Optics 1 · Lecture 1

Fundamental Optics Principles

Vergence, reflection, refraction, and lenses · Dr. Michael J. Collins

Follow along: Lecture handout (PDF) Practice worksheet (PDF)

Introduction

Learning Objectives

Today we will discuss the following principles, and solve problems related to each.

Point Sources & Vergence

Point Sources of Light

If we imagine a small, powerful source of light, we get the physical idealization of a point source.
Light leaves a point source equally in every direction.
We can describe that light two ways: as rays streaming outward, or as wavefronts rippling outward.

Show light as

An idealized point source sends light out equally in every direction. Rays show the direction light travels; wavefronts are the ripples, always perpendicular to the rays.

Point Sources & Vergence

Divergence

From a point source, light diverges: rays stream outward, like light from the sun, while waves ripple outward.
The farther the light travels, the larger and flatter its wavefronts become.

Light is

Light spreading away from a point is diverging. Its wavefronts bulge forward and grow flatter as they travel.

Point Sources & Vergence

Convergence

Using lenses or other means, light can be condensed to focus back onto a point. This is the meaning of convergence.
Converging wavefronts shrink as they close in on the focal point.

Light is

A converging (plus) lens bends parallel rays toward a single point. The wavefronts after the lens curve inward and shrink into the focal point.

Point Sources & Vergence

Measuring Divergence

Sometimes we need to know how far light has diverged from its source.
Taking the reciprocal of the distance (in meters) from the source to the observer gives a unit called the diopter (D).
By convention, divergence is negative and convergence is positive.
Example: a source 4 m away gives the observer $-\tfrac{1}{4\,\text{m}}$ = −0.25 D of divergence.

V = \frac{1}{d\,(\text{m})}

V = -\frac{1}{d} = -\frac{1}{4.00\,\text{m}} = -0.25\,\text{D}

Vergence−0.25 D

Distance d4.0 m

Light is

Vergence is the reciprocal of the distance (in meters) to the point light is diverging from or converging to. Closer to the point means more curved wavefronts and larger vergence.

Point Sources & Vergence

Measuring Convergence

Diopters measure convergence too, only with a positive sign instead of negative.
Example: if a lens focuses light to a point 4 m away, the light leaving the lens has +0.25 D of convergence.

V = +\frac{1}{d} = +\frac{1}{4\,\text{m}} = +0.25\ \text{D}

V = +\frac{1}{d} = +\frac{1}{4.00\,\text{m}} = +0.25\,\text{D}

Vergence+0.25 D

Distance d4.0 m

Light is

Vergence is the reciprocal of the distance (in meters) to the point light is diverging from or converging to. Closer to the point means more curved wavefronts and larger vergence.

Collimation

Collimated Light

As light travels away from its source, it diverges less and less.
At 1 cm from a point source the light diverges by 1/0.01 = 100 D; at 1000 m it diverges by only 1/1000 = 0.001 D.
Far from its source, light is considered collimated: its rays are parallel to each other.
Collimated light has zero vergence.

Move across the chart to read vergence at any distance. Each step to the right is 10× farther and 10× less vergence. Beyond about 6 m the light is treated as collimated: parallel rays, zero vergence.

Collimation

Optical Infinity

When examining patients, we need collimated light to measure prescriptions.
We treat light as collimated beyond a certain distance. In optometry, that distance is 20 feet (6 meters), known as optical infinity.
The true vergence at 6 m is only −1/6 ≈ −0.17 D, small enough to treat as zero.

Vergence at the eye

V = -\frac{1}{d} = -\frac{1}{6.00\,\text{m}} = -0.17\,\text{D}

≈ treated as 0: optical infinity−0.17 D

Chart distance6.00 m

At 6 m (20 ft) light from the chart reaches the eye with only −0.17 D of vergence, close enough to parallel that we treat it as coming from optical infinity.

Reflection

When light passes from one medium toward another, some of it bounces off the second medium. This is reflection.
For example, light in air hitting a still surface of water partially reflects off it.
Light leaves a surface at the same angle it approached, measured from the normal (the line perpendicular to the surface).

