Unit 3 CC & PQ Questions
If a ray of light goes from one medium to another along the normal, it is ....
... undeflected regardless of n
3 properties of image formed by plane mirror
1. Upright image but reversed left to right 2. Same size as object 3. Image distance = object distance
Light reflects from low to high n.... (phase shift?)
1/2 wavelength phase shift upon reflection
PQ 26.4 Suppose that a 21-year-old with normal vision (near point = 25 cm) is standing in front of a plane mirror. How close can he stand to the mirror and still see himself in focus?
12.5 cm Since it is a plane mirror, the object distance will equal the image distance. The image forms at an equal distance behind the mirror. Therefore, it is a virtual image. The person can stand at a distance of half their near point from the mirror, and the other half of the distance is located behind the mirror.
Rayleigh criterion
2 point objects are just barely resolve when 1st dark fringe in diffraction pattern falls directly on the central bright fringe of the other pattern
Ch. 25 CC2 You hold a small hand mirror 0.5 m in front of your face and look at your reflection in full-length mirror 1 m behind you. How far in back of the big mirror do you see the image of your face? SEE PICTURE
2m behind mirror Image from small mirror becomes object for big mirror
PQ 26.2 A diamond (n = 2.42) is lying on a table. At what angle of incidence θ is the light that is reflected from one of the facets of the diamond completely polarized?
67.55 degrees Use Brewster's angle equation because reflected light is completely polarized tan thetaB = 2.42/1
Concave mirror
A mirror that curves inward Real image f = 1/2R
Convex mirror
A mirror that curves outward Virtual image f = -1/2R
Ch. 26 CC3 To shoot a fish with a laser gun, should you aim directly at the image, slightly above, or slightly below? n air = 1, n water = 1.33
Directly at image DRAW A PICTURE Laser is a light wave and will be refracted along same path as light leaving fish and being refracted at water surface into eyes Laser beam = from low to high n = ray bent toward the normal = follows same path as refracted ray which is opposite situation
Ch. 24 CC5 If the magnetic field in an electromagnetic field is doubled, by what factor does the electric field change?
Double E = cB If B is 2x, so is E
Huygens-Fresnel Principle
Every point on a wave front acts as secondary wavelet that spreads out in all directions (acts as point source) with speed equal to the propagation speed of the wave Also, occurs with an obstruction = light hits sides of obstruction and diffracts and this leads to a path difference
Huygen's Principle
Every point on any wave front can be regarded as a new point source of secondary waves that emits uniformly in all directions
_______ of a wave (light) does NOT change when passing from one material to another However, ____ and _____ do change when passing from one material to another
Frequency does NOT change Speed and wavelength do change
The greater the change in n, the ______ change in the direction of the ray
Greater the change in n, greater the change in the direction of the ray = more refraction
Farsightedness
Hyperopic Eye too short or cornea not curved enough Image forms behind retina Corrected with converging lens = near point too far away and this type of lens casts virtual image at near point so can see close objects better
Critical angle & Total Internal Reflection
If incident angle large enough, refracted angle = 90 degrees = critical angle If incident angle is larger than the critical angle, all of the light is reflected at the interface = total internal reflection n1 has to be > n2 for critical angle
Lenses in Combination
Image produced by first lens serves as object for second lens Total magnification = product of all individual magnifications
Ray
Imaginary line along direction of travel of the wave
Constructive interference
Integer shifts of wavelength In phase Add amplitudes (2x) *Can also have partial
Object located between C and F for concave mirror
Inverted, real, larger image
Object located beyond C for concave mirror
Inverted, real, smaller image
Ch. 26 CC4 You are looking at an object (at an angle other than normal incidence) through a pain of glass as shown. Does the object appear to the left or right of its actual location? n glass > n air
Left DRAW A PICTURE From air to glass = low to high n = ray bent toward the normal From glass to air = high to low n = ray bent away from the normal Trace ray that emerges from glass back to spot of object = appears to be to the left of object
Young's Double Slit
Light is wave because of interference pattern that exists on the screen
Human Eye
Many interfaces where refraction occurs = biggest change in n occurs at air-cornea interface Image formed on retina = real, inverted, smaller Image distance fixed Eye changes lens shape to change focal length so can see objects at various distances
Interference is most noticeable when light is.... (2)
