Lab3_Light_and_Telescopes
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343M
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Astronomy
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Dec 6, 2023
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: Eric Chung
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Date
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LIGHT & TELESCOPES
Activity 1: Waves
Astronomers use light to study celestial objects, but the light comes in many forms.
The
most familiar form is visible light, the form of electromagnetic radiation that we can see.
The full spectrum of electromagnetic radiation extends to much longer and too much
shorter wavelengths that we can see, however.
Radio waves, microwaves, infrared,
visible, and ultraviolet light, x-rays, and gamma rays are all forms of electromagnetic
radiation.
Light is an example of a wave, and there are many types of waves in nature.
Examples
of familiar waves are water waves, sound waves, and light waves (and don’t forget
gravity waves!).
Waves are characterized by three parameters: wavelength, frequency
(waves per second), and speed, and these properties are related by a simple
expression. (See module: Nature of Light)
Speed = Wavelength x Frequency
Speed is measured as distance traveled per second.
Wavelength is the length of a wave from peak to peak.
Frequency is the number of waves per second that pass a given point.
Example 1
:
What is the wavelength of a typical sound wave?
The frequency of, say,
middle C is about 256-278 cycles per second depending on the scale, and sound
travels at about 340 meters per second.
F1 = 256 Hz
F2 = 278 Hz
V = 340 m/s
W1 = V/f1 = 1.32 m
W2 = V/f2 = 1.22 m
It’s between 1.328 and 1.223 meters.
Example 2
: A tsunami wave travels at a speed of about 0.2 meters/second, with a
wavelength of about 50,000 m.
What will be the time between peaks of a tsunami wave
coming ashore on the beach?
V = 0.2 m/s
w = 50,000 m
1
f = V/w = 4 x 10^-6 Hz
The time between peaks is 0.000004 Hertz.
Example 3
:
Electromagnetic waves travel with speed of about 300,000 km per second
(3 x 10
8
meters per second).
What is the wavelength of an electromagnetic wave with a
frequency of one billion cycles per second (10
9
cycles per second). What kind of light is
this?
V = 3 x 10^8 m
f = 10^9 Hz
w = V/f = 0.3 m >> radio light (radio wave)
The wavelength is 0.3 meters. This type of light is called radio light, or radio waves.
Example 4
:
Gravity waves also travel at the speed of light (3 x 10
8
meters per second).
Two neutron stars, each with mass equal to two solar masses, orbiting each other with a
separation of 0.63 light seconds and a period of 750 seconds (frequency = 0.0013 per
second), will emit gravity waves.
What is the wavelength of these gravity waves?
V = 3 x 10^8 m/s
f = 0.0013/s
W = S/f = 2.31 x 10^12
It’s 2.308 x 10^12 meters.
Gravitational waves (or gravity waves) are disturbances or ripples in the curvature of
spacetime, generated by accelerated masses, that propagate as waves outward from
their source at the speed of light. In general, gravitational wave frequencies are much
lower than those of the electromagnetic spectrum (a few thousand hertz at most,
compared to some 10
16
to 10
19
Hz for X-rays). Consequently, they have much larger
wavelengths – ranging from hundreds of kilometers to potentially the span of the
Universe.
Activity 2: Analyzing Infrared Images
Examine the three pairs of optical and infrared images of the Old Faithful geyser in
Yellowstone National Park (Old Faithful is the most frequently erupting large geyser in
the park). A geyser is a hot spring which erupts periodically.
These eruptions are
caused by the buildup of hot water and steam trapped by constrictions in the "plumbing
system" of a hot spring.
When enough pressure builds up the geyser erupts.
The three
image pairs are a time sequence from the beginning to the end of the geyser’s eruption.
The infrared images are shown in “pseudocolor” since our eyes cannot see infrared
light.
Color corresponds to temperature with the hottest parts of the image shown as
white light and the coolest parts shown as black.
2
1.
Which regions of the image are the coolest?
From black to blue are the coolest. This is where the geyser is actively erupting.
2.
Which regions are the hottest?
From yellow to white are the hottest. This is where the geyser is not erupting.
3.
Do the infrared images give you information that you cannot get from the visible light
images?
Yes, they let us see temperature differences within images. This allows us to know how
hot the geyser’s eruption is.
4.
Describe the difference between pseudo color and true color.
In what circumstances
would pseudo color be useful?
Pseudo color lets us visualize things that we cannot see with our own eyes. Pseudo
color is helpful when needing to document such aspects like the temperature of the
geyser. True color is the actual colors of objects as seen by our naked eye.
5.
What would astronomers learn from observations of astronomical objects in infrared
light, compared to observations in visible light?
They can learn the temperatures of celestial objects or regions, and cross reference that
with others. Infrared can also be used to study wavelengths, stars, invisible to the
human eye, and more.
Visible Light Images
Infrared Images
3
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The Invisible Yellowstone National Park© Cool Cosmos at Spitzer Science Center
http://coolcosmos.ipac.caltech.edu
Activity 3: Estimating Temperature
Wien’s Law, the wavelength (color) at which an object radiates most strongly is
inversely proportional to the object’s temperature.