\theta_{\text{incidence}} = \theta_{\text{reflection}}

Angle of incidence (from normal)35°

Measure from

θ incidence35°

θ reflection35°

Drag the light source. The angle of reflection always equals the angle of incidence, both measured from the normal.

Worked example 1

Predicting Reflection

A ray approaches a mirror at 32° from the mirror's surface. At what angle does it leave, relative to the surface?
Solution: by the law of reflection, the light leaves at 32° from the surface.

Measure from

Worked example 1: the ray arrives 32° from the mirror’s surface.

Worked example 2

Predicting Reflection

A ray approaches a mirror at 40° from the normal. At what angle does it leave, relative to the mirror's surface?
Solution: the normal is 90° from the surface, so 90° − 40° = 50° from the surface.

Measure from

Worked example 2: the ray arrives 40° from the normal. The normal is 90° from the surface.

Refraction

When light enters one medium from another, the second medium often bends its path. This is refraction.
For example, light traveling from water into air bends, so a straw in a glass looks broken at the surface.

True position

Light from the submerged tip bends away from the normal as it leaves the water. Your brain traces it back in a straight line, so the tip looks higher than it really is.

Refraction

Snell’s Law

Predicting refraction is a little more complicated than predicting reflection.
Measured from the normal, the sine of the angle of incidence times that medium's index of refraction equals the sine of the angle of refraction times its index.
Entering a higher index, light bends toward the normal. Entering a lower index, it bends away.

n_1 \sin\theta_1 = n_2 \sin\theta_2

n₁ sin θ₁ = n₂ sin θ₂

1.000 × sin 40.0° = 1.333 × sin 28.8°(both sides = 0.643)

Angle of incidence θ₁40°

Top medium (n₁)Bottom medium (n₂)

Refraction

Index of Refraction

An index of refraction is a physical property of a pure medium that helps predict how much light bends passing between media.
Air has an index of 1.000; water has an index of 1.333.
Formally, it is the ratio of the speed of light in a vacuum to the speed of light in that medium.

n = \frac{c}{v} \qquad n_{\text{water}} = \frac{c}{v_{\text{water}}} = 1.333

Each pulse travels at v = c / n. The higher the index, the slower light moves through the medium.

Applications

High-Index Lenses

High-index lenses are thinner, lighter lenses for people with stronger prescriptions.
With a higher index than, say, polycarbonate, you need less lens material to bend light the same amount.
Historically they have drawbacks such as color fringing (chromatic aberration) and other visual aberrations.

Prescription−6.00 D

Same −6.00 D prescription, 60 mm lens, drawn to scale. A higher index needs less curvature for the same power, so the lens is thinner.

Applications

Spectacle Materials

Every spectacle material has its own index of refraction.
Common and historic types: glass, CR-39, polycarbonate, Trivex, and high-index plastics.

Index n1.45 – 1.76

Abbe value

CR-39 · n = 1.498 · Abbe 58

Plastic introduced in the 1940s. Lightweight with great optics; the everyday default.

Typical nominal values; vary by manufacturer. Higher Abbe value = less color fringing (chromatic aberration).

Applications

Mirages

On a hot desert road you may see an image rising above the road: distorted, wavy, and often looking like water.
Hot air near the road has a different (lower) index of refraction than the cooler air above it. Light is bent your way, giving a false image.

Road surface temperature55 °C

Air just above a hot road is hotter and less dense, so it has a lower index of refraction than the cooler air above. Light from the sky bends gradually upward toward your eye. Your brain traces it back in a straight line, so you see an inverted image of the sky on the road. It looks like water. (Vertical scale exaggerated.)

Applications

Spear-Fishing

You see a fish directly in front of you. Should you aim your spear at the image, beyond it, or below it?
Answer: below the image. Water's index is greater than air's, so light from the fish bends at the surface, making the fish appear higher than it really is.

Where should you aim the spear?

Fish depth105 cm

You see a fish in front of you. Should you aim at the image, beyond it, or below it? Pick a button or a target.

Applications

Rainbows

Each falling raindrop acts a bit like a prism (we will study prisms in more detail later).
Different colors have slightly different indices of refraction in water, so each drop spreads white light into colors: dispersion, a form of chromatic aberration.
Red is bent least, so it exits at a slightly larger angle (≈42°) than violet (≈40°). Across the sky, red appears on the outside (top) of the bow and violet on the inside.
Rainbows would be complete circles if not for the horizon, as seen from aircraft or in a moonbow.