Monochromatic = same wavelength Coherent = different sources maintain same phase relationship over space and time
Ch. 27 CC5 In a double-slit experiment, what path difference have the waves from each slit traveled to give a minimum at the indicated position? SEE PICTURE
More than 3 wavelengths 3.5 wavelength path difference
Nearsightedness
Myopic Eye too long or cornea too sharply curved Image forms in front of retina Corrected with diverging lens = far point is too close and this type of lens casts virtual image at far point so can see far objects better
Ch. 24 CC3 At a certain point in space, the electric and magnetic fields of an electromagnetic wave at a certain instant are given by: E = 6000 V/m x hat B = 2e-5 T z hat This wave is propagating in the
Negative y-direction Using x (right), y (up), and z (out) coordinate system, Fingers = E = right Palm = B = out of page Thumb = direction of travel = downward or -y direction
Ch. 27 CC2 Two sources S1 and S2 emit coherent, monochromatic light. Point P is 7.3 wavelengths from source S1 and 4.6 wavelengths from source S2. As a result, at point P there is
Neither constructive nor destructive interference. Path difference is 2.7 wavelengths so can't be total constructive or total destructive interference Would be partial destructive interference because closer to half integer multiple of wavelength
PQ 24.3 The drawing shows two sheets of polarizing material, which have their transmission axes crossed. The sheet on the left has its transmission axis aligned vertically. Light is incident from the left, as shown and can be either completely unpolarized or completely polarized along the vertical direction. Which one of the following statements is true concerning the light that exits the polarizer on the right? SEE PICTURE
No light exits the polarizer on the right for either the unpolarized or the polarized incident light. Correct. When the unpolarized light strikes the first polarizer, the light that passes through it is polarized in the vertical direction. When this polarized light strikes the second polarizer, all of it is absorbed, since the two polarizers are crossed. When the polarized light strikes the first polarizer, all of it passes through, since the direction of polarization and the transmission axis are both vertical. When this polarized light strikes the second polarizer, all of it is absorbed, since the two polarizers are crossed. Thus, no light exits the polarizer on the right in either case.
Light reflects from high to low n... (phase shift?)
No phase change upon reflection
Malus's Law Question Two polarizers are oriented such that their transmission axes are perpendicular. A third polarizer is placed between them with its transmission axis at an angle of 45° relative the other polarizers. Unpolarized light is incident on polarizer #1. The amount of light that passes through polarizer #3 will be
Non-zero Horizontal component of light leaving 2nd polarizer will pass through 3rd polarizer
Destructive interference
Odd integer shifts of wavelength Out of phase Cancels or partially cancels each other out *Can also have partial
Ch. 24 CC8 If unpolarized light is incident from the left, in which case will some light get through? SEE PICTURE
Only Case 2 Can't have crossed axes back to back because doesn't let any light through
PQ 26.2 A mixture of red and violet light is incident on a glass-air interface, as the drawing shows. The indices of refraction for red and violet light in glass are nred = 1.520 and nviolet = 1.538. What color(s) of light enters the air? 40.85 degree angle SEE PICTURE
Only Red Correct. The critical angles for the glass-air interface are θc, red = sin−1 (1.000 / 1.520) = 41.14°and θc, violet = sin−1 (1.000 / 1.538) = 40.56°, which is greater than the critical angle for violet light, but less than the critical angle for red light. Therefore, only red light enters the air, the violet light being totally internally reflected. Critical angle for red = 40.56 degrees is less than actual angle of 40.85 degrees so is able to pass into the air Critical angle for violet = 41.14 degrees is greater than actual angle of 40.85 degrees so is totally internally reflected When critical angle > 40.85 degrees = no refraction = total internal reflection
Lens
Optical system with 2 refracting surfaces = thin lens with negligible thickness
Near point
Point nearest the eye where object is placed and still produces sharp image 25 cm for college aged person = unaided eye
PQ 25.1 Which statement regarding real and virtual images is correct? A virtual image is always inverted relative to the object, whereas a real image is never inverted. Rays of light emanate from a virtual image, but only appear to do so from a real image. A virtual image is always larger than the object, whereas a real image is always smaller. Rays of light emanate from a real image, but only appear to do so from a virtual image. A virtual image is always smaller than the object, whereas a real image is always larger.
