Physics Concepts: Ch. 22, 23, 24 & 25

Mirror Rules

1. when the object is located so that the center of curvature lies between the object and a concave mirror surface, the image is real, inverted and reduced in size. 2. when the object is located between the focal point and a concave mirror surface the image is virtual, upright, and enlarged. 3. When the object is in front of a convex mirror, the image is virtual, upright, and reduced in size.

Light from an object passes through a lens and forms a visible image on a screen. If the screen is removed, would you be able to see the image (a) If you remained in your present position and (b) if you could look at the lens along its axis, beyond the original position of the screen?

A. No. the screen is needed to reflect the light toward your eye B. Yes. The light is traveling toward your eye and diverging away from the position of the image, the same as if the object were located at that position. It is kinda like looking through a magnifying glass. You can see a water bottle image from the original object.

22.2 If beam 1 is the incoming beam in Figure, which of the other four beams are due to reflection? Which are due to refraction?

Beams 2 and 4 are reflected; beams 3 and 5 are refracted.

Light can travel from air into water. Some possible paths for the light ray in the water are shown in figure. Which path will the light most likely follow? a) A b) B c) C d) D e) E

C moves toward the normal

A soap film is held vertically in air and is viewed in reflected light as in figure. Explain why the film appears to be dark at the top.

Due to gravity, the soap film tends to sag in its holder, being quite thin at the top and becoming thicker as one moves toward the bottom of the holding ring. Because light reflecting from the front surface of the film experiences a phase change, and light reflecting from the back surface of the film does not, the film must be a minimum of a half wavelength thick before it can produce constructive interference in the reflected light. Thus, the light must be striking the film at some distance from the top of the ring before the thickness is sufficient to produce constructive interference for any wavelength in the visible portion of the spectrum

5. If you want to examine the fine detail of an object with a magnifying glass with a power of +20.0 diopters, where should the object be placed so as to observe a magnified image of the object?

For a lens to operate as a simple magnifier, the object should be located just inside the focal point of the lens. If the power of the lens is +20.0 diopters, its focal length is f = (1.00m)/P = (1.00m)/+20.0 = 0.050 m = 5.00 cm The object should be placed slightly less than 5.00 cm in front of the lens power = 1/f diopters is the unit of power

9. In everyday experience, why are radio waves polarized, whereas light is not?

For regional communication at the Earth's surface, radio waves are typically broadcast from currents oscillating in tall vertical towers. These waves have vertical planes of polarization. Light originates from the vibrations of atoms or electronic transitions within atoms, which represent oscillations in all possible directions. Thus, light generally is not polarized. Polarized light makes wavelengths all go in own single direction.

11. Can a converging lens be made to diverge light if placed in a liquid? How about a converging mirror?

If a converging lens is placed in a liquid having an index of refraction larger than that of the lens material, the direction of refractions at the lens surfaces will be reversed, and the lens will diverge light. A mirror depends only on reflection that is independent of the surrounding material, so a converging mirror will be converging in any liquid. Because the light is going through the water first, and then to the glass which has a lower index of refraction, it will diverge the light and then converge once it passes through. Instead of going towards normal it will go away from normal

13. Why is it so much easier to perform interference experiments with a laser than with an ordinary light source?

If you wish to perform an interference experiment, you need monochromatic coherent light (a laser). To obtain it, you must first pass light from an ordinary source through a prism or diffraction grating to disperse different colors into different directions. Using a single narrow slit, select a single color and make that light diffract to cover both slits for a young's experiment. the procedure is much simpler with a laser because its output is already monochromatic and coherent. You don't have to worry about a diffraction grading.

9. Explain why it is theoretically impossible to see an object as small as an atom regardless of the quality of the light microscope being used

In order for someone to see an object through a microscope, the wavelength of the light in the microscope must be smaller than the size of the object. An atom is much smaller than the wavelength of light in the visible spectrum, so an atom can never be seen with the use of visible light.

