) and a radius of curvature (r). The center of curvature is a point on the optical axis located at a distance equal to the radius of curvature from the vertex of the mirror; in other words, the center of curvature would be the center of the spherically shaped mirror if it were a complete sphere.

REAL WORLD

The passenger-side mirrors in cars are an example of convex mirrors (everything appears smaller and farther away); the small circular mirrors used for applying makeup are an example of concave mirrors (everything appears bigger and closer).

If we were to look from the inside of a sphere to its surface, we would see a concave surface. On the other hand, if we were to look from outside the sphere, we would see a convex surface. For a concave surface, the center of curvature and the radius of curvature are located in front of the mirror. For a convex surface, the center of curvature and the radius of curvature are behind the mirror. Concave mirrors are called converging mirrors and convex mirrors are called diverging mirrors because they cause parallel incident light rays to converge and diverge after they reflect, respectively.

MNEMONIC

Concave is like looking into a cave.

KEY CONCEPT

Concave mirrors are converging mirrors. Convex mirrors are diverging mirrors. The reverse is true for lenses.

There are several important lengths associated with mirrors, as shown in Figure 8.5. The focal length (f) is the distance between the focal point (F) and the mirror. Note that for all spherical mirrors, f=r2 where the radius of curvature (r) is the distance between C and the mirror. The distance between the object and the mirror is o; the distance between the image and the mirror is i.

labeled: object and object distance, image and image distance, focal point and focal length, center of curvature and radius of curvature, principal axis

Figure 8.5. Key Variables in Geometrical Optics The mirror pictured is a concave mirror; light rays are not shown.

There is a simple relationship between these four distances:

1f=1o+1i=2r

Equation 8.3

While it is not important which units of distance are used in this equation, it is important that all values used have the same units as each other.

On the MCAT, you will most often use this equation to calculate the image distance for all types of mirrors and lenses. If the image has a positive distance (i > 0), it is a real image, which implies that the image is in front of the mirror. If the image has a negative distance (i < 0), it is virtual and thus located behind the mirror. Plane mirrors can be thought of as spherical mirrors with infinitely large focal distances. As such, for a plane mirror, r = f = ∞, and the equation becomes 1o+1i=0 or i = –o. This can be interpreted as saying the virtual image is at a distance behind the mirror equal to the distance the object is in front of the mirror.

The magnification (m) is a dimensionless value that is the ratio of the image distance to the object distance:

m=−io

Equation 8.4

By extension, the magnification also gives the ratio of the size of the image to the size of the object. Following the sign convention given later in Table 8.1, the orientation of the image (upright or inverted) can be determined: a negative magnification signifies an inverted image, while a positive value signifies an upright image. If |m| < 1, the image is smaller than the object (reduced); if |m| > 1, the image is larger than the object (enlarged); and if |m| = 1, the image is the same size as the object.

Figure 8.6 shows ray diagrams for a concave spherical mirror with the object at three different points. A ray diagram is useful for getting an approximation of where an image is. On Test Day, ray diagrams can be helpful for a quick determination of the type of image that will be produced by an object some distance from the mirror (real vs. virtual, inverted vs. upright, and magnified vs. reduced). Ray diagrams should be used with caution, however: under the pressure of Test Day, it can be easy to draw them incorrectly. Therefore, it is important to practice drawing ray diagrams to avoid careless errors on Test Day, and it is also important to be familiar with how to solve optics questions mathematically.

When drawing a ray diagram, there are three important rays to draw. For a concave mirror, a ray that strikes the mirror parallel to the axis (the normal passing through the center of the mirror) is reflected back through the focal point (green lines in Figure 8.6 and 8.7). A ray that passes through the focal point before reaching the mirror is reflected back parallel to the axis (red lines). A ray that strikes the mirror at the point of intersection with the axis is reflected back with the same angle measured from the normal (blue lines). In Figure 8.6a, the object is placed beyond F, and the image produced is real, inverted and magnified. In Figure 8.6b, the object is placed at F, and no image is formed because the reflected light rays are parallel to each other. In terms of the mirror equation, we say that the image distance i = ∞ here. For the scenario in Figure 8.6c, the object is placed between F and the mirror, and the image produced is virtual, upright, and magnified.

(a) green and red rays drawn, image is inverted and magnified; (b) green and blue rays drawn, no image; (c) green and red rays drawn and extended, image is upright and magnified

Figure 8.6. Ray Diagrams for Concave (Converging) Mirrors (a) Object is placed beyond F; (b) Object is placed at F; (c) Object is placed between F and the mirror.

KEY CONCEPT

Any time an object is at the focal point of a converging mirror, the reflected rays will be parallel, and thus, the image will be at infinity.

A single diverging mirror forms only a virtual, upright, and reduced image, regardless of the position of the object. The farther away the object, the smaller the image will be. To quickly remember these rules, recall the convenience store security mirrors mentioned at the beginning of the chapter. The ray diagram of a diverging mirror is shown in Figure 8.7.

red, green, and blue lines drawn; image is upright and reduced

Figure 8.7. Ray Diagrams for Convex (Diverging) Mirrors

KEY CONCEPT

To find where the image is (for a mirror), draw the following rays and find a point where any two intersect. This point of intersection marks the tip of the image. If the rays you draw do not appear to intersect, extend them to the other side of the mirror, creating a virtual image.