Rays of light emanate from a real image, but only appear to do so from a virtual image.
Object placed between f and 2f from converging lens
Real, inverted, larger image (projector)
Object placed more than 2f away from converging lens
Real, inverted, smaller image (camera)
Diffuse deflection
Rough surface, reflected rays scattered randomly = vast majority of objects
Ch. 27 CC4 In a double-slit experiment, if the slits are moved further apart, the interference pattern
Shrinks together sin theta = (m * wavelength) / d As d goes up, sin theta goes down
Ch. 26 CC2 To shoot a fish with a gun, should you aim directly at the image, slightly above, or slightly below? n air = 1, n water = 1.33
Slightly below it DRAW A PICTURE Light from fish is refracted at water surface and enters eye Water to air = high to low n = ray bent away from normal Trace refracted ray back in water to see where image of fish appears to be Fish is actually below this image
What slit width is best for interference pattern?
Slit width close to the wavelength of light being used
Specular deflection
Smooth surface, reflected rays parallel
When ray enters medium where n is increased (low to high n), the speed of light _____ and the ray bends ______ the normal
Speed of light decreases Ray bends toward the normal
When ray enters medium where n is decreased (high to low n), the speed of light ____ and the ray bends _____ the normal
Speed of light increases Ray bends away from the normal
Ch. 27 CC3 In a double-slit experiment, when the wavelength of the light is increased, the interference pattern
Spreads out d sin theta = m * wavelength As wavelength goes up, sin theta goes up If infinitely large screen, paths nearly parallel to slit screen = sin approaches 90 degrees
Principal axis of spherical mirror
Straight line drawn through center of curvature and midpoint of mirror
Ch. 24 CC7 If the area, A, of the beam of light is doubled, by what factor does the intensity change?
Unchanged S = c * permittivity of free space * E^2 = c / permeability of free space (u0) * B^2 No A in intensity equation so A unchanged DON'T USE S = PA
Object located inside F for concave mirror
Upright, virtual, larger image
Any object with convex mirrors
Upright, virtual, smaller image
Ch. 24 CC1 If a radio transmitter has a vertical antenna, should a receiver's antenna be vertical or horizontal to obtain the best reception?
Vertical Electrons have entire rod to oscillate = more current and better signal
Object placed f or closer to converging lens
Virtual, upright, large image (magnifying glass)
Diverging lens always forms a _____ image
Virtual, upright, smaller image
Diffraction
Wave nature of light causes interference patterns that blurs shadow and creates fringe pattern around straight outline Waves travel in all in directions after passing through opening EX: sound waves traveling around corners and other barriers
Ch. 24 CC4 In a sinusoidal, traveling electromagnetic wave in a vacuum, the magnetic energy density.. a) is the same at all points in the wave. b) is maximum where the electric field has its greatest value. c) is maximum where the electric field is zero. d) is negative where the electric energy density is positive. e) More than one of the above is true.
b) is maximum where the electric field has its greatest value. E = cB When B is max, so is E Think of oscillating diagram = both at high point at same time
Ch. 27 CC6 A thin layer of turpentine (n = 1.472) lies on top of a sheet of ice (n = 1.309). It is illuminated from above with light whose wavelength in turpentine is 400 nm. Which of the following possible thicknesses of the turpentine layer will maximize the brightness of the reflected light? a) 100 nm b) 200 nm c) 300 nm d) 400 nm e) more than one of the above DRAW PICTURE and WORK OUT LIKE A PROBLEM
e) more than one of the above Ray 1 = low to high n = 1/2 wavelength shift Ray 2 = high to low n = no phase shift One ray shifted "maximize the brightness" = constructive interference Solve for t, plug in values for m until get values listed E --> 100 nm (m=0), 300 nm (m=1)
Mirror Sign Conventions
f + concave mirror f - convex mirror do + object in front of mirror (shiny side) do - object behind of mirror** di + image in front of mirror (shiny side) (real image) di - image behind mirror (virtual image) m + upright image m - inverted image m > 1 image larger than object m < 1 image smaller than object
Lens Sign Conventions
f + converging f - diverging do + object in front of lens do - object behind lens** di + image opposite side as object = real image di - image same side as object = virtual image m + upright image m - inverted image m > 1 image larger than object m < 1 image smaller than object
EX: m for third order dark fringe EX: m for third order bright fringe
m = 2 for third order dark fringe m = 3 for third order bright fringe
PQ 26.1 For the example of refraction illustrated in the drawing θ1 = 53.3° and θ2 = 71.0°. Calculate the ratio n1/n2 of the indices of refraction of the two materials.