Suppose you are observing the interference pattern formed by a Michelson interferometer in a laboratory and a joking colleague holds a lit match in the light path of one arm of the interferometer. Will this match have an effect on the interference pattern?

It could have an effect on the interference pattern?

Try this simple experiment on your own. Take two opaque cups, place a coin at the bottom of each cup near the edge and fill on cup with water. Next, view the cups at some angle from the side so that the coin in water is just visible as shown on the left in figure. Notice that the coin in air is not visible as shown on the right in Figure. Explain this observation

Key: With no water in the cup, light rays from the coin do not reach the eye because they are blocked by the side of the cup. With water in the cup, light rays are bent away from the normal as they leave the water so that some reach the eye. In Figure (a) below, ray a is blocked by the side of the cup so it cannot enter the eye, and ray b misses the eye. In Figure (b), ray a is still blocked by the side of the cup, but ray b refracts at the water's surface so that it reaches the eye. Ray b seems to come from position B, directly above the coin at position A.

11. Figure shows a pencil partially immersed in a cup of water. Why does the pencil appear to be bent?

Light rays coming from parts of the pnecil under water are bent away from the normal as they emerge into the air above. The rays enter the eye at an angles closer to the horizontal; thus, the parts of the pencil under water appear closer to the surface than they actually are, so the pencil appears bent.

13. Why do astronomers looking at distant galaxies talk about looking backward in time?

Light travels through a vacuum at a speed of 3 x 10^8 m/s. Thus, an image we see from a distant star or galaxy must have been generated some time ago. For example, the star Altair is 16 light years away; if we look at an image of Altair today, we know only what Altair looked like 16 years ago. This may not initially seem significant; however, astronomers who look at other galaxies can get an idea of what galaxies looked like when they were much younger. Thus, it does make sense to speak of "looking backward in time".

Suppose reflected white light is used to observe a thin, transparent coating on glass as the coating material is gradually deposited by evaporation in a vacuum. Describe some color changes that might occur during the process of building up the thickness of the coating.

Suppose the index of refraction of the coating is intermediate between vacuum and the glass. When the coating is very thin, light reflected from its top and bottom surfaces will interfere constructively, so you see the surface white and brighter. Once the thickness reaches one-quarter of the wavelength of violet light in the coating, destructive interference for violet light will make the surface look red. Then other colors in spectral order (blue, green, yellow, orange, and red) will interfere destructively, making the surface look red, magenta, and then blue. As the coating gets thicker, constructive interference is observed for violet light and then for other colors in spectral order. Even thicker coatings give constructive and destructive interference for several visible wavelengths, so the reflected light starts looking white again.

A lens with a certain power is used as a simple magnifier. if the power of the lens is doubled, does the angular magnification increase or decreased?

The angular magnification produced by a simple magnifier is m = (25cm)/ f. Note that this is proportional to the optical power of a lens, P = 1 /f, where the focal length f is expressed in meters. Thus, if the power of the lens is doubled, the angular magnification will also double. The simple magnifier achieves angular magnification by permitting the placement of the object closer to the eye than the eye could normally focus.

9. Suppose you are told that only two colors of light (X and Y) are sent through a glass prism and that X is bent more than Y. Which color travels more slowly in the prism?

The color traveling slowest is bent the most. Thus, X travels more slowly in the glass prism. If it has a higher index of refraction it will go slower. Therefore the more bent one will be the slower one through the glass prism.

Lenses used in eyeglasses, whether converging or diverging, are always designed such that the middle of the lens curves away from the eye. Why?

The design is chosen so your eyelashes will not brush against the glass as you blink. A reason involving a little physics is that with this design, when you direct your gaze near the outer circumference of the lens, you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally distorts the direction of the object you are looking at

13. Why does the focal length of a mirror not depend on the mirror material when the focal length of a lens does depend on the lens material?

