Optics

Question 1

What is the law that governs the reflection of a light ray off a flat surface?

Answer 1

θ_i = θ_r

The law of reflection states that: light incident on a reflective surface at some angle will bounce off at an equal angle of reflection, as measured from the normal.

Question 2

What will the angle of reflection be for a light ray incident on a surface, making an angle of 35 degrees to the surface?

Answer 2

θ_r = 55 degrees

Remember: the angles of incidence and reflection must be measured relative to the surface’s normal. 90 - 35 = 55. θ_i and θ_r are shown below.

Question 3

Define:

refraction

Answer 3

Refraction is when a light ray crosses a boundary from one transparent medium to another.

When the light ray changes media it bends either towards or away from the normal, depending on the properties of the respective media.

Question 4

Define:

refractive index

Answer 4

The refractive index of a transparent medium is a unitless ratio of the speed of light in that medium compared to speed of light in a vacuum.

Index of refraction is defined as n = c/v

where:
n = index of refraction of the medium
c = speed of light traveling through a vacuum
v = speed of light traveling through the medium

Question 5

What is the approximate index of refraction of:

• air
• water
• glass

Answer 5

• Air: approximately 1
• Water: approximately 1.33 (or 4/3)
• Glass: varies greatly, but 1.5 and 2 are commonly used values

Since the speed of light can never exceed c, the value for the index of refraction of any medium cannot be less than 1.

Question 6

How is the path of a light ray affected when the ray crosses a boundary from air (low n) into glass (high n)?

Answer 6

When a light ray crosses from air into glass, it bends towards the normal of the surface once inside the glass.

Any boundary crossing into a medium with a higher index of refraction will cause light to bend towards the normal.

Question 7

How is the path of a light ray affected when the ray crosses a boundary from glass into air?

Answer 7

When a light ray crosses from glass into air, it bends away from the normal of the surface in the air.

Any boundary crossing into a medium with a lower index of refraction will cause light to bend away from the normal.

Question 8

Define

Snell’s law

Answer 8

Snell’s law gives the direct relationship between angles of incidence and refraction, from a light ray crossing the boundary between media:

n₁sin(θ₁) = n₂sin(θ₂)

where
n₁ = index of refraction on the incident side
θ₁ = angle of incidence
n₂ = index of refraction on the refraction side
θ₂ = angle of refraction

Question 9

Define

critical angle θ_c

Answer 9

The critical angle θ_c is the angle of incidence in one medium at which the angle of refraction into the new medium would become 90 degrees or higher.

There is only a critical angle for refraction across a boundary at which the index of refraction decreases, such as glass into air.

Question 10

What is the formula for calculating the critical angle θ_c for a refraction from a medium with index n₁ to a medium with index n₂?

Answer 10

θ_c = sin^-1(n₂/n₁)

From Snell’s law:
n₁sin(θ_c) = n₂sin(90)
n₁sin(θ_c) = n₂(1)
sin(θ_c) = n₂/n₁
θ_c = sin^-1(n₂/n₁)

Question 11

Define:

total internal reflection

Answer 11

When a light ray approaches a boundary of decreasing index of refraction at an angle of incidence greater than the critical angle, it undergoes total internal reflection.

The light ray entirely reflects, staying in the incident medium. No light refracts across the boundary.

Question 12

Under what conditions does total internal reflection occur?

Answer 12

For total internal reflection to occur, two conditions must be met:

1. The index of refraction must decrease across the boundary in the direction of light refraction.
2. The angle of incidence of the light ray must exceed the critical angle of the interface.

Question 13

Define:

dispersion

Answer 13

Dispersion is different wavelengths of light refracting at different angles.

The most common example of dispersion tested on the AP Physics exam is white light splitting into a spectrum of colors inside a prism and spreading out at a range of angles.

Question 14

Why is a beam of sunlight separated into a rainbow of different colors as it passes through a prism?

Answer 14

The rainbow occurs due to dispersion.

The white sunlight beam is made up of all the colors of visible light mixed together, as is all white light. The glass of the prism is dispersive, and different colors of light refract differently as they pass through the prism, so they are separated upon exiting the back surface.

Blue light is refracted more than red light in glass.

Question 15

When white light is separated by a prism into its component colors, in what order do they appear?

Answer 15

The separated colors are, in order, red, orange, yellow, green, blue, indigo, violet.

