chapter 9

Question 1

light is what kind of wave?

Answer 1

• electromagnetic wave
  
  • are transverse waves that have the ability to propagate through vacuum, as well as through other media such as air and water.
  • has electrical and magnetic components, with amplitude perpendicular to each other and to the direction that the wave is propogating in

Question 2

naturally occuring light is generally randomly what?

Answer 2

• polarized
  
  • this means that the electric and magnetic fields associated with light waves can be pointing in any direction perpendicular to velocity and perpendicular to each other

Question 3

light can undergo?

Answer 3

• circular polarization
  
  • this describes a state of polarization in which the direction of the electric and magnetic fields rotates steadily over time (right-handed/clockwise or left-handed/counter-clockwise)

Question 4

the face that circular polarized light can be either clockwise or counterclockwise has been utilized to develop a form of spectroscopy known as?

Answer 4

• optical dichroism
  
  • in which the tendency of a molecule to differently absorb clockwise and counterclockwise polarized light is utilized to obtain information about its chirality

Question 5

electromagnetic waves propagate through vacuum at the speed of light which is?

Answer 5

• 3.00 x 10⁸m/s
  
  • fastest speed possible for all forms of conventional matter in the universe

Question 6

the electromagnetic spectrum and what to know:

Answer 6

• the frequency and wavelength are inversely related
• many “things” that we don’t consider light fall into this spectrum
• waves with a high-frequency and short wavelnegth are higher energy so waves with a short-frequency and long wavelength are low energy
• the visible light spectrum is often described in terms of wavelnegth with units of nanometers
• violet light is on the low end of the spectrum (in terms of wavelength) and that red light is on the high end of the spectrum (violet, blue, green, yellow, orange, red) ROYGBIV
• an absorbance peak for a given wavelength of light means that the substance will not appear to be of that colour

Question 7

the equations c = lamda/f and E = hf can be combined to give an expression for energy in terms of wavelength:

Answer 7

• E = hf = hf/lamda

Question 8

what is reflection and refraction?

Answer 8

• both refer to what happens when a wave travelling through one medium encounters another medium
  
  • there are two possibilities: either the wave bounces off the new medium (reflection) or it continues to travel into the new medium, but along a slightly different path (refraction)
    
    • reflection: when a light wave bouncess off a reflective surface, it is reflected symmetrically so the angle of incidence and the angle of reflection are identical but opposite
    • refraction: occurs when a wave travels from one medium to another (changes speed) so we use the refractive index (n) to calculate new speed
      
      • n = c/v_material
        
        • will always be greater than 1, since nothing is faster than c and refractive index of air can be assumed to be 1 (negligible)

Question 9

we define all angles with reference to the “normal line” which is?

Answer 9

• a line perpendicular to the optical interface

Question 10

when light passes from one medium to another and changes speed, it bends, A law known as? relates the refractive index- that is, how much the speed of a light wave changes- to how much it bends

Answer 10

• snell’s law
  
  • n₁sin(theta₁) = n₂sin(theta₂)
    
    • the angle of the light ray to the normal will increase if the light wave is speeding up as it goes from one medium to another and that the angle to the normal will decrease if the light wave is slowing down

Question 11

an important special case occurs when light is moving into a medium with a smaller index of refraction (that is, when n₂ > n₁, or when light is speeding up as it moves into the new medium)

• ex. when light is moving from water to air
  
  • as this happens, Snell;s law tells us that the angle with the normal will increase so the ray of light will be bent further away from the normal. As the angle of the incident ray (theta₁) increases, there will come a point where the anfle of the refracted ray (theta₂) reaches 90°. this is known as the critical anfle

Question 12

As we go beyond the critical angle, the light can no longer refract at all. Instead, all of the light rate are reflected within the original medium. this is known as?

Answer 12

• total internal reflection

Question 13

Another important phenomenon related to refraction is exemplified by how a prism can break up light into its component wavelengths of different colours. this phenomenon is known as?

Answer 13

• dispersion
  
  • the reasion why this takes place is because the speed of light in a non-vacuum medium varies depending on wavelength
  • the speed of a light through a material is related to its wavelength. light with a relatively long wavelength (red light) is not slowed down as much when it enters a prism as light with a relatively short wavelength (purple light) so since it is slowed down more it ebnds more

Question 14

what is diffraction?

Answer 14

• takes place when a wave encounters an obstacle or an apertures (involves a scenario in which a wave hits a barrier that has a small opened. The waves that hit the barrier are simply reflected back, but those that go through the openining don’t simply travel straight though it. Instead, they expand outwards (diffraction)

Question 15

what is single-slit diffraction?