The focal length for a mirror is determined by the law of reflection from the mirror surface. The law of reflection is independent of the material of which the mirror is made and of the surrounding medium. Thus, the focal length depends only on the radius of curvature and not on the material. The focal length of a lens depends on the indices of refraction of the lens material and surrounding medium. Thus, the focal length of a lens depends on the lens material. Lenses will have different index of refraction, and will cause the light to bend.

3. the optic nerve and the brain invert the image formed on the retina. Why don't we see everything upside down?

The image formed on the retina by the lens and cornea is already inverted

The level of water in a clear, colorless glass can easily be observed with the naked eye. The level of liquid helium in a clear glass vessel is extremely difficult to see with the naked eye. Explain. Hint: the index of refraction of liquid helium is close to that of air.

The index of refraction of water is 1.33, quite different from that of air, which has an index of of refraction of 1. The boundary between the air and water is therefore easy to detect, because of the differing refraction effects above and below the boundary. (look at a glass half full of water) The index of refraction of liquid helium, however, happens to be much closer to that of air. Consequently, the refractive difference above and below the helium-air boundary are harder to see. Depends on refractive index - if they are very different in the number - a difference can be seen!

Compare and contrast the eye and a camera. What parts of the camera correspond to the iris, the retina, and the cornea of the eye?

The iris is the shutter that opens and closes the eye. It controls how much light comes in and out of the eye. The retina is the screen in the back of the camera. This screen gathers the image. This is what the retina in the eye does. The cornea of the eye focuses light on the back side of the eye. This is synonymous to the lens of the camera!

1. A lens is used to examine an object across a room. Is the lens probably being used as a simple magnifier? Explain in terms of focal length, the image, and magnification

The observer is not using the lens as a simple magnifier. For a lens to be used as a simple magnifier, the object distance must be less than the focal length of the lens. Also, a simple magnifier produces a virtual image at the normal near point of the eye, or at an image distance of about q=-25cm. With a large object distance and a relatively short image distance, the magnitude of the magnification by the lens would be considerably less than one. Most likely, the lens in this example is part of a lens combination being used as a telescope.

A type of mirage called a pingo is often observed in Alaska. Pingos occur when the light from a small hill passes to an observer by a path that takes the light over a body of water warmer than the air. What is seen is the hill and an inverted image directly below it. Explain how these mirages are formed.

The rays traveling from the hill to just above the water are bent upward by the normal mechanism of refraction, thus forming an inverted image. The rays going directly to the observer's eye pass above the warm air near the water's surface and go directly to the observer's eye, causing the upright image. It inverts the image like a lens would. the image reflects off of the water This mechanism applies to nearly all mirages, not just to specially named ones occurring in Alaska.

Certain sunglasses use a polarizing material to reduce the intensity of light reflected from shiny surfaces, such as water or the hood of a car. What orientation of the transmission axis should the material have to be most effective

The reflected light is partially polarized, with the component parallel to the reflecting surface being the most intense. Therefore, the polarizing material should have its transmission axis oriented in the vertical direction in order to minimize the intensity of the reflected light from horizontal surfaces. Sunglasses made of polarizing material reduce the glare, which is the reflected light. The transmission axes of the lenses are oriented vertically to absorb the strong horizontal component of the reflected light. Because the reflected light is mostly polarized most of the glare can be eliminated without removing most of the normal light.

15. Consider a dark fringe in an interference pattern at which almost no light energy is arriving. Light from both slits is arriving at this point, but the ways cancel. Where does the energy go?

The result of the double slit is to redistribute the energy arriving at the screen. Although there is no energy at the location of a dark fringe, there is four times as much energy at the location of a bright fringe as there would be with only a single narrow slit. The total amount of energy arriving at the screen is twice as much as with a single slit, as it must be according to the law of conservation of energy. the energy goes to the bright fringes

Fingerprints left on a piece of glass such as a windowpane can show colored spectra like that from a diffraction grating. Why?

