Download Physics 101 – HW#10 Solutions – Koskelo

Physics 101 – HW#10 Solutions – Koskelo
1.)
a.) Light slows down when it enters a material. What is actually going on (on a microscopic
scale) that causes this to happen?
The light is actually passed from one atom to the next in the material, which creates a noticeable
decrease in the light’s speed. (see figure 26.7, p. 501 of your text for a good picture of the
process.)
We learned that the index of refraction (n) of a material tells you just how much the speed of light
will decrease in that material. For instance, n is about 1.5 for glass, and n=1.33 for water. In
which material does light travel faster?
The index of refraction of a material (n) gives the factor by which the speed of light is reduced in
that material. The speed of light in a given material is given by the formula:
v = c/n
where v=speed of light in the material, c=speed of light in a vacuum (3·108 m/s) and n is the
index of refraction of the material.
So since n is larger for glass, light will travel slower in glass, and faster in water.
In 2003, physicists at University of Rochester passed light through a special type of ruby which
had an effective index of refraction of about 5 million. How fast does light travel through this
ruby?
v = c/n
so the speed of light in this ruby = 300,000,000 m/s / 5,000,000 = 60 m/s.
This is a huge reduction in light speed. While light normally travels around the Earth in the blink
of an eye, this ruby slows down the light so much that you could beat it in a race with a fast
automobile. (This kind of “slow light” looks like it will have many future applications in the
telecommunications industry.)
2.) Red light of frequency 4.3·1014 Hz travels through a vacuum.
a.) How fast does it travel?
Light in a vacuum (regardless of its frequency) travels at v = 3·108 m/s
This vacuum speed is often labelled ‘c’.
b.) What is its wavelength?
Using the equation v = f λ which relates any wave’s speed, frequency and wavelength, you can
plug in v=3·108 m/s and f=4.3·1014 Hz to get
λ = 7·10-7 m (this wavelength in a vacuum is often called λ0 )
This light then enters a material where the index of refraction n=2.4. When the light passes into
this material,
c.) What is the speed of the light?
Light slows down when it enters a material, by v = c / n. Plugging in c=3·108 m/s and n=2.4, you
get v=1.25 ·108 m/s
d.) What is its wavelength?
Since the wave slows down entering the material, there will be a traffic jam where the
wavelengths crunch up and get reduced, by the formula λn = λ0 / n where λn and λ0 are the
wavelengths of the light in the material and in the vacuum respectively. Plugging in ‘n’ and λ0
from part b: λn =2.9·10-7 m
e.) What is its frequency?
Mathematically:
Using the equation v = f λ using the new v and λ values from parts c & d will actually yield the
exact same frequency value as in the vacuum = 4.3·1014 Hz.
Conceptually:
The light slows down when it enters the material, but the crests of the wave are also closer
together (smaller wavelength) since they bunch up. So it turns out that the frequency at which
crests arrive at you in the material is still the same as it was out in the vacuum.
f.) What color does the light appear to be in the material?
Your eye interprets a light’s frequency (not speed, not wavelength) to determine its color.
Since light does not change its frequency entering a material (though it does change both speed
and wavelength), you will see the same color as before. This should gel with your everyday
experience—a red ball in air looks red underwater as well.
3.)
a.) Light will always take the shortest time between two points, not the shortest distance (this is
known as Fermat’s Principle). Explain how these two things can be different, in the case of
refraction.
The shortest path between point A and point B is a straight line, but if this path has two different
materials within it (in which light has two different speeds), the quickest way for light to get from
point A to point B is not necessarily the shortest distance.
In a situation involving two different materials, it turns out that the fastest way for light to get
from point A to point B is to spend a little more time in the material in which it can move faster
(smaller n), and a little less time in the material in which it moves slower (larger n). This is what
gives rise to the light bending (refracting) when it passes from one material to another.
(Remember the David Hasselhoff example from lecture.)
b.) Light passes from air into water. Fill in the following:
The light will SLOW DOWN as it passes into the water.
The light will bend TOWARDS the normal (of the interface between the materials.)
Light travels faster in air (n=1) than in water (n=1.33), which makes intuitive sense since water is
a lot more crowded (molecule-wise) and it’s harder for light to travel in it.
As a result, the light spends more time in the air, and less in the water, which means the light will
bend towards the normal when it enters the water:
c.) Light passes from an n=1.9 material into an n=1.4 material. Fill in the following:
The light will SPEED UP when it enters the second material.
The light will bend AWAY FROM the normal (of the interface between the materials.)
This is the opposite of the above. The light travels slower in a n=1.9 material than in a n=1.4
material, so the light will spend want to spend less time in the first material and more in the
second, to minimize the time spent getting from point A to point B. This means the light will
bend away from the normal.