Sign Conventions for Mirrors

Table 8.1 provides the sign convention for single mirrors. Note that on the MCAT, for almost all problems involving mirrors, the object will be placed in front of the mirror. Thus, the object distance o is almost always positive.

Table 8.1 Sign Convention for a Single Mirror

SYMBOL POSITIVE NEGATIVE

o Object is in front of mirror Object is behind mirror (extremely rare)

i Image is in front of mirror (real) Image is behind mirror (virtual)

r Mirror is concave (converging) Mirror is convex (diverging)

f Mirror is concave (converging) Mirror is convex (diverging)

m Image is upright (erect) Image is inverted

KEY CONCEPT

The focal length of converging mirrors (and converging lenses) will always be positive. The focal length of diverging mirrors (and diverging lenses) will always be negative.

MNEMONIC

Image types with a single lens or mirror (assuming o is positive): UV NO IR

Example: An object is placed 6 cm in front of a concave mirror that has a 10 cm radius of curvature. Determine the image distance, the magnification, whether the image is real or virtual, and whether it is inverted or upright.

Solution: Use the optics equation:

1f=1o+1i=2r1i=2r−1o1i=210 cm−16 cm=12−1060=260i=602=+30 cm

A positive value for i signifies that the image is in front of the mirror and is therefore real. For a single lens or mirror with o > 0, a real image will always be inverted.

After determining i, the magnification m can be calculated as:

m=−io=−30 cm6 cm=−5

The negative sign on the magnification confirms that the image is inverted, and the fact that |m| > 1 indicates that the image is enlarged.

Refraction

Refraction is the bending of light as it passes from one medium to another and changes speed. The speed of light through any medium is always less than its speed through a vacuum. Remember that the speed of light in a vacuum, c, is equal to 3.00×108 ms. The speed of light in air is just slightly lower that this value; on the MCAT, it is appropriate to use 3.00×108 ms for the speed of light in air.

REAL WORLD

When a pencil (or any straight object) is dipped into a glass of water at an angle, it looks impossibly bent where it intersects the surface of the water because the light reflecting off of the portion of the pencil under water is refracted.

Snell’s Law

When light is in any medium besides a vacuum, its speed is less than c. For a given medium

n=cv

Equation 8.5

where c is the speed of light in a vacuum, ν is the speed of light in the medium, and n is a dimensionless quantity called the index of refraction of the medium. The index of refraction of a vacuum is 1, by definition; for all other materials, the index of refraction will be greater than 1. For air, n is essentially equal to 1 because the speed of light in air is extremely close to c. The indices of refraction for a number of common media are shown in Table 8.2. These values are provided only for reference; they need not be memorized.

Table 8.2 Indices of Refraction of Common Media

MEDIUM INDEX OF REFRACTION (n)

Vacuum 1 (by definition)

Air 1.0003

Ice 1.31

Water 1.33

Acetone 1.36

Ethanol 1.36

Cornea (human) 1.37–1.40

Lens (human) 1.39–1.41

Glass (various types) 1.48–1.93

Diamond 2.42

Refracted rays of light obey Snell’s law as they pass from one medium to another:

n1 sin θ1 = n2 sin θ2

Equation 8.6

where n1 and θ1 refer to the medium from which the light is coming and n2 and θ2 refer to the medium into which the light is entering. Note that θ is once again measured with respect to the normal, as shown in Figure 8.8.

incident ray leaving water and entering air as a refracted ray bent further from the normal

Figure 8.8. Snell’s Law

From Snell’s law, we can see that when light enters a medium with a higher index of refraction (n2 > n1), it bends toward the normal (sin θ2 < sin θ1; therefore, θ2 < θ1), as shown in Figure 8.9. Conversely, if the light travels into a medium where the index of refraction is smaller (n2 < n1), the light will bend away from the normal (sin θ2 > sin θ1; therefore, θ2 > θ1).

a pencil in water appears bent because of the water’s higher refractive index; when light travels from a medium with low refractive index to one with higher refractive index, it bends toward the nornal

Figure 8.9 Refraction of Light from Air into Water

KEY CONCEPT

Remember that when light enters a medium with a higher index of refraction, it bends toward the normal. When light enters a medium with a lower index of refraction, it bends away from the normal.

Example: A penny sits at the bottom of a pool of water (n = 1.33) at a depth of 3.0 m. If an observer 1.8 m tall stands 30 cm away from the edge, how close to the side can the penny be and still be visible?

Solution: First, draw a picture of the situation:

description given in preceding text

Note that the light is coming from the water (n1 = 1.33) and going into the air (n2 ≈ 1), so the light is bent away from the normal (θ2 > θ1). We need to find the angle that the light rays make with the normal to the water’s surface:

tan θ2=0.3 m1.8 m=0.167θ2=tan−10.167=9.5°

Using Snell’s law, we can solve for θ1:

sinθ1=(n2n1)sinθ2=(11.33)sin9.5°=0.1651.33θ1=sin−1(0.1651.33)=7.1°

Now, we can find x using trigonometry:

x = (3 m) × tan θ1 = 3 tan 7.1° = 3 × 0.124 = 0.37 m = 37 cm

Note that you will not be expected to calculate precise values of trigonometric functions or inverse trigonometric functions on Test Day. This question is provided mainly as an opportunity to see the application of Snell’s law.