The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the finger touches a smooth surface, the oils from the skin will be deposited on the surface in the pattern of the closely spaced ridges. The clear spaces between the lines of deposited oil can serve as the slits in a crude diffraction grating and produce a colored spectrum of the light passing through or reflecting from the glass surface. It reflects the light of different wavelengths at different angles.

1. Why does the arc of a rainbow appear with red on top and violet on the bottom?

The spectrum of the light sent back to you from a drop at the top of the rainbow arrives such that the red light (deviated by an angle of 42 degrees) strikes the eye while the violet light (deviated by 40 degrees) passes over your head. Thus, the top of the rainbow looks red. At the bottom of the rainbow, violet arrives at your eye and red light is deviated toward the ground. Thus, the bottom part of the rainbow appears violet. It all has to do with the angle of the arc.

Holding your hand at arm's length, you can readily block direct sunlight from your eyes. Why can you not block sound from your ears this way?

The wavelength of light is extremely small in comparison to the dimensions of your hand, so the diffraction of light around obstacles the size of your hand is totally negligible. However, sound waves have wavelengths that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.

If a cylinder of solid glass or clear plastic is placed above the words LEAD OXIDE and viewed from the side, as shown in figure, the word LEAD appears inverted, but the word OXIDE does not. Explain.

The wavelength should not effect whether the image is inverted or not. My guess is that the word OXIDE has a horizontal axis of symmetry thus its inverse is identical to the original image and thus does not appear inverted.

7. In dispersive materials, the angle of refraction for a light ray depends on the wavelength of the light. Does the angle of reflection from the surface of the material depend on the wavelength? Why or why not?

There is no dependence of the angle of reflection on wavelength, because the light does not enter deeply into the material during reflection - it reflects from the surface

9. In a Jules Verne novel, a piece of ice is shaped into a magnifying lens to focus sunlight to start a fire. Is that possible?

This is a possible scenario. When light crosses a boundary between air and ice, it will refract in the same manner as it does when crossing a boundary of the same shape between air and glass. Thus, a converging lens may be made from ice as well as glass. However, ice is such a strong absorber of infrared radiation that it is unlikely you will be able to start a fire with a small ice lens.

7. Suppose you want to a converging lens to project the image of two trees onto a screen. One tree is a distance x from the lens; the other is at 2x as in figure. You adjust the screen so that the near tree is in focus. If you now want the far tree to be in focus, do you move the screen toward or away from the lens?

We consider the two trees to be two separate objects. The far tree is an object that is farther from the lens than the near tree. Thus, the image of the far tree will be closer to the lens than the image of the near tree. The screen must be moved closer to the lens to put the far tree in focus. Since the tree is farther away, the lines converge closer, therefore you need to bring the screen closer to the lens to focus it.

A virtual image is often described as an image through which light rays don't actually travel, as they do for a real image. Can a virtual image be photographed?

Yes, a virtual image can be photographed. A virtual image is an image formed when the reflected rays from a point on an object always diverge. Virtual image cannot be captured on a plain screen, however if a camera in line with the reflected rays then it can be photographed.

11. Would it be possible to place a non-reflective coating on an airplane to cancel radar waves of wavelength 3 cm?

Yes. In order to do this, first measure the radar reflectivity of the metal of your airplane. Then choose a light, durable material that has approximately half the radar reflectivity of the metal in your plane. Measure its index of refraction, and place onto the metal a coating equal in thickness to one-quarter of 3 cm, divided by that index. Sell the plane quickly, and then you can sell the supposed enemy new radars operating at 1.5 cm, which the coated metal will reflect with extra high efficiency.

1. Tape a picture of yourself on a bathroom mirror. Stand several centimeters away from the mirror. Can you focus your eyes on both the picture taped to the mirror and your image in the mirror at the same time? So where is the image of yourself?

