Chapter 8 Light & Optics

Question 1

Electromagnetic waves are transverse waves because

Answer 1

• 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

Question 2

Electromagnetic spectrum from lowest frequency to highest

Answer 2

-Radiowaves, microwaves, infrared, visible light, ultraviolet, and x ray

Question 3

Speed of Light

Answer 3

-Electromagnetic waves vary in frequency and wavelength, but in a vacuum & in air all electromagnetic waves travel at the same speed

Question 4

Speed of Light Equation

Answer 4

c=fλ
c= speed of light in a vacuum/air
f=frequency
λ=wavelength

Question 5

Visible Region

Answer 5

• only part of the spectrum that is perceived as light by the human eye
• different wavelengths perceived as different colors
• violet at one end (400nm) and red at the other (700nm)
• Light that contains all the colors in equal intensity is perceived as white
• Object that appears red is one that absorbs all the colors of light except red

Question 6

Blackbody

Answer 6

• refers to an ideal absorber of all wavelengths of light

- would appear completely black if it were at a lower temperature than its surroundings

Question 7

Rectilinear Propagation

Answer 7

-When light travels through a homogeneous medium, it travels in a straight line

Question 8

Theory of Geometrical Optics

Answer 8

• The behavior of light at the boundary of a medium or interface between two media
• Explains reflection and refraction

Question 9

Reflection

Answer 9

The rebounding of incident light waves at the boundary of a medium
-light waves that are reflected are not absorbed into the second medium; they bounce off the boundary and travel back through the first medium

Question 10

Real vs Virtual images created by a mirror

Answer 10

• Real if the light converges at the position of the image
• Virtual if the light only appears to be coming from the position of the image but doesn’t actually converge there
• Distinguishing feature of real images: ability of the image to be projected onto a screen

Question 11

Plane mirrors

Answer 11

• flat, reflective surfaces cause neither convergence nor divergence of reflected light rays
• since the light doesn’t converge, plane mirrors will always create virtual images
• Create the appearance of light rays originating behind the mirrored surface

Question 12

Spherical mirrors

Answer 12

• Center of curvature: a point on the optical axis located at a distance equal to the radius of curvature from the vertex of the mirror
• Concave(converging) mirror: center of curvature and radius of curvature are located in front of the mirror
• For all spherical mirrors: f=r/s (r=radius of curvature, the distance between C and the mirror)

Question 13

Focal Length (f)

Answer 13

The distance between the focal point (F) and the mirror

-

Question 14

Image distance equation

Answer 14

1/f=1/o+1/i=2/r
f=focal length
o=distance between object and mirror
i=distance between image and mirror
r=radius of curvature
-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 located behind the mirror
-plane mirrors: r=f=infinity and the equation becomes 1/o+1/i=o

Question 15

magnification (m) and equation

Answer 15

-Dimensionless value that is the ratio of the image distance to the object distance
m=-i/o
-Also gives the ratio of the size of the image to the size of the object
-negative magnification=inverted image
-positive magnification=upright image
-|m|1 image is larger than object. |m|=1 image is the same size as the object

Question 16

Refraction

Answer 16

• The bending of light as it passes from one medium to another and changes speed
• Speed of light through any medium is always less than its speed through a vacuum

Question 17

Numerical value for the speed of light through a vacuum/air

Answer 17

3.00x10^8 m/s

Question 18

Snell’s Law Equation

Answer 18

n=c/v
-c=speed of light in a vacuum (3.00 x10^8 m/s)
-v=speed of light in a medium
-n=dimensionless quantity called index of refraction of the medium
index of refraction of a vacuum/air=1 and all other materials will be greater than 1

Question 19

Snell’s Law

Answer 19

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

Question 20

Snell’s Law equation for when light is passing from one medium to another

Answer 20

n1sinθ1=n2sinθ2

-n1 and θ1 refer to the medium where the light is coming and n2 θ2 refer to the medium where the light is entering

Question 21

Critical Angle

Answer 21

• Refracted angle θ2 equals 90 degrees.
• when light travels from a medium with a higher index of refraction to a medium with a lower index of refraction (like water to air) the refracted angle is larger than the incident angle and the refracted light ray bends away from the normal

Question 22

Critical Angle Equation

Answer 22

θc=sin^-1(n2/n1)

-θ2=90 degrees

Question 23

Total internal reflection

Answer 23

-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 (greater than 90 degrees)

Question 24

Lenses

Answer 24

• refract light while mirrors reflect it.

- has two focal points with one on each side

Question 25

Lensmaker's equation

Answer 25

- used where thickness cannot be neglected - focal length in relation to the curvature of the lens surfaces - 1/f=(n-1)(1/r1-1/r2) - n=the index of refraction of the lens material - r1=the radius of curvature of the first lens surface - r2=the radius of the curvature of the second lens

Question 26

Lens power equation

Answer 26

P=1/f | -P has the same sign as f, therefore positive for a converging lens and negative for a diverging lens

Question 27

Hyperopia

Answer 27

Causes convergence of light to correct farsightedness

Question 28

Myopia

Answer 28

Causes divergence of light to correct nearsightedness

Question 29

Spherical aberration

Answer 29

- 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 - Parallel rays are not perfectly reflected or refracted through the focal point, leading to blurriness at the periphery of the image

Question 30

Dispersion

Answer 30

When various wavelengths of light separate from each other | -most common example: splitting of white light into its component colors using a prism (rainbow)

Question 31

Chromatic aberration

Answer 31

Dispersive effect within a spherical lens

Question 32

Diffraction

Answer 32

The spreading out of light as it passes through a narrow opening or around an obstacle -light that spreads out diffracts

Question 33

Location of dark fringes Equation

Answer 33

asinθ=nλ a=width of slit θ=angle between the line drawn from the center of the lens to the dark fringe and the axis of the lens n=integer indicating the number of the fringe λ=wavelength of the incident wave

Question 34

Interference

Answer 34

- The displacements of waves add together when waves interact with each other - Regions of constructive interference between the two light waves appear as bright fringes (maxima), regions where light waves interfere destructively appear as dark fringes (minima)

Question 35

Appearance of dark fringes equation

Answer 35

dsinθ=(n+1/2)λ d=distance between the two slits θ=angle between the line drawn from the midpoint between the two slits to the dark fringe n=integer indicating the number of the fringe λ=wavelength of the incident wave

Question 36

Diffraction gratings

Answer 36

- Consist of multiple slits arranged in patterns - can create colorful patterns similar to a prism as the different wavelengths interfere in characteristic patterns (like organization of the grooves on a CD act like a diffraction grating, creating an iridescent rainbow pattern on the surface of the disk

Question 37

x-ray diffraction

Answer 37

Uses the bending of light rays to create a model of molecules -often combined with protein crystallography during protein analysis, dark and light fringes do not take on a linear appearance, but a complex two dimensional image

Question 38

Plane-polarized light

Answer 38

Light in which the electric fields of all the waves are oriented in the same direction (electric field vectors are parallel) - unpolarized light has a random orientation of its electric field vectors (sunlight and light bulbs) - Plane polarized: stereoisomers, 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