Total Internal Reflection

When light travels from a medium with a higher index of refraction (such as water) to a medium with a lower index of refraction (such as air), the refracted angle is larger than the incident angle (θ2 > θ1); that is, the refracted light ray bends away from the normal. As the incident angle is increased, the refracted angle also increases, and eventually, a special incident angle called the critical angle (θc) is reached, for which the refracted angle θ2 equals 90 degrees. At the critical angle, the refracted light ray passes along the interface between the two media. The critical angle can be derived from Snell’s law if θ2 = 90°, such that

θc=sin−1(n2n1)

Equation 8.7 Total internal reflection, a phenomenon in which all the light incident on a boundary is reflected back into the original material, results with any angle of incidence greater than the critical angle, θc, as shown in Figure 8.10.

increasing incident angle leads to increasing refracted angle until critical angle is reached, where refracted angle is 90 degrees; then total internal reflection occurs at any further incident angle

Figure 8.10. Total Internal Reflection At the incident angle ofθc, the refracted angle is equal to 90°; at incident angles above 90°, total internal reflection occurs.

KEY CONCEPT

Total internal reflection occurs as the light moves from a medium with a higher refractive index to a medium with a lower one.

Example: From the previous example, suppose another penny is 3 m away from the edge. Will a light ray going from this penny to the edge of the pool emerge from the water?

Solution: The angle made by the second penny’s light ray is

θ1=tan−1(oppositeadjacent)=tan−1(3 m3 m)=tan−1(1)=45°

Find the critical angle:

θc=sin−1(n2n1)=sin−1(11.33)=sin−1(0.75)=48.8°

It is not expected on the MCAT that we know the inverse sin of 0.75. However, we do know that sin45°=22≈0.71. Since sin (θc) is greater than 0.71, the critical angle must be greater than 45°. Therefore, θc > θ1 and the light ray will emerge from the pool.

Lenses

There is an important difference between lenses and mirrors, aside from the fact that lenses refract light while mirrors reflect it. When working with lenses, there are two surfaces that affect the light path. For example, a person wearing glasses sees light that travels from an object through the air into the glass lens (first surface). Then the light travels through the glass until it reaches the other side, where again it travels out of the glass and into the air (second surface). The light is refracted twice as it passes from air to lens and from lens back to air.

Thin Spherical Lenses

On the MCAT, lenses generally have negligible thickness. Because light can travel from either side of a lens, a lens has two focal points, with one on each side. The focal length can be measured in either direction from the center. For thin spherical lenses, the focal lengths are equal, so we speak of just one focal length for the lens as a whole.

Figure 8.11a illustrates that a converging lens is always thicker at the center, while Figure 8.11b illustrates that a diverging lens is always thinner at the center. The basic formulas for finding image distance and magnification for spherical mirrors also apply to lenses. The object distance o, image distance i, focal length f, and magnification m, are related by the equations 1f=1o+1i and m=−io.

(a) convex lens causing parallel light rays to refract to second focal point or light from first focal point to be refracted as parallel light rays; (b) concave lens causing parallel light rays to refract from first focal point or light traveling toward second focal point to be refracted as parallel light rays

Figure 8.11. Ray Diagrams for Single Lenses (a) Convex (converging) lenses; (b) Concave (diverging) lenses.

REAL WORLD

Converging lenses (reading glasses) are needed by people who are “farsighted.” Diverging lenses (standard glasses) are needed by people who are “nearsighted.”

Real Lenses

For lenses where the thickness cannot be neglected, the focal length is related to the curvature of the lens surfaces and the index of refraction of the lens by the lensmaker’s equation:

1f=(n−1)(1r1−1r2)

Equation 8.8

where n is the index of refraction of the lens material, r1 is the radius of curvature of the first lens surface, and r2 is the radius of curvature of the second lens surface.

The eye is a complex refractive instrument that uses real lenses. The cornea acts as the primary source of refractive power because the change in refractive index from air is so significant. Then, light is passed through an adaptive lens that can change its focal length before reaching the vitreous humor. It is further diffused through layers of retinal tissue to reach the rods and cones. At this point, the image has been focused and minimized significantly, but is still relatively blurry. Our nervous system processes the remaining errors to provide a crisp view of the world.

KEY CONCEPT

To find where the image is (for a lens), draw the following rays and find a point where any two intersect. This point of intersection marks the tip of the image. If the rays you draw do not appear to intersect, extend them to the same side of the lens from which the light came, creating a virtual image.

Sign Conventions for Lenses

Note that the sign conventions change slightly for lenses. For both lenses and mirrors, positive magnification represents upright images, and negative magnification means inverted images. Also, for both lenses and mirrors, a positive image distance means that the image is real and is located on the real (R) side, whereas a negative image distance means that the image is virtual and located on the virtual (V) side. Table 8.3 summarizes the sign conventions for single lenses.

Table 8.3 Sign Convention for a Single Lens

SYMBOL POSITIVE NEGATIVE

o Object is on same side of lens as light source Object is on opposite side of lens from light source (extremely rare)

i Image is on opposite side of lens from light source (real) Image is on same side of lens as light source (virtual)

r Lens is convex (converging) Lens is concave (diverging)

f Lens is convex (converging) Lens is concave (diverging)

m Image is upright (erect) Image is inverted

KEY CONCEPT

It is important to realize that concave mirrors and convex lenses are both converging and thus have similar properties. Convex mirrors and concave lenses are both diverging and also have similar properties.