This is easily remembered by the mnemonic ROYGBIV (Roy Gee Biv).

Question 16

Define:

the focal point of an optic

Answer 16

An optic’s focal point is the location where all rays parallel to the principal axis of the optic will cross after reflecting from/refracting through the optic.

An optic with a short focal point is a strong optic, bending light severely, while an optic with a long focal point is a weak optic, affecting light rays less significantly.

Question 17

Define:

the image of an optical system

Answer 17

The image of an optical system is an optical reproduction of a physical object, formed when light rays transmitted by the object are caused to converge at a specific point by an optic or series of optics.

Many optics problems will involve calculating the location of the system’s image, given a particular object and lens or mirror.

Question 18

What are the characteristics of the parallel ray in an optical system?

Answer 18

The parallel ray is a light ray which reflects off the top of an object, and travels parallel to the principal axis of the optic.

After reflecting off/refracting through the optic, the parallel ray crosses the principal axis at the focal point, and can be used to help locate the image.

Question 19

What are the characteristics of the focal ray in an optical system?

Answer 19

The focal ray is a light ray which reflects off the top of an object, and travels through the focal point of the optic.

After reflecting off/refracting through the optic, the focal ray emerges parallel to the principal axis of the optic, and can be used to help locate the image.

Question 20

What are the characteristics of the center ray in an optical system?

Answer 20

The center ray is a light ray which reflects off the top of an object, and impacts the optic directly in the center.

The center ray is undeflected by reflecting off/refracting through the optic. It travels in a straight line, and can be used to help locate the image.

Question 21

What makes a mirror converging?

Answer 21

A mirror is converging if any inbound light ray which is parallel to the mirror’s principal axis crosses the axis after reflecting off the mirror.

Converging mirrors are concave in shape.

Question 22

What is the radius of curvature of a converging mirror?

Answer 22

The radius of curvature of a mirror is the radius of the sphere of which the mirror is a small portion.

Question 23

How does the radius of curvature of a converging mirror compare to the mirror’s focal length?

Answer 23

For all spherical mirrors, the focal length is one-half the radius of curvature.

Question 24

When is the image of an optical system real?

Answer 24

An optical system projects a real image when light rays reflected off/transmitted through the optic cross after the reflection/transmission.

Examples of real images include the images on movie screens and the image your eye produces on your retina.

Question 25

When is the image of an optical system virtual?

Answer 25

An optical system projects a virtual image when light rays reflected off/transmitted through the optic are diverging away from one another.

Examples of virtual images include the images created by flat mirrors and magnifying glasses.

Question 26

An optical system consists of an object and a single converging mirror. If the object is a distance between f and 2f from the optic, what are the properties of the final image?

Answer 26

The image is further from the optic than the object is, and the image is real, magnified, and inverted.

Find where the parallel, focal, and center rays cross after reflecting off the mirror to locate the image. Since the rays cross after reflection, the image is real.

Question 27

An optical system consists of an object and a single converging mirror. If the object is at the focal length f, what are the properties of the final image?

Answer 27

No image is formed.

The parallel and center rays are parallel to one another after reflecting off the mirror. Since the rays never cross after reflection, no image is formed.

Question 28

An optical system consists of an object and a single converging mirror. If the object is inside the focal length f, what are the properties of the final image?

Answer 28

The image is further from the optic than the object is, and the image is virtual, magnified, and upright.

The parallel and center rays are diverging after reflecting off the mirror. To locate the image, the rays much be projected behind the mirror to the point where they cross. Since only the virtual rays cross, the image is virtual.

Question 29

What is the image equation (also known as the lens equation) used to calculate?

Answer 29

The image equation relates an optical system’s focal length to its object and image distance.

The image equation reads: 1/f = 1/o + 1/i

where
f = focal length of the optic (lens or mirror)
o = distance between optic and object
i = distance between optic and image

Question 30

What is the focal length of a converging mirror?

Answer 30

The focal length is the distance from the mirror to the focal point. Rays parallel to the optical axis cross at the focal point after reflecting off the mirror.

The focal length of any spherical optic is one-half its radius of curvature.

Question 31

What is the magnification of an optical system?

Answer 31

A optical system’s magnification is the ratio of the size of the image to the size of the object.

Magnification can be calculated from the image and object distance of the optical system. The magnification m is calculated as: m = -(i/o)

Question 32

What does the sign of the magnification of an optical system indicate about the image?