Sky view

Inside each drop, light refracts, reflects off the back, and refracts again. Each color bends a slightly different amount, so red leaves at about 42° and violet at about 40° from the sunlight. Red comes from drops higher in the sky, so in a primary rainbow red is on the outside (top) and violet is on the inside. Without the horizon in the way, the bow is a full circle. (Dispersion and band width exaggerated.)

A detour

Iris: from rainbows to eyes

Painting-style image of Iris, Greek goddess of the rainbow — **Iris** (Ἶρις): Greek goddess of rainbows, and the namesake of…

Close-up of a human eye showing a brown iris — …the **human iris**, the colored ring of the eye.

Lenses & Focal Points

Positive Lenses & Focal Points

A lens that converges light to a point is a positive lens. By convention its focal point is placed to its right.
Example: a lens that converges collimated light to a point 25 cm away is a +4.00 D lens, because 1/0.25 m = 4 D.

P = \frac{1}{f\,(\text{m})}

Incoming U0.00collimated

Lens P+4.00diopters

Outgoing V+4.00converging

Focal length+25.0 cmf = 1/P

U + P = V \;\Rightarrow\; (0.00) + (+4.00) = +4.00\,\text{D}

Lens power P+4.00 D

A plus lens converges collimated light to a real focal point F′ on its right, at

f = 1/P

. A +4.00 D lens focuses at 25 cm.

Lenses & Focal Points

Negative Lenses & Focal Points

A lens that diverges light is a negative lens. By convention its focal point is placed to its left.
Instead of a real focus, a negative lens makes collimated light appear to diverge from a point a certain distance in front of the lens.
Example: if the light seems to come from 25 cm before the lens, it is a −4.00 D lens, because 1/−0.25 m = −4 D.

Incoming U0.00collimated

Lens P-4.00diopters

Outgoing V-4.00diverging

Focal length−25.0 cmf = 1/P

U + P = V \;\Rightarrow\; (0.00) + (-4.00) = -4.00\,\text{D}

Lens power P−4.00 D

A minus lens diverges collimated light. Trace the rays backward (dashed) and they seem to come from a virtual focal point F′ on the lens's left. A −4.00 D lens makes light seem to come from 25 cm in front of it.

Lenses & Focal Points

Two Focal Points: Positive Lenses

Every lens has two focal points. For a positive lens, the one on the right is where collimated light focuses.
The one in front (left) is where a light source produces collimated light beyond the lens.
Example: a light at the front focal point of a +5.00 D lens (20 cm away) reaches the lens with 1/−0.20 m = −5 D of divergence.
The lens adds +5 D, producing zero vergence: collimated light.

V = U + P = (-5) + (+5) = 0\ \text{D}

Incoming U-5.00diopters

Lens P+5.00diopters

Outgoing V0.00collimated

Focal length+20.0 cmf = 1/P

U + P = V \;\Rightarrow\; (-5.00) + (+5.00) = 0.00\,\text{D}

Lens power P+5.00 D

Put a point source at the front focal point F of a plus lens. The light arrives with vergence

U = -P

, and the lens adds exactly

+P

, so it leaves collimated.

Lenses & Focal Points

Two Focal Points: Negative Lenses

For a negative lens, the focal point on the left is the imaginary point from which collimated light appears to emanate.
The one on the right is where, if converging light were heading toward it, the lens would cancel that convergence.
Example: +5.00 D of converging light reaching a −5.00 D lens emerges collimated.
That light would have focused 20 cm right of the lens, so that point is its right-hand focal point.

V = U + P = (+5) + (-5) = 0\ \text{D}

Incoming U+5.00diopters

Lens P-5.00diopters

Outgoing V0.00collimated

Focal length−20.0 cmf = 1/P

U + P = V \;\Rightarrow\; (+5.00) + (-5.00) = 0.00\,\text{D}

Lens power P−5.00 D

Light converging toward a point 20 cm to the right (+5.00 D) meets a −5.00 D lens. The vergences cancel, so the light leaves collimated. That point is the minus lens's right-hand focal point.