If the n of the first material was instead 2.1, then the amount by which the light was bent would
be MORE THAN BEFORE.
If the n values of the two materials were 2.1 & 1.4 (instead of 1.9 & 1.4), that would mean that
the n values would have a bigger difference—which would mean that the difference in speeds of
light in each material would be bigger. This would result in a more extreme bending of light,
away from the normal—since the light would be that much more inclined to get the heck out of
the n=2.1 material (very slow) into the n=1.4 material (faster).
(Can you see the general relationship between how much the n values differ, and how much the
light is bent?)
In general, the bigger the differences in the two materials’ n values, the more extreme the
bending (either towards the normal, when going from lower to higher n, or away from the
normal, when going from higher n to lower n.)
4.) Place a glass test tube in water and you can see the tube. Place it in clear soybean oil, and
you may not be able to see it. What does this tell you about the speed of light in the oil and in the
glass?
The reason why you can see something transparent (glass) inside something else that’s
transparent (water) is because they have different n values—so light refracts when passing
through the water into/out of the glass. Your eyes recognize that the light passing through has
been bent (distorts the shadows and objects you can see behind the transparent material), and you
recognize that there’s something in the water.
But, soybean oil has a nearly identical n value as water, so when light passes from the oil into/out
of the test tube, it doesn’t refract at all. Therefore, you can’t tell that there is something there. In
order to see the clear test tube within a clear fluid, light needs to somehow interact differently
with the two materials.
5.)
a.) If while standing on a bank you wished to spear a fish out in front of you, would you aim
above, below, or directly at the observed fish to make a direct hit?
You should aim below the observed fish. Your eyes are being deceived as to the real location of
the fish, because the light coming from the fish is refracting (away from the normal) as it passes
into the air and to your eye (see figure 28.27, p. 540 in your text). Hence, you perceive the fish as
being shallower than it really is.
So if you throw your spear (which will travel in a straight line) directly at where you perceive the
fish to be, the spear will pass above its head. You need to aim lower than that.
If instead you zapped the fish with a laser, would you aim above, below, or directly at the
observed fish?
You should aim directly at the observed fish. The laser light will refract as it enters the water,
following the exact reverse path as the light coming from the fish to your eye. Interestingly, this
approach allows you to be completely ignorant of refraction, and hit the fish by dumb luck, just
by aiming at it.
6.) When your eye is submerged in water, is the bending of light rays from water to your eyes
more, less, or the same as in air?
Remember the discussion from the very end of #5— the more the difference in the n values, the
more the bending. Let’s compare:
normal eye usage: light goes from air (n=1) into eye fluid (n>1... the fluid in the eye is very close
to water.)
underwater eye usage: light goes from water (n=1.33) into eye fluid (n=about the same).
In this case, the n difference is much less, so the bending of light going from water into your eye
is less.
This is why your eyes go out of focus underwater: because the light rays don’t bend the same as
usual, when passing into your eye—hence the image formed by your eyeball won’t fall quite in
the right spot inside your eye (see #12)...
Why will goggles allow a swimmer under water to focus more clearly on what he is looking at?
... but you can use goggles (filled with air) to make sure that light enters your eye from air, as
you’ve evolved to do. Your eyes have evolved to account for this air-to-fluid refraction...
Design a pair of goggles for a fish to wear, so he can see clearly when out of water.
...of course, fish have evolved to utilize light entering their eyes directly from water, so if a fish
were to come into the air and didn’t want to see blurry (like we do underwater), the fish would
need to wear goggles filled with water!
7.) Concave mirror: You stand at a reasonably far away distance from a concave mirror (caves
in towards you).Your reflected image will be:
Upside down. The nature of the reflections off a concave mirror, as shown in class, will be to
bend the light rays inward, so the top light rays coming from the object get bounced out near the
bottom of the mirror, and vice versa. This results in an upside down image. Try it by looking on
the concave side of a shiny spoon (side the food goes on).
Interestingly though, if you bring an object close enough to the concave mirror, the light rays
from the source spread out so much that the concave mirror cannot bend them inward any more.
In this case, the concave mirror acts more like a plane mirror and the image will be right side up.
This is why I needed to say “a reasonably far distance” in the statement of the problem.
Convex mirror: You stand in front of a convex mirror (bends away from you).
Your reflected image will be:
Right side up. The nature of the reflections off a convex mirror, as shown in class, will be to
bend the light rays outwards, so the top light rays coming from the object stay on top, and the
bottom light rays from the object stay at the bottom. This will mean the image created will be in
the same orientation (not flipped) to the original object’s. Try it by looking at the convex side of a
shiny spoon (opposite the side the food goes in).