The designations of real and virtual are often a point of confusion for students because they are on opposite sides when comparing mirrors and lenses. To identify the real side (R), remember that the real side is where light actually goes after interacting with the lens or mirror. For mirrors, light is reflected and, therefore, stays in front of the mirror. Hence, for a mirror, the real side is in front of the mirror, and the virtual side is behind the mirror. For lenses, the convention is different: because light travels through the lens and comes out on the other side, the real side is on the opposite side of the lens from the original light source, and the virtual side is on the same side of the lens as the original light source. Although the object of a single lens is on the virtual side, this does not make the object virtual. Objects are real, with a positive object distance, unless they are placed in certain multiple lens systems in which the image of one lens becomes the object for another (a scenario which is very rarely encountered on the MCAT).

Focal lengths and radii of curvature have a simpler sign convention. For both mirrors and lenses, converging species have positive focal lengths and radii of curvature, and diverging species have negative focal lengths and radii of curvature. Remember that lenses have two focal lengths and two radii of curvature because they have two surfaces. For a thin lens where thickness is negligible, the sign of the focal length and radius of curvature are given based on the first surface the light passes through.

Power

Optometrists often describe a lens in terms of its **power (P). This is measured indiopters**, where f (the focal length) is in meters and is given by the equation

P=1f

Equation 8.9

P has the same sign as f and is, therefore, positive for a converging lens and negative for a diverging lens. People who are nearsighted (can see near objects clearly) need diverging lenses, while people who are farsighted (can see distant objects clearly) need converging lenses. Bifocal lenses are corrective lenses that have two distinct regions—one that causes convergence of light to correct for farsightedness (hyperopia) and a second that causes divergence of light to correct for nearsightedness (myopia) in the same lens.

MCAT EXPERTISE

The MCAT does expect students to understand what myopia and hyperopia are, as well as their corresponding ray diagrams and correction strategies.

Multiple Lens Systems

Lenses in contact are a series of lenses with negligible distances between them. These systems behave as a single lens with equivalent focal length given by

1f=1f1+1f2+1f3+⋅⋅⋅+1fn

Equation 8.10

Because power is the reciprocal of focal length, the equivalent power is

P = P1 + P2 + P3 + ··· + *P**n*

Equation 8.11

REAL WORLD

The human eye typically has an optical power of around 60 diopters. Most contact lens wearers have prescriptions between 0.25 and 8 diopters (both positive and negative). An eye that requires a +8 or −8 prescription still has an optical power that is about 87% of the typical value.

A good example of lenses in contact is a corrective contact lens worn directly on the eye. In this case, the cornea of the eye (a converging lens) is in contact with a contact lens (either converging or diverging, depending on the necessary correction), and their powers would be added.

For lenses not in contact, the image of one lens becomes the object of another lens. The image from the last lens is considered the image of the system. Microscopes and telescopes are good examples of these systems. The magnification for the system is

m = m1 × m2 × m3 × ··· × mn

Equation 8.12

MCAT EXPERTISE

Because the MCAT is a timed test, you will not be given a massive multiple lens system and be expected to make calculations, although the testmaker might test a multiple-lens system conceptually.

Example: An object is 15 cm to the left of a thin diverging lens with a 45 cm focal length as shown below. Find where the image is formed, if it is upright or inverted, and if it is real or virtual. What is the radius of curvature, assuming the lens is symmetrical and is made of glass with a non negligible thickness and an index of refraction of 1.50?

concave lens with two radii of curvature labeled, object, and object distance labeled

Solution: The image distance (i) is found using the equation

1f=1o+1i→1i=1f−1o

Because the lens is diverging, the focal length has a negative sign (f = −45 cm). The object, like any object in a single-lens system, has a positive sign (o = +15 cm). Now we can solve for i:

1i=1f−1o=1−45 cm−115 cm=−1−345=−445i=−454=−11.25 cm

The negative sign indicates that the image is on the same side of the light source and is virtual. Remember that for a single lens or mirror, virtual images are always upright.

The thickness of a lens is usually negligible, but we are told otherwise in this question. To determine the radii of curvature, we use the lensmaker’s equation. Because the lens is symmetrical, the radii are equal but opposite in sign. As the light progresses from left to right, the first surface of the lens is concave (r1 < 0), and the second surface of the lens is convex (r2 > 0).

1f=(n−1)(1r1−1r2)=(n−1)(−1r−1r)=(n−1)(−2r)r=−2f(n−1)=−2(−45 cm)(1.5−1)=2(45)(0.5)=45 cm

Spherical Aberration

Spherical mirrors and lenses are imperfect. They are therefore subject to specific types of errors or aberrations. Spherical aberration is a blurring of the periphery of an image as a result of inadequate reflection of parallel beams at the edge of a mirror or inadequate refraction of parallel beams at the edge of a lens. This creates an area of multiple images with very slightly different image distances at the edge of the image, which appears blurry. This phenomenon can be seen in Figure 8.12.

parallel light rays converged to a focal point, but with imperfect refraction so that rays don’t quite meet at the focal point

Figure 8.12. Spherical Aberration Parallel rays are not perfectly reflected or refracted through the focal point, leading to blurriness at the periphery of the image.