Answer 32

The sign of the magnification indicates the direction of the image.

If m > 0, the image is upright.
If m < 0, the image is inverted.

Question 33

What does the magnitude of the magnification of an optical system indicate about the image?

Answer 33

The magnitude of the magnification indicates the size of the image.

If |m| > 1, the image is magnified, larger than the object.
If |m| < 1, the image is reduced, smaller than the object.

Question 34

For an optical system, if the object and image distance are o = 30 cm and i = 60 cm, respectively, what is the magnification of the image?

Answer 34

m = -2

Remember, m = -i/o; so for this system:
m = -i/o = -(60/30) = -2
So the image is inverted, and twice as tall as the object.

Question 35

For an optical system, if the object and image distance are o = 20 cm and i = -10 cm, respectively, what is the magnification of the image?

Answer 35

m = +½

Remember, m = -i/o; so for this system:
m = -i/o = -((-10)/20) = +½
So the image is upright, and half as tall as the object.

Question 36

What makes a mirror diverging?

Answer 36

A mirror is diverging if any inbound light ray which is parallel to the principal axis points away from the axis after reflecting off the mirror.

All diverging mirrors are convex in shape.

Question 37

What does an optic’s focal length indicate about its properties?

Answer 37

An optic’s focal length indicates the converging or diverging nature of the optic.

If f > 0, the optic is converging.
If f < 0, the optic is diverging.
If f = infinity, the optic is planar.

Question 38

For an optical system made up of a single mirror, if the object distance is o = 30 cm and the image distance is i = -30 cm, what type is the mirror?

Answer 38

f = infinity, and the mirror is planar.

Given i and o, f can be calculated as:
1/f = 1/o + 1/i = 1/30 + 1/(-30) = 0
f = infinity

Planar mirrors produce virtual images, and i = -o.

Question 39

For an optical system made up of a single mirror, if the object distance is o = 15 cm and the image distance is i = 30 cm, what type is the mirror?

Answer 39

f = 10 cm, and the mirror is converging.

Given i and o, f can be calculated as:
1/f = 1/o + 1/i = 1/15 + 1/30 = 3/30
f = 30/3 = 10 cm

Question 40

What is the radius of curvature of a diverging mirror?

Answer 40

The radius of curvature of a diverging mirror is the radius of the sphere of which the mirror is a small portion.

Question 41

How does the radius of curvature of a diverging mirror compare to the mirror’s focal length?

Answer 41

For all spherical mirrors, the focal length is one-half the radius of curvature.

Question 42

What is the focal length of a diverging mirror?

Answer 42

The focal length is the distance from the mirror to the focal point. Virtual extensions of rays parallel to the principal axis cross at the focal point after reflecting off the mirror.

The focal length of any spherical optic is one-half its radius of curvature.

Question 43

An optical system consists of an object and a single diverging mirror. What are the properties of the final image?

Answer 43

The image is closer to the optic than the object is, and the image is virtual, reduced, and upright.

Virtual extensions of the parallel and center rays cross behind the mirror after reflection; this is where the image is located. Since only the virtual rays cross after reflection, the image is virtual.

Question 44

For an optical system made up of a single mirror, if the object distance is o = 30 cm and the image distance is i = -15 cm, what type is the mirror?

Answer 44

f = -30 cm, and the mirror is diverging.

Given i and o, f can be calculated as:
1/f = 1/o + 1/i = 1/30 + 1/(-15) = -1/30
f = -30 cm

Question 45

What makes a lens converging?

Answer 45

A lens is converging if any inbound light ray which is parallel to the optical axis crosses the axis after refracting through the lens.

Converging lenses are convex in shape.

Question 46

What is the focal length of a converging lens?

Answer 46

The focal length is the distance from the lens to the focal point. Rays parallel to the principal axis cross at the focal point after refracting through the lens.

Question 47

An optical system consists of an object and a single converging lens. If the object is a distance between f and 2f from the optic, where are the properties of the final image?

Answer 47

The image is further from the optic than the object is, and the image is real, magnified, and inverted.

Find where the parallel, focal, and center rays cross after refracting through the lens to locate the image. Since the rays cross after refraction, the image is real.

Question 48

An optical system consists of an object and a single converging lens. If the object is at the focal length f, what are the properties of the final image?

Answer 48

No image is formed.

