Chapter 8: Optics Flashcards

Question

Plane mirrors can be thought of as spherical mirrors with infinitely large focal distances what does this mean for the nature of the image and the distance from o to i?

Answer 1

As such, for a plane mirror, r=f=infinity and the equation becomes 1/o+1/i=0, or o=-i. This can be interpreted as saying the virtual image is a distance behind the mirror equal to the distance the object is in front of the mirror.

Answer 2

Magnification is the ratio of the image distance to the object distance (and is therefor dimensionless). Magnification also gives us the ratio of the size of the image to the size of the object. m=-i/o

Answer 3

Draw perpendicular ray and ray to normal.

Answer 4

For a concave mirror, a ray that strikes the mirror parallel to the axis is reflected back through the focal point (GREEN LINES). A ray that passes through the focal point before reaching the mirror is reflected back parallel to the axis (RED LINES). A ray that strikes the mirror at the point of intersection with the axis is reflected back with the same angle measured from the normal (BLUE LINES).

Answer 5

If the object is placed beyond F, the image will be real, inverted, and magnified. If the object is placed at F , no image is formed because the reflected light rays are parallel to each other. If the object is placed between F and the mirror, the image produced is virtual, upright, and magnified.

Answer 6

If the object is placed beyond the focal point, the image will be real, inverted, and magnified. If the object is placed on the focal point, the image will be at infinity. If the object is placed between the focal point and the mirror, the image will be virtual upright and magnified.

Answer 7

A single diverging mirror forms only a virtual, upright, and reduced image, regardless of the position of the object. The farther away the object, the smaller the image will be.

Answer 8

A security mirror is a convex, diverging optical system. Parallel light ray that hit the mirror or reflected in multiple directions, which allows the observer to see a large field vision, even if the image is somewhat distorted, and the objects in the image are closer than they appear. A passenger side mirror of a car is also a convex mirror, allowing the driver to see a wider view of the other vehicles behind the car (images looking closer than they appear).

Answer 9

We really only need to draw one way a diagram for convex (diverging) mirrors as the image will always be the virtual, upright, and reduced regardless of the position of the object.

Answer 10

A ray that strikes the mirror parallel to the axis, a ray that passes through the focal point before reaching the mirror, a ray that strikes the mirror at the point of intersection with the axis.

Answer 11

The focal length of converging mirrors and converging lenses will always be positive. The focal length of diverging mirrors and diverging lenses will always be negative.

Answer 12

A convex mirror is diverging because when light rays hit its curved surface, they spread out or diverge away from each other. A concave mirror is converging because when light rays hit its curve surface, they converge the light rays toward a focal point. Convex mirrors always produce virtual images, meaning the image appears to be behind the mirror and cannot be projected onto a screen.

Answer 13

For clarification: i is positive for real images (in front of mirror) and negative (behind mirror) for virtual images. r is positive for concave (in front of mirror) and negative for convex (behind mirror). (r is related to the focal point, f=r/2) f is positive for concave/converging mirror (in front of mirror) and negative for convex/diverging mirror (behind mirror) f=r/2 Positive m: upright; negative m: inverted The object distance o is almost always positive (which means that the object is placed in front of the mirror)

Answer 14

For a single mirror or lens with o>0, a real image will always be inverted. We we needed to extrapolate this based on our knowledge of concave mirrors. We needed to use the optics equation and recognize the variables given to solve for i (image distance).

Answer 15

Upright images are always Virtual NO images formed when the object is a focal length away. Inverted images are always Real

Answer 16

Inverted Upright Larger Smaller

Answer 17

Refraction is the bending of light as it passes from one medium to another and changes speed. The speed of light through any medium is always less than its speed through a vacuum, but in air is really really really close to 3×10^8 m/s.

Answer 18

The index of refraction is a dimensionless quality that describes the ratio between the speed of light in a vacuum, and the speed of light in a medium. The index of refraction of a vacuum, by definition, is one. For all other materials, the index of refraction will be greater than one.