You will not be able to focus your eyes on both the picture and your image at the same time. To focus on the picture, you must adjust your eyes so that an object several centimeters away is in focus. Thus, you are focusing on the mirror surface. But your image in the mirror is as far behind the mirror as you are in front of it. Thus, you must focus your eyes beyond the mirror, twice as far away as the picture to bring the image into focus,

1. Your automobile has two headlights. What sort of interference pattern do you expect to see from them? Why?

You will not see an interference pattern from the automobile headlights for two reasons. The first is that the headlights are not coherent sources and are therefore incapable of producing sustained interference. Also, the headlights are so far apart in comparison to the wavelengths emitted that, even if they were made into coherent sources, the interference maxima and minima would be too closely spaced to be observable.

22.1 Which part of Figure, (a) or (b), better shows specular reflection of light from the roadway?

a

25.2 Suppose you are observing a binary star with a telescope and are having difficulty resolving the two stars. Which color filter will better help resolve the stars? a) blue b) red c) neither because colored filters have no effect on resolution

a

If a Young's experiment carried out in air is repeated under water, would the distance between bright fringes (a) increase, (b) decrease, or (c) remain the same?

a The separation decreases because the wavelength of the light decreases in the water. This is because water has a higher index of refraction. Because the wavelength decrease while all else stays the same, angle theta must have to increase.

7. Figure shows rays with wavelength lambda incident from above onto thin films surrounded by air. (a) Will the film in figure appear bright due to constructive interference or dark due to destructive interference? Indicate your answer with B for bright or D for dark. (b) Repeat part (a) for figure. (c) Repeat part (a) for figure

a) Dark - destructive interference (it goes from low to high therefore it is destructive, and then it travels 1.5 lambda twice so you get a hole number so it doesn't change phases) b) Bright - constructive interference (it goes from low to high therefore it is destructive but then it travels 1.25 twice so you get 2.5 a half number so it does change phases and becomes constructive) c) Dark - destructive intereference 2nt = (m+0.5)lambda - constructive 2nt = m x lambda - destructive if the film is placed between two different media, one of lower refractive index of the film and one of higher refractive index, then the equations are revered.

5. Sodium's emission lines at 589.0 nm and 589.6 nm pass through a diffraction grating and form two m=+1 maxima on a viewing screen. Would the spacing between the two lines increase, decrease, or remain unchanged if the grating is exchanged (a) for one having fewer lines per millimeter or (b) for one with twice the total number of lines? What if (c) the intensity of the light is doubled, or (d) the maxima are viewed in second order?

a) Decrease. (both theta angles will decrease because you are dividing them by a greater distance). b) Unchanged (you're adding more lines but not per mm. Just more lines at the same distance so the interference and spacing will stay the same) c) Unchanged (intensity doesn't matter; peaks are bigger but the distances should be the same!) d) Increased. (theta increases, therefore the spacing between the two lines will double) If you increase m then you increase the fringes because m and d are directly proportional

A plane monochromatic light wave s incident on a double slit as illustrated in figure. if the viewing screen is moved away from the double slit, what happens to the separation between the interference fringes on the screen? a) it increases b) it decreases c) it remains the same d) it may increase or decrease, depending on the wavelength of the light e) more information is required

a) It increases; this is because although theta stays the same, the distance between the two angles gets a little bigger.

5. Determine whether each of the following statements is true or false. a) the angle theta in snell's law is measured between the ray and a line perpendicular to the surface. b) the speed of light in a material increases as the material's index of refraction increases c) the ratio v/lamba of a photon's speed to it's wavelength has the same value for any index of refraction n d) photons of blue light have a higher energy than photons of red light. e) a photon's energy depends on its brightness

a) T normal is perpendicular, and the angle is always measured in respect to the normal. b) F the speed of light will decrease as it passes through a material with a greater index of refraction c) T yes, the ratio will stay the same d) T red has a bigger wavelength and a smaller frequency, so it has a smaller energy of photons. e) F no, it depends on its frequency

5. Construct ray diagrams whether each of the following statements is true (T) or false (F). a) for any object in front of a diverging lens, the image is virtual and in front of the lens. b) A converging lens always forms a real image and diverging lens always forms a virtual image c) For an object located at the focal point of any lens, the image has a magnification equal to one.

a) T b) F; converging can either form a real or virtual c) F? M = -q/p it could be one but it doesn't have to be.