8.) The location where the image is formed by a lens depends on how far away the object is.
Explain the differences in how a camera and a human eye make an image fall in the desired
location.
In a camera, you literally move the lens back and forth, so that the image will always fall on the
right spot (film in the back of the camera)—even though different object distances will make the
image fall at different distances behind the lens.
A more distance object’s light rays arrive at the camera lens essentially parallel, so the
converging lens can converge them over a shorter distance. The film can be closer to the lens.
A near object’s light rays are spreading out more, so the converging lens takes a longer distance
over which to converge them. The film must be further back from the lens. Next time you take a
photo of a nearby object with an automatic camera, note that the lens moves outwards from the
camera, to create more distance between the lens and film.
We obviously can’t zoom the lens of our eye back and forth. Instead, our eye muscles (called
ciliary muscles) squeezes the lens of our eye into a different shape, when we look at nearby and
far away objects. (When these muscles get tired, they’re no longer able to push the lens into the
tight shape required to focus on near objects—perhaps that is the case when you are reading this
right now!)
A far-sighted person (one who cannot see near) needs glasses with a converging lens, because
without them, the image formed by the lens in their eye FALLS BEHIND THE BACK OF THE
EYEBALL.
A converging lens (thicker in the middle) helps to “converge” the light, so if the image formed by
the eyeball is too far back (past the back of the eyeball—most likely because this person’s eyeball
is somewhat short), a converging lens can help the image fall closer than before—directly on the
back of the eyeball, where it should be, so we can see it clearly.
A near-sighted person (one who cannot see far) needs glasses with a diverging lens, because
without them, the image formed by the lens in their eye IS NOT FAR ENOUGH BACK IN
THEIR EYEBALL.
A diverging lens (thinner in the middle) helps to “diverge” the light a little, so if the image
formed by the eyeball is not far enough back in the eyeball (most likely because this person’s
eyeball is too elongated), a diverging lens can help the image fall further back than before—right
on the back of the eyeball, where it should be so we can see it clearly.
A person who is both far and near sighted needs to wears bifocals. Bifocals have one type of lens
on the top half of the glasses, and the other type of lens on the bottom half. Which lens is on
which half? (Hint: You can often see people who wear bifocals staring down their noses to read a
book.)
Someone who is reading a book is clearly looking through the bottom half of the bifocals—so the
bottom half of the glasses is helping them to see NEAR.This requires a converging lens, as we
figured out above. Vice versa for the top lens. So:
top lens: diverging lens (can see far now - corrects nearsightedness)
bottom lens: converging lens (can see near now - corrects farsightedness)
You could just as easily build bifocals with the top and bottom lens switched—but then someone
who wore them would have to hold a book above their head when reading, and conversely, tilt
their head back and look down their nose to look into the distance!
9.)
a.) No, we wouldn’t. If all the colors contained in the white light refracted the same way, they
would stay together and we would only see white light, not colors separated out into a rainbow.
b.) This must mean that violet:
· experiences THE LARGEST index of refraction of all the colors.
When passing from a lower to higher n material, all light bends towards the normal, but violet
gets bent the most towards it. Conversely, when passing from a higher to lower n material, all
light bends away from the normal, but violet gets bent the most away from it. So violet always
gets bent the most extremely, in any given situation.
This must mean that violet sees the largest value for the index of refraction (n) than any of the
other colors--- since violet gets bent the most, it must experience the n value most different
(greatest) in the raindrop, as compared to n=1 for air.
· travels the SLOWEST of all the colors when passing through the raindrop.
Largest n means slowest speed (v = c/n).
· if you're on the ground looking at the rainbow, you will have to look MORE HORIZONTALLY
to see violet.
Since violet is bent more extremely, its path (as shown in lecture) will be bent more sharply away
from the original direction of the light—in fact, come closest to going exactly 180 o (also
horizontal) from its original direction.... violet comes out only 40o shy of exactly opposite its
original path, whereas red (not bent as much) comes out 42o shy of exactly the opposite of its
original path.
· hence violet will always appear at the BOTTOM of a rainbow.
... so the sharper bend in violet’s path will cause it to come out more horizontally, and you will
need to look more horizontally (40o) in the sky to see this color emanating from water droplets.
That’s the bottom of the rainbow. You will need to look higher in the sky (at 42o) to see the top of
the rainbow, which is red.
Also, as a point of interest, rainbows are circular—if the horizon wasn’t blocking your view of it
(such as when in an airplane), you would see violet from all water droplets that were 40o to you,
including below the horizon (and red everywhere from water droplets that were 42o to you.) So
there is no such thing as the end of a rainbow!