Wrap-up

Key Ideas

Vergence

V = \frac{1}{d\,(\text{m})}

− diverging, + converging

Optical infinity

6\ \text{m} = 20\ \text{ft} \Rightarrow V \approx 0

treated as collimated

Reflection

\theta_i = \theta_r

measured from the normal

Snell’s Law

n_1 \sin\theta_1 = n_2 \sin\theta_2

Index of refraction

n = \frac{c}{v}

Lens power

P = \frac{1}{f\,(\text{m})}

Vergence equation

V = U + P

exiting = entering + power

Ready to practice?Open interactive worksheet Worksheet PDF

Optics 1 · Lecture 1

Fundamental Optics Principles

Vergence, reflection, refraction, and lenses · Dr. Michael J. Collins

Follow along: Lecture handout (PDF) Practice worksheet (PDF)

Introduction

Learning Objectives

Today we will discuss the following principles, and solve problems related to each.

Point Sources & Vergence

Point Sources of Light

If we imagine a small, powerful source of light, we get the physical idealization of a point source.
Light leaves a point source equally in every direction.
We can describe that light two ways: as rays streaming outward, or as wavefronts rippling outward.

Show light as

An idealized point source sends light out equally in every direction. Rays show the direction light travels; wavefronts are the ripples, always perpendicular to the rays.

Point Sources & Vergence

Divergence

From a point source, light diverges: rays stream outward, like light from the sun, while waves ripple outward.
The farther the light travels, the larger and flatter its wavefronts become.

Light is

Light spreading away from a point is diverging. Its wavefronts bulge forward and grow flatter as they travel.

Point Sources & Vergence

Convergence

Using lenses or other means, light can be condensed to focus back onto a point. This is the meaning of convergence.
Converging wavefronts shrink as they close in on the focal point.

Light is

A converging (plus) lens bends parallel rays toward a single point. The wavefronts after the lens curve inward and shrink into the focal point.

Point Sources & Vergence

Measuring Divergence

Sometimes we need to know how far light has diverged from its source.
Taking the reciprocal of the distance (in meters) from the source to the observer gives a unit called the diopter (D).
By convention, divergence is negative and convergence is positive.
Example: a source 4 m away gives the observer $-\tfrac{1}{4\,\text{m}}$ = −0.25 D of divergence.

V = \frac{1}{d\,(\text{m})}

V = -\frac{1}{d} = -\frac{1}{4.00\,\text{m}} = -0.25\,\text{D}

Vergence−0.25 D

Distance d4.0 m

Light is

Vergence is the reciprocal of the distance (in meters) to the point light is diverging from or converging to. Closer to the point means more curved wavefronts and larger vergence.

Point Sources & Vergence

Measuring Convergence

Diopters measure convergence too, only with a positive sign instead of negative.
Example: if a lens focuses light to a point 4 m away, the light leaving the lens has +0.25 D of convergence.

V = +\frac{1}{d} = +\frac{1}{4\,\text{m}} = +0.25\ \text{D}

V = +\frac{1}{d} = +\frac{1}{4.00\,\text{m}} = +0.25\,\text{D}

Vergence+0.25 D

Distance d4.0 m

Light is

Vergence is the reciprocal of the distance (in meters) to the point light is diverging from or converging to. Closer to the point means more curved wavefronts and larger vergence.

Collimation

Collimated Light

As light travels away from its source, it diverges less and less.
At 1 cm from a point source the light diverges by 1/0.01 = 100 D; at 1000 m it diverges by only 1/1000 = 0.001 D.
Far from its source, light is considered collimated: its rays are parallel to each other.
Collimated light has zero vergence.

Collimation

Optical Infinity

When examining patients, we need collimated light to measure prescriptions.
We treat light as collimated beyond a certain distance. In optometry, that distance is 20 feet (6 meters), known as optical infinity.
The true vergence at 6 m is only −1/6 ≈ −0.17 D, small enough to treat as zero.

Vergence at the eye

V = -\frac{1}{d} = -\frac{1}{6.00\,\text{m}} = -0.17\,\text{D}

≈ treated as 0: optical infinity−0.17 D

Chart distance6.00 m

At 6 m (20 ft) light from the chart reaches the eye with only −0.17 D of vergence, close enough to parallel that we treat it as coming from optical infinity.