n1 / n2 = 1.179 Solve Snell's Law of Refraction for ratio n1 / n2 = sin theta 2 / sin theta 1
Ch. 26 CC1 Parallel light rays cross interfaces from medium 1 into medium 2 and then into medium 3. What can we say about the relative sizes of the index of refraction of these media? SEE PICTURE
n2 > n1 > n3 From 1 to 2 = low to high n because refracted ray is bent toward the normal = then 2 > 1 From 2 to 3 = high to low n because refracted ray is bent away from the normal = then 2 > 1 Greater the change in n, greater the refraction = 3 is refracted more than 1 so greater change = then n3 is lowest value
PQ 26.1 The drawings show two examples in which a ray of light is refracted at the interface between two liquids. In each example the incident ray is in liquid A and strikes the interface at the same angle of incidence. In one case the ray is refracted into liquid B, and in the other it is refracted into liquid C. The dashed lines denote the normals to the interfaces. Rank the indices of refraction of the three liquids in descending order (largest first). SEE PICTURE
nC, nA, nB Correct. When the light is refracted into liquid B it is bent away from the normal, so that nA > nB. When the light is refracted into liquid C it is bent toward the normal, so that nC > nA. Therefore, we conclude that nC >nA > nB.
Ch. 24 CC2 In a vacuum, one color of red light has a wavelength of 700 nm and one color of violet light has a wavelength of 400 nm. This means that in a vacuum, the red light Determine frequency, wavelength, speed
none of above Red light has longer wavelength and lower frequency than violet light All light travels at speed of light
PQ 24.2 An electromagnetic wave is traveling in a vacuum. The magnitudes of the electric and magnetic fields of the wave are _____________, and the electric and magnetic energies carried by the wave are _____________.
proportional (but not equal) to each other, equal Correct. The magnitudes of the electric and magnetic fields of the wave are proportional to each other, according to E = cB. The wave carries equal amounts of electric and magnetic energy.
PQ 25.1 An object is placed at a known distance in front of a mirror whose focal length is also known. You apply the mirror equation and find that the image distance is a negative number. This result tells you that __________
the image is virtual. Correct. A negative image distance means that the image is behind the mirror and, hence, is a virtual image.
Wave front
the locus of all adjacent points on a wave that are in phase
Index of refraction
the ratio of the speed of light in a vacuum to the speed of light in a medium n = c / v
For any wave, v = ____ * _____
v = wavelength * frequency Frequency always the same, v is always less than c, so wavelength is also reduced So, wavelength 2 = ( n1 / n2 ) * wavelength 1
PQ 25.2 Suppose you hold up a small convex mirror in front of your face. Which answer describes the image of your face?
virtual, upright Correct. A convex mirror always produces a virtual, upright image.