4
From the Wien’s Law,
Wavelength(max) =
2,900,000 / Temperature
Wavelengths are measured in nanometers, and temperatures in degrees Kelvin.
1-
Using the infrared camera or a thermometer, you can measure the temperatures
of materials available in the room such as ice, walls, water, hot water, skin,
incandescent light bulb. Their temperatures are already given. Using Wien’s Law,
you can determine the wavelength at which each material emits the most thermal
radiation. Fill in your answers in the table below.
2-
The Sun is the brightest in green light, about 500 nanometers.
What is its
temperature? Fill in your answers in the table below.
3-
3-
What is the source of the light that we
see as visible light?
The sun, stars, supernovae, nebula, gas, dust, planets, asteroids. Etc.
4-
What is the source of the light that the infrared camera sees?
5
Material
Temperature
(K)
Peak
Wavelength
(nm)
Ice
273.15
W =
2,900,000/273.15
= 10617 nm
infrared
Walls
296.95
9,766 infrared
Skin
306.15
9,472 infrared
Hot
Water
333.15
8,705 infrared
Light
Bulb
2810.93
1032 infrared
The Sun
5788
501
Sun, dust, star, gas.
5-
A microwave telescope “sees” microwave light coming from all directions in the
sky.
This microwave signal is fairly strong, accounting for about 1% of the “noise”
detected by a television antenna set “off channel.”
A plot of the brightness of the
microwave emission vs. wavelength is shown below.
From Wien’s law and the
temperature, estimate the wavelength maximum at which the microwave
emission is brightest.
w = 2,900,000/2.7 = 1074074.074 nm = 1.07 m
6-
The brightness of the microwave radiation in a small patch of sky is shown below
in pseudocolor. What is the range of temperature observed in the microwave
radiation?
T = 2.721 K – 2.729 K (scientific paper)
How much would this difference in temperature shift the wavelength of the peak
of the spectrum shown above?
W = 1065785 nm – 1062660 nm = 1.066 – 1.063 m
(1 nm = 1*10^-9 m)
6
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Activity 4: Telescopes
To find the minimum size a telescope needs to be resolve detail of a particular angular
extent is
D
>
0.025
λ
/
α
α
is the separation angle in arc seconds
λ
is the wavelength of light in nanometers (nm)
D is the diameter of the telescope in centimeters (cm)
1-
The diameter of a person’s pupil when fully adapted to the dark is typically about 8
millimeters. This is the diameter of the aperture of the eye. By contrast, some of the
largest telescopes have aperture diameters of about 8 meters. How does their light-
gathering power compare? (Hint: calculate the area of a circle)
Eye = Pi * R^2 = 3.14 * (8/2)^2 = 50.27 mm^2
Telescope = 50,270 mm^2
Ratio= 1:1000
2-
Compare the light-gathering power of a telescope with a 10-centimeter (about 4-
inch) diameter mirror to that of the human eye. (Take the diameter of the pupil of the
eye to be about 5 millimeters and assume a wavelength in the middle of the visible
500 nm.)
w = 500 nm
D telescope = 10 cm
D eye= 5mm = 0.5 cm
Telescope – alpha > 0.025 * 500/10 = 1.25 arcsec
Eye – alpha > 0.025 * 500/0.5 = 25 arcsec
Ratio = 1:20
So the telescope can therefore detect objects as much as 20 times smaller than your
eye can discern.
7
3-
Can the unaided human eye resolve a crater on the Moon whose angular diameter
is 2 minutes of arc (= 120 seconds of arc)? (Take the diameter of the pupil of the
eye to be about 5 millimeters and the wavelength of the light to be 500 nanometers.)
w = 500 nm
D eye = 5mm = 0.5 cm
Eye – alpha > 0.025 * 500/0.5 = 25 arcsec
That would be the minimum thing we would be able to see if crater on moon 120 arcsec
> 25 arcsec, Answer = Yes
4-
Determine the resolving power of a 25-meter radio telescope observing 10-
centimeter radio waves. Compare this to its resolving power for 1-meter radio
waves. (Remember to convert units for the equation above)
w1 = 10 cm = 10^8 nm
w2 = 1 m = 10^9 nm
alpha 1 > 0.025 * w1/ D = 1,000 arcsec
alpha 2 > 0.025 * w2/ D = 10,000 arcsec
ratio = 1:10
5-
The Very Large Array is a radio interferometer observatory in New Mexico with
twenty-seven 25-meter telescopes. In its widest arrangement, it has the resolving
power of a telescope 36 kilometers in diameter. What is better about this
arrangement than a single 36-kilometer diameter telescope? What is lacking
compared to a single 36-kilometer dish?
Array vs Single Dish (36 km diameter)
Advantages: Better resolving power and individual telescopes can be used to
observe different parts of the sky simultaneously.
Disadvantages: Some images are incomplete and takes up a lot more space.
6-
Is the Dobsonian telescope a refracting or reflecting telescope?
Reflecting telescope.
7-
On the “8-inch Dobsonian” telescopes, which part measures 8 inches? What
advantages does an 8-inch telescope have over, say, a 4-inch telescope?
The diameter of the mirror is 8 inches. It’s clearer, it will go deeper and show more, and
is brighter.
8