Answer 15

• characterized by a massive intensity peak at the center of the diffraction pattern, followed by alternating areas of gradually weaker intensity peaks and areas of darkness as you move up or down from the center
  
  • most useful to characterize single-slit diffraction patterns in terms of locating the minima, or areas that exhibit destructive interference. this equation gives the location of the mth minimum for light waves with wavelength lamda in a single-slit setup with an aperture of length A:
    
    • Asin(theta) = m(lamda)
      
      • A and sin(theta) are inversely related so a wider aperture will produce narrower, more closely separated areas of intensity anf vice versa

Question 16

diffraction setups are not limited to single slits. another example of diffraction patterns occurs in a double-slit arrangement:

Answer 16

• we assume that the width of each of the 2 apertures is negligible compared to the distance bewteen them (D)
• we assume that the optical screen is separated from the screen with 2 slits by a much greater distance than D
  
  • double-slit diffraction is characterized by a much more even distribution of minima and maxima than single slit.
  • The formula for the nth maximum is given as follows: Dsin(theta) = n(lamda)
    
    • this equation implies that the maxima occur at whole-number multiplies of wavelength so the minima occur at half-wavelgnths: Dsin(theta) = (n + 1/2)•lamda

Question 17

Combining thousands (or more) tiny slits into a small area results in?

Answer 17

• a diffraction grating

Question 18

what is thin-film interference?

Answer 18

• It occurs when light waves reflect off both the top and bottom boundaries of a substance that forms a thin film

Question 19

what are mirrors?

Answer 19

• substances from which light rays only reflect, without any significant absorbance or refraction

Question 20

there are 3 geometrical types of mirrors we need to know:

Answer 20

• plane (linear)
• convex
• concave

Question 21

what is the first principle of geometric optics?

Answer 21

• visible objects can be treated as sources of light waves
  
  • when light rays emanating from an object are reflected from a mirror, we can perceive them as an image. A real image is formed in the plane in which the reflected light waves converge again, and a virtual image is formed when the reflected light waves don’t actually converge in a phyiscal plane, but our perceptual apparatus reconstructs an image based on where it appears they were coming from

Question 22

What is a plane mirror?

Answer 22

• a ray of light that hits a plane mirror perpendicularly will be bounced back in the same direction, and for rays of light that hit the plant mirror on an angle, they will reflect at the same but opposite angle to the nromal

Question 23

curved mirrors come in 2 forms: concave and convex

Answer 23

• convex mirrors scatter incident light rays outwards and concave mirrors cause incident light rays to converge on each other

Question 24

plane mirrors and convex mirrors must?

Answer 24

• form virtual images
  
  • shows that parallel light rays that hit a plane or convex mirror will never intersect with each other after they’ve reflected. Instead, when we perceive the parallel or diverging light rays reflected by plane and convex mirrors, respectively, our brains “work backward” to determine what their source would have been, and this is what we perceive as a virtual image

Question 25

concave mirrors are capable of forming?

Answer 25

• real images, because they cause physically real light waves to converge on each other

Question 26

• We can visualize curved mirrors as being pieces of the circumference of some imagined shere. that sphere has a radius (r) and we can define a point known as the focal point (f)

Answer 26

• Imagine parallel rays hitting a curved mirror. The point at which the reflection of these parallel incident rays converge is defined as the focal point
  
  • in a concave mirror, the focal point is on the same side as the object, whereas in a convex mirror, the focal point is where the virtual rays would meet on the other side of the mirror.
  • Interestingly, the focal point occurs at half of the radius: f = r/2

Question 27

Plane mirrors can be considered as spherical mirrors with an infinte radius, meaining that their focal point is likewise defined as extending out to infinity and teherefore not existing to any meaningful extent.

Answer 27

• the distance if an object (o) is is the location of an object relative to the mirror and the distance of the image (i) is where the image of that object is formed
  
  • the image is not generally formed at the focal point

Question 28

There are 2 ways that you can determine the location of an image:

Answer 28

• by using an equation or by tracing ray lines

Question 29

An equation known as the thin lens equation is our master equation for optics:

Answer 29

• relates f, o and i as follows:
  
  • 1/o + 1/i = 1/f
    
    • the sign conventions are straightforward when applying this to mirrors: positive values correspond to being in front of the shiny part of the mirror, and negative values correspond to being behind the mirror. Therefore concave mirros have a focal length of f > 0 and convex mirros have a focal length of f < 0

Question 30

Magnification can be defined in terms of i and o as follows:

Answer 30

m = - i/o

Question 31

for ray diagrams:

Answer 31

• we want to trace 2 specific rays in most cases. One ray extends parallel from the object, and when it is reflected, will travel through the focal point of the mirror (real or virtual manner). The second ray is drawn from the top of the object, but at an angle such that either it intersects the focal point (or, if the object is closer to the mirror than the focal point, such that it can be extended backwards to the focal point). this second rau will be reflected from the mirror in a line parallel to the nromal
  
  • having drawn these rays, the image can be found very simply: it will be wherever these 2 rays intersect, and they will intersect at a point that tells you how large the image is

Question 32

For a plane mirror, the focal distance is defined as infinity, and we can approximaye 1/infinite as 0. Therefore we can apply the thin lens equation as follows:

Answer 32

• 1/0 + 1/i = 1/f → 1/o + 1/i + 1/infinte → 1/o + 1/i = 0
  
  • this means that o = -i. Substituting this into the magnification equation, we get:
    