REAL WORLD

If you remember back to conic sections from your precalculus class, it should be no surprise that spherical mirrors and lenses do not focus light perfectly. Parabolas are perfect reflectors, meaning that parallel light rays are reflected perfectly through the focal point. This is used in extracorporeal shock wave lithotripsy, in which a parabolic mirror is positioned with a kidney stone at the focal point. Sound waves are reflected off of the mirror and create enough vibration in the kidney stone to shatter it.

Chromatic aberration, discussed below, is predominantly seen in spherical lenses.

Dispersion

As discussed earlier, the speed of light in a vacuum is the same for all wavelengths. However, when light travels through a medium, different wavelengths travel at different speeds. This fact implies that the index of refraction of a medium affects the wavelength of light passing through the medium because the index of refraction is related to the speed of the wave by n=cv. It also implies that the index of refraction itself actually varies with wavelength. When various wavelengths of light separate from each other, this is called dispersion. The most common example of dispersion is the splitting of white light into its component colors using a prism.

If a source of white light is incident on one of the faces of a prism, the light emerging from the prism is spread out into a fan-shaped beam, as shown in Figure 8.13. This occurs because violet light has a smaller wavelength than red light and so is bent to a greater extent. Because red experiences the least amount of refraction, it is always on top of the spectrum; violet, having experienced the greatest amount of refraction, is always on the bottom of the spectrum. Note that as light enters a medium with a different index of refraction, the wavelength changes but the frequency of the light does not.

white light dispersed into component colors in a prism

Figure 8.13. Dispersion in a Prism Due to their different speeds while inside the prism, the various wavelengths of light are refracted to different degrees.

Chromatic Aberration

Chromatic aberration, shown in Figure 8.14, is a dispersive effect within a spherical lens. Depending on the thickness and curvature of the lens, there may be significant splitting of white light, which results in a rainbow halo around images. This phenomenon is corrected for in visual lenses like eyeglasses and car windows with special coatings that have different dispersive qualities from the lens itself.

parallel light rays refracted to different degrees, depending on color of light

Figure 8.14. Chromatic Aberration Light dispersion within the glass lens leads to the formation of a rainbow halo at the edge of the image.

MCAT CONCEPT CHECK 8.2

Before you move on, assess your understanding of the material with these questions.

**Mirrors Symbol Positive Negative o i r f m Lenses Symbol Positive Negative *o i r f m***

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

8.3 Diffraction

LEARNING OBJECTIVES

After Chapter 8.3, you will be able to:

light passing through lens and double slits, forming alternating bright and dark fringes

Diffraction refers to the spreading out of light as it passes through a narrow opening or around an obstacle. Interference between diffracted light rays lead to characteristic fringes in slit–lens and double-slit systems. Diffraction and interference are significant evidence for the wave theory of light.

Single Slit

Although it is usually safe to assume that nonrefracted light travels in a straight line, there are situations where light will not actually travel in a straight-line path. When light passes through a narrow opening (an opening with a size that is on the order of light wavelengths), the light waves seem to spread out (diffract), as is shown in Figure 8.15. As the slit is narrowed, the light spreads out more.

incident wave passes through narrow slit and emanates out in a wide arc

Figure 8.15. Diffraction Light emerges from a narrow slit in a wide arc, not a narrow beam.

Slit–Lens System

If a lens is placed between a narrow slit and a screen, a pattern is observed consisting of a bright central fringe with alternating dark and bright fringes on each side, as shown in Figure 8.16. The central bright fringe (maximum) is twice as wide as the bright fringes on the sides, and as the slit becomes narrower, the central maximum becomes wider. The location of the dark fringes (minima) is given by the formula

a sin θ = nλ

Equation 8.13

where a is the width of the slit, θ is the angle between the line drawn from the center of the lens to the dark fringe and the axis of the lens, n is an integer indicating the number of the fringe, and λ is the wavelength of the incident wave. Note that bright fringes are halfway between dark fringes.

incident wave passing through slit with width a and lens, forming bright central fringe (n = 0) and dark fringe (n = 1) at angle theta

Figure 8.16. Single-Slit Diffraction with Lens

Multiple Slits

When waves interact with each other, the displacements of the waves add together in a process called interference, as described in Chapter 7 of MCAT Physics and Math Review. In his famous double-slit experiment, Thomas Young showed that the diffracted rays of light emerging from two parallel slits can interfere with one another. This was a landmark finding that contributed to the understanding of light as a wave. Figure 8.17 shows the typical setup for Young’s double-slit experiment. When monochromatic light (light of only one wavelength) passes through the slits, an interference pattern is observed on a screen placed behind the slits. Regions of constructive interference between the two light waves appear as bright fringes (maxima) on the screen. Conversely, in regions where the light waves interfere destructively, dark fringes (minima) appear.

light passing through lens and double slits separated by distance d, forming alternating bright and dark fringes; screen is distance D from slits; first bright fringe at angle theta and distance y from axis

Figure 8.17. Young’s Double-Slit Experiment

BRIDGE

Light is similar to other waveforms; it is affected by constructive and destructive interference when light passes through a slit and a lens, and when light passes through multiple slits. Interference also occurs with sound waves, as discussed in Chapter 7 of MCAT Physics and Math Review.