Plane wave
waves whose crests form of pattern of straight line wave fronts Far away from source, radii of spheres have become very large, section of sphere can be considered a flat plane
Resolving Power
Diffraction through an aperture = patterns overlap too much and can't resolve differences between 2 objects
Diverging lens
AKA concave Thicker at edges than center Corrects nearsightedness
Converging lens
AKA convex Center thicker than edges Corrects farsightedness
Interference
Any situation in which 2+ waves overlap in space
Diffraction increases as what ratio increases
As (wavelength / width of opening) ratio increases Best when have small width that is close to the wavelength of light being used
Brewster's Law for polarizing angle
At Brewster's angle, reflected beam is completely polarized perpendicular to line of incidence AND the refracted beam is partially polarized parallel to the line of incidence Reflected ray and refracted ray 90 degrees from each other Light that lies parallel to line of incidence is not reflected at all but completely refracted
Bigger the aperture, bigger the _____ _______
Bigger the aperture, bigger the resolving power EX: fly eyes
Bright spots ______ in brightness as get _____ from central bright spot
Bright spots decrease in brightness as get further away from central bright spot
Ch. 25 CC1 You are standing in front of a mirror at the point P shown. There is a light bulb behind a screen that you cannot see directly. As you look in the mirror, where does the image of the light bulb appear? SEE PICTURE
C = where virtual image appears do is from lightbulb to E, di is from E to C Even if walk towards the mirror, still location C because object hasn't moved. But this does change the angle of incident and angle of reflection but still same virtual image location
Ch. 24 CC9 If unpolarized light is incident from the left, and the first polarizer was removed which case(s) would let some light through? SEE PICTURE
Cases 1 and 2 Can't have crossed axes back to back because doesn't let any light through
Accommodation
Ciliary muscles change focal length of lens
Ch. 27 CC1 Two sources S1 and S2 emit coherent, monochromatic light. Point P is 7.3 wavelengths from source S1 and 4.3 wavelengths from source S2. As a result, at point P there is
Constructive interference Path difference is even integer multiple of wavelength
Astigmatism
Cornea surface not spherical, curved more in one area than another Corrected with non-symmetrical lens
PQ 25.1 You hold the words "TOP DOG" in front of a plane mirror. What does the image of these words look like? SEE PICTURE
Correct. Letters and words held up to a mirror are reversed left-to-right and right to-left.
PQ 24.3 The drawing shows two sheets of polarizing material, each of which has its transmission axis aligned vertically. Light is incident from the left, as shown, and can be either completely unpolarized or completely polarized along the vertical direction. In either case, the average intensity of the incident light is the same, and light exits the polarizer on the right. Which one of the following statements is true concerning the intensity of the exiting light? SEE PICTURE
The intensity of the exiting light is greater when the incident light is completely polarized. Correct. When the incident light is completely unpolarized, half of its intensity is absorbed by the polarizer on the left, and half passes through. The half that passes through is completely polarized along the vertical direction, which is the same as the transmission axis of the second polarizer. Thus, the second polarizer absorbs none of the light, and the intensity of the exiting light is half that of the incident light. When the incident light is completely polarized along the vertical direction to begin with, it passes through both sheets of material with none of its intensity being absorbed. In this case the exiting light has the same intensity as the incident light. Thus, the exiting light has a greater intensity when the incident light is polarized.
Far point
The location of the farthest object on which the fully relaxed eye can focus. Infinity with unaided eye
Ch. 24 CC6 If the electric field in the beam of light is doubled, by what factor does the intensity change?
Times four S = c * permittivity of free space * E^2 Square of E = 4x
Ch. 26 CC5 Parallel rays of red light that are directed at a converging lens (n=1.47) are focused at a point P on the principle axis to the right of the lens when the lens is surrounded by air (n=1). If the lens is surrounded by water (n=1.33) instead of air, where will the red parallel rays be focused relative to point P? SEE PICTURE
To right of point P In water, change in n is much lower so refraction is also lower Rays will refract less in water and converge at a point to the right of P
Thin-film interference
Two light waves enter eye = one ray has traveled extra distance (2t) through a film so interference occurs between 2 waves EX: soap bubbles = constructive interference for given thickness results in a color = varying thickness creates rainbow pattern EX: gasoline on water EX: bug coloring due to interference not pigmentation EX: anti-reflective coating
Principle of Linear Superposition
When two or more waves overlap in the same space and time, the amplitudes combine to form a new, single wave
PQ 27.2 Light passes through a single slit. If the width of the slit is reduced, what happens to the width of the central bright fringe? a. The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. b. The central bright fringe becomes narrower, because angle that locates the first dark fringe on either side of the central bright fringe becomes smaller. c. The central bright fringe becomes narrower, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. d. The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes smaller. e. The width of the central bright fringe does not change, because it depends only on the wavelength of the light and not on the width of the slit.
a. The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. Correct. The width of the central bright fringe is determined by the angle θ that locates the first (m = 1) dark fringe on either side of the central bright fringe. This angle is given by sin θ = 1 λ/W, where λ is the wavelength and W is the width of the slit. If the width of the slit decreases, the angle increases.