Answer 19

Refracted rays of Lido base no laws they passed from one medium to another.

Answer 20

When light enters a medium with higher index of refraction, it bends toward the normal. When light enters a medium with lower index of refraction, it bends away from the normal.

Answer 21

When light travels from a medium with a higher index of refraction (such as water) to a medium with a lower index of refraction (such as air), the refracted angle is larger than the incident angle. That is, the refracted light ray bends away from the normal. As the incident angle is increased, the refracted angle also increases, and eventually, a special incident angle called the critical angle is reached for which the reflected angle equals 90°. At the critical angle, the refracted light ray passes along the interface between the two media.

Answer 22

Aside from the fact that lenses transmit light while mirrors reflect it, there are two surfaces that affect the light path with lenses. The light is reflected twice as it passes from air to lens and from lens back to air.

Answer 23

This is sort of a trick question. Farsighted means nearby objects are blurry. Reading glasses are needed by people who are “farsighted” and are converging lenses. Nearsighted means far away objects are blurry. Standard glasses are needed by people who are “nearsighted” and are diverging lenses.

Answer 24

Because light can travel from either side of lens, lens has two focal points, with one on each side. The focal lengths can be measured in either direction from the center. For thin spherical lenses, the focal links are equal, so we speak of just one focal length for the lens as a whole.

Answer 25

Similarity is that the distance between the focal point and mirror, f, are all equal.

Answer 26

Nearsightedness is when the converging lens of your eye focuses images in front of your retina. If you place a diverging lens in front of your eye, which moves the rays apart and refocuses the image on your retina.

Answer 27

Farsighted means near objects are blurry. Need converging lens in front of eye. Lens of eye converges rays behind the retina. Nearsighted means far objects are blurry. Need diverging lens in front of eye. Lens of eye converges rays in front of retina.

Answer 28

Converging lenses are thicker in the center, diverging lenses are thinner in the center.

Answer 29

The image will always be virtual and upright.

Answer 30

The Real side of a mirror is in front of the mirror. The Real side for a lens is the other side of the lens from the light source.

Answer 31

This equation demonstrates how to calculate f for a lens where thickness cannot be ignored (a real lens). It depends on the index of refraction for the lens, and the difference of the inverse of radius of curvature of first and second lens.

Answer 32

Spherical mirrors and lenses are imperfect and subject to specific types of errors or aberrations. Spherical aberration is 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.

Answer 33

Dispersion is where various wavelengths of light separate from each other, like light through a prism. The index of refraction of a medium affects the wavelength of light passing through the medium because the index of refraction is related or the speed of the wave by n=c/v. It also implies that the index of refraction itself actually varies with the wavelength.

Answer 34

When light passes through a prism, IT REFRACTS, and the light disperses into various wavelengths and therefor colors. Violet light has the smallest wavelength and therefor experiences the greatest amount of refraction among the colors of the visible spectrum. Red light has the longest wavelength and therefor experiences the least amount of refraction among the colors of the visible spectrum. Because of this, the pattern that emerges is the red comes out on top and violet emerges on the bottom.

Answer 35

As light enters a medium with a different index of refraction, the wavelength changes but the frequency of the light does not.

Answer 36

Chromatic aberration is a dispersive effect within a spherical lens.

Answer 37

True. In optics, incident angles are always measured relative and to the normal.

Answer 38

Light will bend toward normal when going from a medium with low n to a medium with high n. Light will bend away from normal when going from a medium with high n to a medium of low n.

Answer 39

Dispersion is the separation of light by wavelength when passing through a medium. Dispersion is the tendency for different wavelengths of light to experience different degrees of refraction in a medium, leading to separation of light into the visible spectrum (a rainbow). Aberration is a blurring of an image due to imperfections in a lens. Aberration (spherical or chromatic) is the alteration or distortion of an image as a result of an imperfection in the optical system.