Construct ray diagrams to determine whether each of the following statements is true (T) or false (F). (a) for an object at a concave mirror's center of curvature, the image is real and inverted. (b) As an object approaches the focal point of a concave mirror, the image size shrinks to zero. (c) For an object in front of a convex mirror, the image is always virtual and upright.

a) True b) True c) True When the object is in front of a convex mirror, the image is virtual, upright, and reduced in size.

Large telescopes are usually reflecting rather than refracting. List some reasons for this choice.

a) a lens must have two precision surfaces; a mirror needs only one b) Lenses absorb light, while mirrors do not c) Lenses are subject to chromatic aberration d) Heavy lenses, which can only be supported at their edges, tend to deform under their own weight

The top row of Figure shows three ray diagrams for an object O in front of a concave mirror and the bottom row shows three ray diagrams for an object O in front of a convex mirror. In each diagram, one ray is drawn correctly and the other is drawn incorrectly. For (a)-(f), determine whether the red (R) or blue (B) ray is drawn correctly.

a) blue b) red c) red d) red e) red f) blue

A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the following quantities increases, decreases, or remains unchanged. Indicate you answers with I, D, or U. a) the ray's angle with the normal b) the light's wavelength c) the light's frequency d) the light's speed e) the photon energy

a) decreases the refracted ray is bent toward the normal ray b) decrease v = lamba x f if V decreases as it passes through a medium, the wavelength will also decrease c) U As light travels from one medium to another, its frequency doesn't change d) decrease the velocity of light will decrease as it passes through a medium e) U this is because E = hf f stays the same and so does E

A CCD camera is equipped with a lens with constant focal-length. As the f-number is decreased, determine whether the following quantities increase, decrease, or remain unchanged. a) the aperture b) the depth of field c) the intensity of light reaching the sensor d) the appropriate exposure time

a) increases increasing the setting from one f number to a higher f number decreases the area of aperture. b) decreases a high value for the f-number allows for large depth of field and means that objects at a wide range of distances from the lens form reasonably sharp images on the sensor. c) increases d) decreases; if f number is given as a description of lens speed - if it decreases, the exposure time will also decrease

3. The top row of Figure shows three ray diagrams for an object O in front of a converging lens and the bottom row shows three ray diagrams for an object O in front of a diverging lens. In each diagram, one ray is drawn correctly and the other is drawn incorrectly. For (a)-(f), determine whether the red (R) or blue (B) ray is drawn correctly.

a) red b) blue c) red d) red e) blue f) blue

Choose the option from each pair that makes the following statement correct. For a farsighted person, the [(a) near point; (b) far point] is always located farther than [(c) 1 m; (d) 25 cm] from the eye and the corrective lens is [(e) converging; (f) diverging].

a, d, e For a farsighted person, the near point is always located farther than 25 cm from the eye and the corrective lens is converging. Near point is closer farther because they can't see near their eyes.

23.6 An object is placed to the left of a converging lens. Which of the following statements are true, and which are false? (a) the image is always to the right of the lens, (b) the image can be upright or inverted, (c) the image is always smaller or the same size as the object.

a. F b. T c. F

23.3 T or F: (a) The image of an object placed in front of a concave mirror is always upright (b) The height of the image of an object placed in front of a concave mirror must be smaller than or equal to the height of the object. (c) The image of an object placed in front of a convex mirror is always upright and smaller than the object

a. False Concave can be a real, inverted image or a virtual, upright image. b. False The height of the image can be enlarged if the object is between the focal point and the mirror. c. True Convex mirror will always be upright, virtual and reduced in size.