Reflection

When light passes from one medium toward another, some of it bounces off the second medium. This is reflection.
For example, light in air hitting a still surface of water partially reflects off it.
Light leaves a surface at the same angle it approached, measured from the normal (the line perpendicular to the surface).

\theta_{\text{incidence}} = \theta_{\text{reflection}}

Angle of incidence (from normal)35°

Measure from

θ incidence35°

θ reflection35°

Drag the light source. The angle of reflection always equals the angle of incidence, both measured from the normal.

Worked example 1

Predicting Reflection

A ray approaches a mirror at 32° from the mirror's surface. At what angle does it leave, relative to the surface?
Solution: by the law of reflection, the light leaves at 32° from the surface.

Measure from

Worked example 1: the ray arrives 32° from the mirror’s surface.

Worked example 2

Predicting Reflection

A ray approaches a mirror at 40° from the normal. At what angle does it leave, relative to the mirror's surface?
Solution: the normal is 90° from the surface, so 90° − 40° = 50° from the surface.

Measure from

Worked example 2: the ray arrives 40° from the normal. The normal is 90° from the surface.

Refraction

When light enters one medium from another, the second medium often bends its path. This is refraction.
For example, light traveling from water into air bends, so a straw in a glass looks broken at the surface.

True position

Light from the submerged tip bends away from the normal as it leaves the water. Your brain traces it back in a straight line, so the tip looks higher than it really is.

Refraction

Snell’s Law

Predicting refraction is a little more complicated than predicting reflection.
Measured from the normal, the sine of the angle of incidence times that medium's index of refraction equals the sine of the angle of refraction times its index.
Entering a higher index, light bends toward the normal. Entering a lower index, it bends away.

n_1 \sin\theta_1 = n_2 \sin\theta_2

n₁ sin θ₁ = n₂ sin θ₂

1.000 × sin 40.0° = 1.333 × sin 28.8°(both sides = 0.643)

Angle of incidence θ₁40°

Top medium (n₁)Bottom medium (n₂)

Refraction

Index of Refraction

An index of refraction is a physical property of a pure medium that helps predict how much light bends passing between media.
Air has an index of 1.000; water has an index of 1.333.
Formally, it is the ratio of the speed of light in a vacuum to the speed of light in that medium.

n = \frac{c}{v} \qquad n_{\text{water}} = \frac{c}{v_{\text{water}}} = 1.333

Each pulse travels at v = c / n. The higher the index, the slower light moves through the medium.

Applications

High-Index Lenses

High-index lenses are thinner, lighter lenses for people with stronger prescriptions.
With a higher index than, say, polycarbonate, you need less lens material to bend light the same amount.
Historically they have drawbacks such as color fringing (chromatic aberration) and other visual aberrations.

Prescription−6.00 D

Same −6.00 D prescription, 60 mm lens, drawn to scale. A higher index needs less curvature for the same power, so the lens is thinner.

Applications

Spectacle Materials

Every spectacle material has its own index of refraction.
Common and historic types: glass, CR-39, polycarbonate, Trivex, and high-index plastics.

Index n1.45 – 1.76

Abbe value

CR-39 · n = 1.498 · Abbe 58

Plastic introduced in the 1940s. Lightweight with great optics; the everyday default.

Typical nominal values; vary by manufacturer. Higher Abbe value = less color fringing (chromatic aberration).

Applications

Mirages

On a hot desert road you may see an image rising above the road: distorted, wavy, and often looking like water.
Hot air near the road has a different (lower) index of refraction than the cooler air above it. Light is bent your way, giving a false image.

Road surface temperature55 °C

Applications

Spear-Fishing

You see a fish directly in front of you. Should you aim your spear at the image, beyond it, or below it?
Answer: below the image. Water's index is greater than air's, so light from the fish bends at the surface, making the fish appear higher than it really is.

Where should you aim the spear?

Fish depth105 cm

You see a fish in front of you. Should you aim at the image, beyond it, or below it? Pick a button or a target.