    • m = -i/o = -(i/-i) = 1
      
      • this just confirms that the image formed by a plane mirror is virtual, upright, at the same distance as the object, and does not show any magnification

Question 33

for convex mirrors, for which f < 0. we can rearrange the thin les equation to solve for where the image will be:

Answer 33

• 1/i = 1/f - 1/o
  
  • dealing with objects placved in front of mirrors so i < 0 meaning that the image will be virtua;
    
    • for the magnification equation i < 0 means -i so we have m = -(-1/o) or + 1/o so magnification will be positive so it will be upright

Question 34

for concave mirros, for which f > 0. we can rewrie the thin lens equation to solve for i:

Answer 34

• 1/i = 1/f - 1/o
  
  • o > 0 (the object is infront of the mirror). Since f > 0 and o > 0, solving for i involves subtracint a positive number from another positive numer.
    
    • 3 cases that we need to analyse:
      
      • o > f menas that the object is more distant than the focal point so i is positive and that we have a real image. the magnification formula will yeild a negative sign indicating the image is inverted
      • o = f means that the object is placed at the focal point. i would be considered infinite so no image is formed. the magnification formula would yield a meaningless value here as well
      • and o < f this means that the object is closer to the mirror than the focal point which is less than 0 so i is negative indicating the image is now virtual and the magnification formula will show that the image is upright

Question 35

real and virual images summary:

Answer 35

• | real images| virtual images|

Formed by: real rays | back-traced rays

Located: in front of the mirror | behind the mirror

Orientation: inverted | upright

Sign of i: positive | negative

type of mirror: concave, only if f > o | plane, convex or concave if o < f

Question 36

what is the basic difference between lenses and mirrors?

Answer 36

• light passes through lenses, which are typically made of glass. light is refracted both upon entering a lens and upon exiting a lens, and lenses differ in terms of whether they cause refracted rays to converge or diverge
  
  • convex lenses cause rays to converge
  • concave lenses cause them to diverge

Question 37

we can define the focal length in the same way for lenses and apply the thin lens equation

Answer 37

• 1/o + 1/i = 1/f
  
  • if we do need to take into account for the thickness of the lens, the index of refraction of the elns becomes important. We use the lensmaker’s equation to allow for the focal length (f) to be determined in terms of index of refraxtion of the lens (n) and the radii of each face of the lens (r₁ and r₂):
    
    • 1/f = (n - 1) (1/r₁ - 1/r₂)

Question 38

Understanding the sign conventions of how the thin lens equation is applied to lenses is very important. The basic commonality in the sign conventions as applied to mirrors and lenses is that?

Answer 38

• real images are formed by real light rays, and that we use positive signs to refer to real images
  
  • the main difference is that light rays pass through lenses but are reflected by mirrors, meaning that real images will be on the other side of the lens from the sourve of light, while virtual imahes will be on the same side. Based on this convention, f will be positive for convex (converging) lenses, while it will be negative for concave (diverging lenses). this is also a point of similarity between the two

Question 39

Since lenses are used ofr a wide range of practical applications, it’s helpful to have a measure of how “powerful” a lens is, or how much it bends light. The power of a lens (P) is defined as:

Answer 39

• the inverse of the focal length in meters
  
  • P = 1/f
    
    • the units of power are diopters (m^-1)
      
      • focal length must be in meters

Question 40

diopeters follow the same sign conventions as f so:

Answer 40

• a positive value corresponds to a converging lens and a negative value to a diverging lens
  
  • a lens with a very small focal length by definition must bend light considerably so it will have a large value in diopters, which directly indicates that the lens is powerful

Question 41

Rays may not meet up prefectly after refracting through a spherical lens or being reflected by a spherical mirror. this is known as?

Answer 41

• spherical aberration because it is characterized in a difference in the refraction parrerns of light rays that hit close to the cneter of a spherical lens compared to those that hit on the edges

Question 42

It is also possible to build systems consisting of more than one lens.

• In such systems, the image generated by one lens serves as the object of another lens, whcih in turn henerates a second image. the total magnification of such systems is calculated by?

Answer 42

• multiplying the individual magnifications of each lens

Question 43

light is first refracted by?

Answer 43

• the cornea
  
  • acts as a lens with a fixed focal length; that is it refracts light considerably, but its activity as a lens cannot be modified
    
    • the lens of the eye is located a bit more deeply. works to ensure that imahes are focuse donto the retina to be processed by the NS

Question 44

In many peopoe, the focusin system of the lens does not work completely correctly.

• in near sighted indiciduals (myopia), the lens refracts light too much, making the image focus in front of the retina. In this condition, individuals can see near objects clearly, but not far objects. How is it corrected?
• Farsightedness is the opposite condition, in which the lens does not bend light sufficiently, meaning that the image would be formed behind the retina. In this condition, an individual can see far-off objects clearly, but near objects may seem blurry. how is it treated?

Answer 44

• Near-sightedness (myopia) is corrected by using a diverging lens, in the form of glasses or a contact lens. to spread out the light waves just enough to hit the retina correctly
• farsightedness is treated by using a converging lens that makes the light rays converge closer, on the restina itself