Count the number of 180 phase reversal for the interfering rays in a) none - constructive b) one - destructive c) two - constructive

an electromagnetic wave undergoes a phase change of 180 degrees upon reflection from a medium that has an index of refraction higher than the one in which the wave was traveling.

23.5 In figure, the blue object arrow is replaced by one that is much taller than the lens. How many rays from the object will strike the lens?

an infinite number

23.1 In the over head view of figure, the image of the stone seen by observer 1 is at C. Where does the observer 2 see the image: at A, B, C, D, E, or not at all?

at C

13. Suppose a microscope's resolution is diffraction limited, which one of the following changes would provide the greatest improvement to its resolution? a) observing at a longer wavelength through a smaller aperture. b) observing at a shorter wavelength through a larger aperture. c) decreasing the object distance d) using a CCD sensor instead of a standard eyepiece

b

15. An object represented by a gray arrow, is placed in front of a plane mirror. Which of the diagrams in figure best describes the image, represented by the pink arrow?

b

22.3 A material has an index of refraction that increases continuously from top to bottom. Of the three paths shown in figure, which path will a light ray follow as it passes through the material?

b

23.4 A clear plastic sandwich bag filled with water can act as a crude converging lens in air. if the bag is filled with air and placed under water, is the effective lens (a) converging or (b) diverging

b

24.4 Suppose Young's experiment is carried out in air, and then, in a second experiment, the apparatus is immersed in water. In what way does the distance between bright fringes change? (a) They move farther apart. (b) They move closer together. (c) There is no change.

b

24.5 In a single-slit diffraction experiment, as the width of the slit is made smaller, does the width of the central maximum of the diffraction pattern (a) become smaller, (b) become larger, or (c) remain the same?

b

3. A plane monochromatic light wave is incident on a double slit as illustrated in figure. As the slit separation decreases, what happens to the separation between the interference fringes on the screen? a) it decreases b) it increases c) it remains the same d) it may increase or decrease, depending on the wavelength of the light e) more information is required

b as the slit separation decreases, the separation between the interference fringes on the screen increase. This is because y = lambda x L x m/d. If m and lambda stay the same while d decreases, sine theta will increase meaning that the distance will increase

7. Choose the option from each pair that makes the following statement correct. For a nearsighted person, the [(a) near point; (B) far point] is always located closer than [(c) infinity; (d) 25 cm] from the eye and the corrective lens is [(e) converging; (f) diverging].

b, c, f for a nearsighted person, the far point is always located closer than infinity from the eye and the corrective lens is diverging. Far point is closer because they can't see far away.

14. A person spear fishing from a boat sees a stationary fish a few meters away in a direction about 30 degrees below the horizontal. To spear the fish, and assuming the spear does not change direction when it enters the water, should the person a) aim above where he sees the fish b) aim below the fish c) aim precisely at the fish

b. aim below the fish If you are using a bow and arrow to spear the fish you should realize that the arrow will not deviate as it enters the water; it will continue along the original straight line path from the bow. Therefore, in order to strike the fish you should aim below where the fish appears to be. WE are seeing the fish at 30 degrees. This means the fish is 60 degrees to the normal. Use n1sintheta1=n2sintheta2. This will n1sintheta1/1.33 = (do the arc sin of this) to find theta 2. We get 40 degrees. Therefore you need to aim below where the fish actually appears to be. if n1>n2 then it will refract to be a bigger angle, theta 1 is less than theta 2. if n2> n1 then theta 1 is > theta 2. Because air is a smaller n than water, the angle in the air will be bigger than the actual angle in the water in relation to normal!