PQ 26.3 An object is situated to the left of a lens. A ray of light from the object is close to and parallel to the principal axis of the lens. The ray passes through the lens. Which one of the following statements is true? a. The ray passes through the lens without changing direction, no matter whether the lens is converging or diverging. b. The ray passes through a focal point of the lens only if the lens is a converging lens. c. The ray passes through a focal point of the lens, no matter whether the lens is converging or diverging. d. The ray crosses the principal axis at a distance from the lens equal to twice the focal length, no matter whether the lens is converging or diverging. e. The ray passes through a focal point of the lens only if the lens is a diverging lens.
b. The ray passes through a focal point of the lens only if the lens is a converging lens. Correct. For a converging lens, rays that are parallel to and close to the principal axis are bent toward the axis by the lens and pass through the focal point on the far side of the lens. For a diverging lens, rays that are parallel to and close to the principal axis are bent away from the axis by the lens. After being bent they appear as if they originated at the focal point on the near side of the lens. They do not actually pass through that focal point, however.
PQ 27.1 A series of bright fringes appears on the viewing screen of a Young's double-slit experiment. Suppose you move from one bright fringe to the next one farther out from the central bright fringe. Does the difference in path lengths of the light waves increase or decrease, and by what amount does it change? Express your answer in terms of the wavelength λ of the light. a. difference in path length increases by 1/2λ b. difference in path length increases by λ c. difference in path length increases by 2λ d. difference in path length decreases by 1/2λ e. difference in path length decreases by λ
b. difference in path length increases by λ Correct. The difference in path lengths of the light waves increases by one wavelength as one moves from one bright fringe to the next one farther out.
PQ 24.1 An electromagnetic wave travels in a vacuum. The wavelength of the wave is tripled. How is this accomplished?
by reducing the frequency of the wave by a factor of three Correct. The wavelength λ, frequency f, and speed c of an electromagnetic wave are related according to c = λf, where c is the same for any electromagnetic wave traveling in a vacuum and is independent of λ and f. Since c is constant, λ and f are inversely proportional. When f is reduced by a factor of three, λ increases by a factor of three.
PQ 27.3 What happens to the diffraction pattern when the number of lines per centimeter of a diffraction grating is increased? a. The number of principal maxima that can be seen on a screen increases. b. Nothing happens to the diffraction pattern. c. The distance between two adjacent principal maxima increases. d. An identical diffraction pattern can be created by using light of a longer wavelength.
c. The distance between two adjacent principal maxima increases. Correct. As the number of lines per centimeter increases, the separation d between adjacent slits becomes smaller. The angle θ that specifies a principal maxima of a diffraction grating is given by sin θ = mλ/d, where λ is the wavelength and m = 0, 1, 2, 3, ... . For fixed values of m and the wavelength, this relation shows that as d decreases, θ increases. For a fixed screen location, this means that the distance between two adjacent principal maxima also increases.
PQ 27.1 Light of wavelength 600 nm in vacuum is incident nearly perpendicularly on a thin film whose index of refraction is 1.5. The light travels from the top surface of the film to the bottom surface, reflects from the bottom surface, and returns to the top surface, as the drawing indicates. What is the total (down-and-back) distance traveled by the light inside the film? Express your answer in terms of the wavelength λfilm of the light within the film. a. 3λfilm b. 2λfilm c. 12λfilm d. 6λfilm e. 4λfilm SEE PICTURE = 1200 nm width of film
d. 6λfilm Correct. The down-and-back distance traveled by the light wave in the film is 2400 nm. The wavelength of the light within the film is λfilm = λvacuum/nfilm = 600 nm / 1.5 = 400nm. Thus, the down-and-back distance is equivalent to (2400 nm)λ400 nm = 6λfilm.
PQ 26.3 What kind of image can a single converging lens NOT produce? a. A real image that is inverted and is larger with respect to the object. b. A virtual image that is upright and is larger with respect to the object. c. A real image that is inverted and is smaller with respect to the object. d. A virtual image that is upright and smaller with respect to the object
d. A virtual image that is upright and smaller with respect to the object Correct. A converging lens can produce an upright virtual image, if the object is within the focal point of the lens. However, this image is larger (not smaller) than the object.