Answer 40

A handy way to determine which way light will refract is to imagine a marching band entering a plot of sand. The line of musicians nearer to the plot of sand will slow down first, the line of musicians will reach the sand later. Look it up if you need an example.

Answer 41

Light diffraction refers to the spreading out of light as it passes through a narrow opening or around an obstacle. Think “diff” is diffuse

Answer 42

Dark fringes are minima and bright fringes are called maxima.

Answer 43

n+0.5 as bright fringes (maxima) are halfway between dark fringes (minima)

Answer 44

Need to recognize that we aren’t given the hypotenuse. We can make an assumption that the distance between the screen and the slits is approximately equal to the hypotenuse, and can substitute sintheta=tantheta. Recognize that when calculating deltan, the 1/2 cancels. Is math from there.

Answer 45

Diffraction gratings consist of multiple slits arranged in patterns. They create colorful patterns similar to a prism as the different wavelengths interfere in characteristic patterns. Thin films cause interference patterns because light waves reflecting off the external surface interfere with light waves reflecting off the internal surface of the film.

Answer 46

Diffraction through a single slit does not create characteristic fringes when projected on a screen, although the light does spread out. When a lens is introduced into the system, the additional refraction of light causes constructive destructive interference, creating fringes.

Answer 47

Fringes result from constructive and destructive interference between light rays.

Answer 48

The image formed during double slit diffraction contains fringes because light rays constructively and destructively interfere. A single slit forms an image of a wide band of light, spread out from its original beam.

Answer 49

True. Maxima and minimum alternate in a diffraction pattern. A maximum is equidistant between two minima, and a minimum is equidistant between two maxima.

Answer 50

Plane polarized light contains light waves with parallel electric field vectors. Circularly polarized light selects for a given amplitude and has a continuously rotating electric field direction.

Answer 51

Plain polarization has no effect on the wavelength (or frequency or speed) of light. Polarization does affect the amount of light passing through medium and light intensity.

Answer 52

Two ways to do this problem. Draw a ray diagram. If done properly, you will get a real inverted magnified image. Drill and kill this until you know this. Use sign convention. If the object is at the center of curvature, its distance is 2f. Plug into the optics equation. i ends up positive, and therefor real. FOR SINGLE MIRRORS OR LENSES, ALL REAL IMAGES ARE INVERTED. (We can also solve for m and end up with a negative number)

Answer 53

This is a plug and play question. Knowing the equation for double slit, we solve for wavelength. We got stuck in algebra land there and needed to refer to the answer sheet. Get used to this variety of algebra.

Answer 54

We solved this problem analytically. First off. We know that the frequency of light doesn’t change when going from one medium to another, this eliminates C and D. Secondly, we know that as light moves from a lower index of refraction to a higher index index of fraction (n increases), the angle of fraction will be smaller than the incident angle (closer to the normal) so sayeth Snells law.

Answer 55

Light will not pass through the third polarizer (D). Plain polarized light is light in which the electric fields of all the waves are oriented in the same direction. Light passing through the first two polarizers will only contain rays with their electric field vectors in the same direction. When it reaches the third polarizer, however, the light will not be able to pass through because all the light rays will be oriented in the direction dictated by the first and second polarizers.

Answer 56

Pay attention to the sign convention here. If f>o, then 1/f - 1/o is negative because a larger denominator means a smaller number and therefore i is negative given 1/f = 1/o + 1/i ……. 1/i = 1/f - 1/o. Negative i means virtual, and virtual images are upright.

Answer 57

When light passes through a narrow opening, the light wave spread out. As the slit narrows, the light wave spread out even more.

Answer 58

Draw diagram. The angle given is respect to the normal you know that the incident angle must be 60°. You know that the reflected beam will have an angle of 60° relative to the normal (law of reflection). We then know the reflected beam will make an angle of 30° with the plain of the glass. If the reflected and refracted beams are perpendicular to each other, the reflected beam will make a 60° angle with the plain of the glass. We can solve for the reflected angle: 30°. Plug and play.