15. If you increase the aperture diameter of a camera by a factor of 3, how is the intensity of the light striking the film affected? a) it increases by a factor of 3 b) it decreases by a factor of 3 c) it increases by a factor of 9 d) it decreases by a factor of 9 e) increasing the aperture doesn't affect the intensity

c

23.2 A person spearfishing from a boat sees a fish located 3 m from the boat at an apparent depth of 1 m. To the spear fish, should the person (a) at, (b) above, or (c) below the image of the fish?

c

24.1 In a two slit interference pattern projected on a screen, are the fringes equally spaced on the screen (a) everywhere, (b) only for large angles, or (c) only for small angles

c

22.4 As light travels from a vacuum (n=1) to a medium such as glass (n>1), which of the following properties remains the same: the (a) wavelength, (b) wave speed, or (c) frequency?

c as light travels from one medium to another, its frequency doesn't change

Light in medium A undergoes a total internal reflection as it reaches the interface with medium B. Which of the following statements must be true? a) nB<nA b) nB>nA c) all light rays that undergo a total internal reflection travel along the interface between the two materials d) light traveling in the opposite direction, from B into A, cannot undergo a total internal reflection

c at the critical angle it will just travel along the surface of the two materials.

24.3 A young's double slit experiment is performed with three different colors of light: red, green, and blue. Rank the colors by distance between adjacent bright fringes from smallest to largest. (a) red, green, blue, (b) green, blue, red, (c) blue, green, red

c blue has a smaller wavelength than red. the smaller the wavelength, the smaller the distance is between the fringes.

25.1 Two campers wish to start a fire during the day. One camper is nearsighted and one is farsighted. Whose glasses should be used to focus the Sun's rays onto some paper to start the fire? a) either camper's b) the nearsighted camper's c) the farsighted camper's

c farsighted glasses will converge. Converging all of the rays into one spot will likely start a spark

15. A light ray containing both blue and red wavelength is incident at an angle on a slab of glass. Which of the sketches in Figure represents the most likely outcome? a) A b) B c) C d) D e) none of these

c the higher index of refraction means that the wavelength is reduced. Blue since it has lower wavelength it will have a bigger index of refraction thus being closer to normal. Red has a higher wavelength thus it will have a smaller index of refraction and be farther away from normal.

24.2 If the distance between the slits is doubled in Young's experiment, what happens to the width of the central maximum? (a) the width is doubled, (b) the width is unchanged, (c) the width is halved

c y bright = (lambda)Lm/d It is halved since we are doubling the distance between the slits. central maxima is referring to the bright maximum

Figure shows light from material A with index of refraction nA entering materials B and C with indices of refraction nB and nC. Rank the three indices of refraction from largest to smallest a) nA,nB,nC b) nB,nC,nA c) nC,nA,nB d) nB,nA,nC e) nC,nB,nA

d B is the largest, A is the next largest, C is the next largest. if n1 is greater than n2, then sin theta 2 is greater than sin theta 1

11. A patient has a near point of 1.25 m. Is she nearsighted or farsighted? Should the corrective lens be converging or diverging?

farsighted; converging

The index of refraction

n = c/v n = (speed of light in vacuum)/(speed of light in a medium) n will always be a number greater than or equal to 1 because v is always less than c. n is equal to one for vacuum As light travels from one medium to another, its frequency doesn't change v = f(lamba)

3. A light ray travels through three parallel slabs having different indices of refraction. The rays shown are only the refracted rays. Rank the materials according to the size of their indices of refraction, from largest to smallest.

nA>nC>nB A has the largest refraction because it is the more vertical - largest angle C has the next largest refraction because it is bigger angle than B

Colors in order of DECREASING wavelength

red orange yellow green blue violet

24.6 If laser light is reflected from a phonograph record or a compact disc, a diffraction pattern appears. The pattern arises because both devices contain parallel tracks of information that act as a reflection diffraction grating. Which device, record or compact disc, results in diffraction maxima that are farther apart?

the compact disc

During LASIK eye surgery (laser- assisted in situ keratomileusis), the shape of the cornea is modified by vaporizing some of its material. If the surgery is performed to correct for nearsightedness, how does the cornea need to be reshaped?

the cornea needs to become more concaved to diverge light