The positions of dark fringes (minima) on the screen can be found from the equation

dsinθ=(n+12)λ

Equation 8.14

where d is the distance between the two slits, θ is the angle between the line drawn from the midpoint between the two slits to the dark fringe and the normal, n is an integer indicating the number of the fringe, and λ is the wavelength of the incident wave. Note that bright fringes are halfway between dark fringes.

Example: In a double-slit experiment, what is the linear distance y between the sixth and eighth minima on the screen? (Note: The wavelength λ is 550 nm, the slits are separated by a distance of 0.14 mm, and the screen is 70 cm from the slits.)

Solution: The position of a dark fringe (minimum) is given by

dsinθ=(n+12)λ

We do not know the value of sinθ(oppositehypotenuse). However, for small angles, sin θ ≈ tan θ. This is because the length of the hypotenuse is very close to the length of the adjacent side. We do know the value of tanθ(oppositeadjacent), so we can substitute it into the equation and still get very close to the correct answer:

d sin θ=(n+12)λd tan θ≈(n+12)λ d(yD)≈(n+12)λ            y≈(n+12)λDdy8−y6≈(n8+12)λDd−(n6+12)λDd         Δy≈(Δn)λDd=(2)(550×10−9 m)(0.7 m)0.14×10−3 m=550×10−5 m=5.5 mm

Diffraction gratings consist of multiple slits arranged in patterns. Diffraction gratings can create colorful patterns similar to a prism as the different wavelengths interfere in characteristic patterns. For example, the organization of the grooves on a CD or DVD act like a diffraction grating, creating an iridescent rainbow pattern on the surface of the disc. Thin films may also cause interference patterns because light waves reflecting off the external surface of the film interfere with light waves reflecting off the internal surface of the film, as shown in Figure 8.18. Common examples of thin films are soap bubbles or oil puddles in wet parking lots. Note that the interference here is not between diffracted rays, but between reflected rays.

two waves bouncing off internal and external surfaces of film, respectively, and partially constructively interfering

Figure 8.18. Thin Film Interference Interference patterns, C, occur as light waves reflecting off the external surface of the film, B, interfere with light waves reflecting off the internal surface of the film, A. Note that there would be a small degree of refraction as well, although this is not shown in the image.

X-Ray Diffraction

X-ray diffraction uses the bending of light rays to create a model of molecules. X-ray diffraction is often combined with protein crystallography during protein analysis. Dark and light fringes do not take on a linear appearance, but rather a complex two-dimensional image. An example of an x-ray diffraction pattern is shown in Figure 8.19.

concentric rings of blue light of various intensity with a "starry sky" appearance

Figure 8.19. X-Ray Diffraction Pattern

BRIDGE

X-ray diffraction and protein crystallography are commonly used to analyze the structure of proteins. These techniques, as well as a number of other protein assays, are discussed in Chapter 3 of MCAT Biochemistry Review.

MCAT CONCEPT CHECK 8.3

Before you move on, assess your understanding of the material with these questions.

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

8.4 Polarization

LEARNING OBJECTIVES

After Chapter 8.4, you will be able to:

Plane-Polarized Light

Plane-polarized (or linearly polarized) light is light in which the electric fields of all the waves are oriented in the same direction (that is, their electric field vectors are parallel). It follows that their magnetic field vectors are also parallel, but convention dictates that the plane of the electric field identifies the plane of polarization. Unpolarized light has a random orientation of its electric field vectors; sunlight and light emitted from a light bulb are prime examples. One of the most common applications of plane-polarized light on the MCAT is in the classification of stereoisomers, as discussed in Chapter 2 of MCAT Organic Chemistry Review. The optical activity of a compound, due to the presence of chiral centers, causes plane-polarized light to rotate clockwise or counterclockwise by a given number of degrees relative to its concentration (its specific rotation). Remember that enantiomers, as nonsuperimposable mirror images, will have opposite specific rotations.

REAL WORLD

Plane-polarized light is used to diagnose a number of diseases. Amyloidosis, caused by the buildup of various forms of misfolded proteins, is diagnosed by biopsy and staining the tissue with Congo red stain; a bright “apple green” color is seen under plane-polarized light. Gout (the precipitation of monosodium urate crystals) and pseudogout (the precipitation of calcium pyrophosphate crystals) are differentiated by their precipitate colors under polarized light: monosodium urate appears yellow and calcium pyrophosphate appears blue when the axis of the crystal is aligned with a polarizer.

There are filters called polarizers, often used in cameras and sunglasses, which allow only light with an electric field pointing in a particular direction to pass through. If one passes a beam of light through a polarizer, it will only let through that portion of the light parallel to the axis of the polarizer. If a second polarizer is then held up to the first, the angle between the polarizers’ axes will determine how much light passes through. When the polarizers are aligned, all the light that passes through the first polarizer also passes through the second. When the second polarizer is turned so that its axis is perpendicular, no light gets through at all.

KEY CONCEPT

The electric fields of unpolarized light waves exist in all three dimensions: the direction of the wave’s propagation is surrounded by electric fields in every plane perpendicular to that direction. Polarizing light limits the electric field’s oscillation to only two dimensions.

Circular Polarization

Circular polarization is a rarely seen natural phenomenon that results from the interaction of light with certain pigments or highly specialized filters. Circularly polarized light has a uniform amplitude but a continuously changing direction, which causes a helical orientation in the propagating wave, as shown in Figure 8.20. The helix has average electrical field vectors and magnetic field vectors that lie perpendicular to one another, like other waves, with maxima that fall on the outer border of the helix.

helical-shaped electric field

Figure 8.20. Circularly Polarized Light

MCAT CONCEPT CHECK 8.4

Before you move on, assess your understanding of the material with these questions.