Applications

Rainbows

Each falling raindrop acts a bit like a prism (we will study prisms in more detail later).
Different colors have slightly different indices of refraction in water, so each drop spreads white light into colors: dispersion, a form of chromatic aberration.
Red is bent least, so it exits at a slightly larger angle (≈42°) than violet (≈40°). Across the sky, red appears on the outside (top) of the bow and violet on the inside.
Rainbows would be complete circles if not for the horizon, as seen from aircraft or in a moonbow.

Sky view

A detour

Iris: from rainbows to eyes

Lenses & Focal Points

Positive Lenses & Focal Points

A lens that converges light to a point is a positive lens. By convention its focal point is placed to its right.
Example: a lens that converges collimated light to a point 25 cm away is a +4.00 D lens, because 1/0.25 m = 4 D.

P = \frac{1}{f\,(\text{m})}

Incoming U0.00collimated

Lens P+4.00diopters

Outgoing V+4.00converging

Focal length+25.0 cmf = 1/P

U + P = V \;\Rightarrow\; (0.00) + (+4.00) = +4.00\,\text{D}

Lens power P+4.00 D

A plus lens converges collimated light to a real focal point F′ on its right, at

f = 1/P

. A +4.00 D lens focuses at 25 cm.

Lenses & Focal Points

Negative Lenses & Focal Points

A lens that diverges light is a negative lens. By convention its focal point is placed to its left.
Instead of a real focus, a negative lens makes collimated light appear to diverge from a point a certain distance in front of the lens.
Example: if the light seems to come from 25 cm before the lens, it is a −4.00 D lens, because 1/−0.25 m = −4 D.

Incoming U0.00collimated

Lens P-4.00diopters

Outgoing V-4.00diverging

Focal length−25.0 cmf = 1/P

U + P = V \;\Rightarrow\; (0.00) + (-4.00) = -4.00\,\text{D}

Lens power P−4.00 D

Lenses & Focal Points

Two Focal Points: Positive Lenses

Every lens has two focal points. For a positive lens, the one on the right is where collimated light focuses.
The one in front (left) is where a light source produces collimated light beyond the lens.
Example: a light at the front focal point of a +5.00 D lens (20 cm away) reaches the lens with 1/−0.20 m = −5 D of divergence.
The lens adds +5 D, producing zero vergence: collimated light.

V = U + P = (-5) + (+5) = 0\ \text{D}

Incoming U-5.00diopters

Lens P+5.00diopters

Outgoing V0.00collimated

Focal length+20.0 cmf = 1/P

U + P = V \;\Rightarrow\; (-5.00) + (+5.00) = 0.00\,\text{D}

Lens power P+5.00 D

Put a point source at the front focal point F of a plus lens. The light arrives with vergence

U = -P

, and the lens adds exactly

+P

, so it leaves collimated.

Lenses & Focal Points

Two Focal Points: Negative Lenses

For a negative lens, the focal point on the left is the imaginary point from which collimated light appears to emanate.
The one on the right is where, if converging light were heading toward it, the lens would cancel that convergence.
Example: +5.00 D of converging light reaching a −5.00 D lens emerges collimated.
That light would have focused 20 cm right of the lens, so that point is its right-hand focal point.

V = U + P = (+5) + (-5) = 0\ \text{D}

Incoming U+5.00diopters

Lens P-5.00diopters

Outgoing V0.00collimated

Focal length−20.0 cmf = 1/P

U + P = V \;\Rightarrow\; (+5.00) + (-5.00) = 0.00\,\text{D}

Lens power P−5.00 D

Light converging toward a point 20 cm to the right (+5.00 D) meets a −5.00 D lens. The vergences cancel, so the light leaves collimated. That point is the minus lens's right-hand focal point.

Wrap-up

Key Ideas

Vergence

V = \frac{1}{d\,(\text{m})}

− diverging, + converging

Optical infinity

6\ \text{m} = 20\ \text{ft} \Rightarrow V \approx 0

treated as collimated

Reflection

\theta_i = \theta_r

measured from the normal

Snell’s Law

n_1 \sin\theta_1 = n_2 \sin\theta_2

Index of refraction

n = \frac{c}{v}

Lens power

P = \frac{1}{f\,(\text{m})}

Vergence equation

V = U + P

exiting = entering + power

Ready to practice?Open interactive worksheet Worksheet PDF