The parallel and center rays are parallel to one another after refracting through the lens. Since the rays never cross after refraction, no image is formed.

Question 49

An optical system consists of an object and a single converging lens. If the object is inside the focal length f, what are the properties of the final image?

Answer 49

The image is further from the optic than the object is, and the image is virtual, magnified, and upright.

The parallel and center rays are diverging after refracting through the lens. To locate the image, the rays much be projected behind the lens to the point where they cross. Since only the virtual rays cross, the image is virtual.

Question 50

For an optical system made up of a single lens, if the object distance is o = 20 cm. and the image distance is i = 20 cm, what type is the lens?

Answer 50

f = 10 cm, and the lens is converging.

Given i and o, f can be calculated as:
1/f = 1/o + 1/i = 1/20 + 1/20 = 1/10
f = 10 cm

Question 51

What makes a lens diverging?

Answer 51

A lens is diverging if any inbound light ray which is parallel to the principal axis points away from the axis after refracting through the lens.

Diverging lenses are concave in shape.

Question 52

What is the focal length of a diverging lens?

Answer 52

The focal length is the distance from the lens to the focal point. Virtual extensions of rays parallel to the principal axis cross at the focal point after refracting through the mirror.

Question 53

An optical system consists of an object and a single diverging lens. What are the properties of the final image?

Answer 53

The image is closer to the optic than the object is, and the image is virtual, reduced, and upright.

Virtual extensions of the parallel and center rays cross behind the lens after refraction; this is where the image is located. Since only the virtual rays cross after refraction, the image is virtual.

Question 54

For an optical system made up of a single lens, if the object distance is o = 30 cm. and the image distance is i = -15 cm, what type is the lens?

Answer 54

f = -30 cm, and the lens is diverging.

Given i and o, f can be calculated as:
1/f = 1/o + 1/i = 1/30 + 1/(-15) = -1/30
f = -30 cm

Question 55

Define:

a diopter

Answer 55

A diopter is a unit of strength for an optical element.

It is calculated by taking the reciprocal of the focal length: d = 1/f
Focal lengths are typically reported in m, so 1 diopter is 1 m^-1.

Question 56

Which lens bends light more severely, a 2 diopter lens or an 8 diopter lens?

Answer 56

The 8 diopter lens bends light more.

The shorter a lens’ focal length, the more quickly light converges on the optical axis after passing through the lens. The focal length of the 2 diopter lens is 1/2 m, while that of the 8 diopter lens is 1/8 m. The 8 diopter lens, with its shorter focal length, bends light more.

Question 57

If a 2 diopter lens is combined with a 3 diopter lens, what is the approximate strength of the combined lens system?

Answer 57

The combined lens system has a strength of approximately 5 diopters.

A convenience of the diopter system is that it is roughly additive; when multiple elements are combined, their overall diopter rating is roughly the sum of the strength of each individual element.

Question 58

Define:

chromatic aberration of a lens

Answer 58

Chromatic aberration is the characteristic colored halo around an image created by a lens.

This occurs due to dispersion. Different colors of light will focus at different points, so an image created by a lens of an object illuminated by white light will appear blurred.

Question 59

Define:

spherical aberration of an optic

Answer 59

Spherical aberration is the characteristic blurriness of an image created by a real spherical lens.

Spherical optics focus light rays well, so long as the light rays are near the center of the optic. Light rays near the optic’s edge, however, will not focus at exactly the same point, so images including light rays from the optic’s edge will appear blurred.

Question 60

How can the overall magnification of a system of multiple optics be calculated?

Answer 60

The overall magnification of a system of multiple optics is simply the product of the magnification of each of the individual optics.

Question 61

The object of an optical system with two lenses is 15 cm tall. If the magnification of the lenses is m₁ = +2 and m₂ = +1/3, respectively, what is the overall height of the final image?

Answer 61

The final image is 10 cm tall

The overall magnification of the system is the product of the two magnification.

m_tot = m₁m₂ = (2) (1/3) = 2/3

So the final image is upright, and 2/3 the height of the original object; 10 cm overall.

Question 62

If a microscope has a 10x eyepiece and a 40x objective, how large will a 10 µm object appear?

Answer 62

The object will appear to be 4 mm.

The overall magnification is (10) (40) = 400x, so the final image size is:
10x10^-6 * 400 = 4x10^-3 m