_________________________________

_________________________________

_________________________________

Conclusion

This chapter illuminated the key behaviors and characteristics of light and optical systems. First, we described the nature of the electromagnetic (EM) wave, noting that we can only perceive light in the visible range (400 nm–700 nm). We then focused on geometrical optics to consider the reflective and refractive behaviors of light, noting the ways in which mirrors reflect light to produce images and lenses refract light to produce images. We acknowledged the fact that light doesn’t always travel in straight-line pathways but can bend and spread out through diffraction. We examined the pattern of interference that occurs when light passes through a double slit, as demonstrated in Young’s double-slit experiment. Finally, we wrapped up with a discussion on plane-polarized and circularly polarized light. In this chapter, we considered the properties that support the wave theory of light. In the next chapter, we’ll explore the photon and properties that support the particle theory of light, as well as other atomic and nuclear phenomena.

GO ONLINE

You’ve reviewed the content, now test your knowledge and critical thinking skills by completing a test-like passage set in your online resources!

CONCEPT SUMMARY

Electromagnetic Spectrum

Geometrical Optics

**Converging Systems Diverging Systems o relative to *f*** o > 2f o = 2f 2f > o > f o = f o < f all object distances

image real, inverted, reduced real, inverted, same real, inverted, magnified no image virtual, upright, magnified

virtual, upright, reduced

Diffraction

Polarization

ANSWERS TO CONCEPT CHECKS

**8.1**

**8.2**

-

**Mirrors Symbol Positive Negative *o*** Object is in front of mirror Object is behind mirror (extremely rare)

i Image is in front of mirror (real) Image is behind mirror (virtual)

r Mirror is concave (converging) Mirror is convex (diverging)

f Mirror is concave (converging) Mirror is convex (diverging)

m Image is upright (erect) Image is inverted

**Lenses Symbol Positive Negative *o*** Object is on same side of lens as light source Object is on opposite side of lens from light source (extremely rare)

i Image is on opposite side of lens from light source (real) Image is on same side of lens as light source (virtual)

r Lens is convex (converging) Lens is concave (diverging)

f Lens is convex (converging) Lens is concave (diverging)

m Image is upright (erect) Image is inverted

**8.3**

**8.4**

SCIENCE MASTERY ASSESSMENT EXPLANATIONS

1. C

It is unnecessary to memorize the entire electromagnetic spectrum for Test Day; however, it is important to know that the visible spectrum runs from 400–700 nm. We can calculate the wavelength of this light ray:

c=fλλ=cf=3×108ms5×1014 Hz=6×10−7 m=600 nm

This wavelength falls within the visible spectrum and has a yellow-orange color.

2. D

The color of an object is determined by the wavelength of light reflected by that object. Since anthocyanin pigment appears red, it must reflect red light. The frequency of light can be calculated using the equation c = fλ. Red light has a wavelength of 700 nm, which can be converted to 700 × 10-9 m. Plugging in values yields: f = (3 × 108) / (700 × 10-9) = 4.2 × 1013 Hz, matching (D). Note that a close eye for units would enable the immediate elimination of (A).

3. A

One could solve this question with a ray diagram, but be wary about using ray diagrams on Test Day. It is easy to make small mistakes that cause the light rays not to intersect. Therefore, solve the question using the sign convention. If the object is at the center of curvature, its distance is 2f. We can plug into the optics equation:

1f=1o+1i→1i=1f−1o=1f−12f=12f   i=2f=r

Because i is positive, the image is real. For single mirrors or lenses, all real images are inverted.

4. B

To calculate the wavelength, use the formula: d sin(θ) = (n + 1/2)λ. Rearrange the formula to solve for wavelength: λ = d sin (θ) / (n + 1/2). Since the distance is given in millimeters, first convert to meters; d = 3 × 10–4 m. Since the question asks for the second dark fringe, we have n = 2. Plugging in the given values yields: λ = (3 × 10–4)sin (30o) / (2 + 1/2) = 6 × 10–5 m, matching choice (B).

5. A

This question contains two parts—we have to determine the frequency and the angle of refraction of the light ray. The first part, however, is straightforward because the frequency of a light ray traveling from one medium to another does not change. Because the frequency must be 5 × 1014 Hz, we can eliminate (C) and (D). For the angle of refraction, we can either calculate it or determine it using logic. First, the light ray goes from air into crystal; that is, from a low index of refraction to a higher one. According to Snell’s law, the angle of refraction will be smaller than the incident angle (closer to the normal). When the light ray moves from crystal to chromium, it again goes from a lower index of refraction into a higher one, thus making the angle of refraction even smaller, eliminating (B). This question could also be answered by calculation using Snell’s law, but the calculations are time consuming and unnecessary.

6. D

Plane-polarized light is light in which the electric fields of all the waves are oriented in the same direction. Light passing through the first two polarizers will only contain rays with their electric field vectors in the same direction. When it reaches the third polarizer, however, the light will not be able to pass through because all the light rays will be oriented in the direction dictated by the first and second polarizers.

7. C

Plane polarizers only allow one specific orientation of the electric field of light to pass through. When plane polarizers are perpendicular to each other, no light can pass through the second polarizer, which supports (C) as the correct answer. Note that if two polarizers are aligned, all of the light that passes through the first will also pass through the second; there is no "twice as plane polarized," eliminating (B).

8. B

The image produced by a convex lens can be either real or virtual. It is real if the object is placed at a distance greater than the focal point, and virtual if the object is placed at a distance less than the focal point (between the focal point and the lens). Remember that for a single mirror or lens, an image that is real must be inverted and one that is virtual must be upright. In this question, the object is placed in front of the focal point, so the image must be virtual and, therefore, upright. We could also determine this from the optics equation. If f > o, then 1f−1o is negative, and i is therefore negative (virtual).

9. D

This question is testing our understanding of total internal reflection. As the laser beam travels from water to air—that is, from a higher to a lower index of refraction—the angle of refraction increases. At the critical angle (θc), the angle of refraction becomes 90°; at this point, the refracted ray is parallel to the surface of the water. When the incident angle is greater than the critical angle, all the light is reflected back into the water. The question is asking for the critical angle:

θc=sin−1n2n1=sin−111.33=sin−10.75

The inverse sine of 0.75 must be slightly higher than 45°(sin45°=22=0.707). 48.59° is the exact answer.

10. C

This question is testing our understanding of diffraction. When light passes through a narrow opening, the light waves spread out; as the slit narrows, the light waves spread out even more. When a lens is placed between the narrow slit and the screen, a pattern consisting of alternating bright and dark fringes can be observed on the screen. As the slit becomes narrower, the central maximum (the brightest and most central fringe) becomes wider. This can also be seen in the equation for the position of dark fringes in a slit–lens setup (a sin θ = nλ). As a, the width of the slit, decreases, sin θ must increase because nλ is constant for a given fringe. If sin θ increases, θ necessarily increases, implying that the fringes are spreading farther apart.

11. D

All images produced by plane mirrors will be virtual, so statement III is true. The same goes for diverging species (convex mirrors and concave lenses), so statement II is true. Converging species (concave mirrors and convex lenses) can produce real or virtual images, depending on how far the object is from the species, so statement I is also true.

12. A

First, the color of the light is irrelevant here; the ratio would be the same even if the specific color were not mentioned. Second, recall Snell’s Law: n1 sin θ1 = n2 sin θ2. Although we don’t know the value of n for either medium, you do know the simple relationship n=cv. Replacing n in Snell’s law, and canceling out c from both sides, we get:

cv1 sin θ1=cv2 sin θ2 sin30°v1=sin45°v2         12v1=22v2            v2=v12

13. D

The overall magnification of a system of multiple lenses is simply the product of each lens’s magnification. In this case, that is 10 × 40 = 400.

14. D

Drawing a diagram is best here. Because the angle given is with respect to the normal, you know that the incident angle must equal 60°. You know that the reflected beam will have an angle of 60° relative to the normal. Therefore, the reflected beam will make an angle of 30° with the plane of the glass. If the reflected and refracted beams are perpendicular to each other, the refracted beam will make a 60° angle with the plane of the glass. θrefracted is therefore 30° relative to the normal.

incident and reflected angle are 60 degrees; refracted angle is 30 degrees

Using n1 sin θ1 = n2 sin θ2, we have

1 sin 60°=n2 sin 30°          32=n2(12)           3=n2

15. D

Light can be split into its component colors by dispersion, such as that through a prism, eliminating (A). Diffraction by a diffraction grating will also separate colors by their wavelengths, eliminating (B). The refraction of light within a thin film also leads to light dispersion as the different colors are refracted at slightly different angles in the film, eliminating (C). A mirror with significant aberration could lead to a separation of light into its component colors, but we are told that this is an ideal mirror. Thus, (D) is the correct answer.

GO ONLINE

Consult your online resources for additional practice.

EQUATIONS TO REMEMBER

(8.1) Speed of light from frequency and wavelength: c = fλ

(8.2) Law of reflection: θ1 = θ2

(8.3) Optics equation: 1f=1o+1i=2r

(8.4) Magnification: m=−io

(8.5) Index of refraction: n=cv

(8.6) Snell’s law: n1 sin θ1 = n2 sin θ2

(8.7) Critical angle: θc=sin−1(n2n1)

(8.8) Lensmaker’s equation: 1f=(n−1)(1r1−1r2)

(8.9) Power: P=1f

(8.10) Focal length of multiple lens system: 1f=1f1+1f2+1f3+⋅⋅⋅+1fn

(8.11) Power of multiple lens system: P = P1 + P2 + P3 + ··· + *P**n*

(8.12) Magnification of multiple lens system: m = m1 × m2 × m3 × ··· × *m**n*

(8.13) Positions of dark fringes in slit–lens setup: a sin θ = nλ

(8.14) Positions of dark fringes in double-slit setup: dsinθ=(n+12)λ

SHARED CONCEPTS

Behavioral Sciences Chapter 2

Sensation and Perception

Biochemistry Chapter 3

Nonenzymatic Protein Function and Protein Analysis

Organic Chemistry Chapter 2

Isomers

Organic Chemistry Chapter 11

Spectroscopy

Physics and Math Chapter 7

Waves and Sound

Physics and Math Chapter 9

Atomic and Nuclear Phenomena

← Waves and Sound All chapters Atomic and Nuclear Phenomena →