Velocity Related Problem and Its Solution

Chapter 5
Solution.1
Velocity of light in air = 186,282 miles / sec or 299,792 km/s
Light speed = 670,616,629 mph.
Hence the lights can transverse 7.5 times round the Earth in a second. (Redd)
Solution.2
Highest energy to lowest energy Sequence:
a. Gamma rays
b. X-rays
c. Ultraviolet rays
d. Red visible light
e. Blue visible light
f. Infrared rays
g. Radio waves
Solution.3
(a) All the mentioned light waves above are having same speed in the medium of air (c=
3 ×108 m/s). Consequently, the proportion of the velocity of ultraviolet and
infrared waves is 1:1.
(b) In the air, all mentioned light waves travel on similar velocity without any difficulty
considering the wavelength or regularity. Gamma emission has a minor wavelength
and higher frequency, radio emission has a minor frequency and higher wavelength.
So, gamma rays, X-rays, and microwaves which travel in the air have similar speeds
but different wavelengths.
(c) For Yellow visible light, the wavelength is 570-590nm, and for Blue visible light, the
wavelength is 450-495nm. So yellow visible light is the fastest through the vacuum.

Solution.4
Astrophysicists utilize the Doppler Effect to study the gesticulation of matters across the
World, from nearby solar spheres to the development of detached galaxies. Doppler shift is
the modification in the distance of a wave (sound, light, etc.) owing to the comparative
gesticulation of source and receiver. Things heartrending away have their released
wavelengths elongated.
Doppler Effect:
The Doppler’s effect is the increase and decrease of the frequency in different waveforms
such as sound, light etc. due to the movement of source and observer towards each other. The
impact of this effect can be observed by the instant change in the pitch of the siren.
Solution.5
(a) The net electromagnetic radiation of sun at Earth (1 AU) = 1361 Watts per square
meter
For mercury:
The net electromagnetic radiation of sun = 0.4 AU
The net electromagnetic radiation of sun received by mercury = 1361 × 0.4 w/m2
The net electromagnetic radiation of sun received by mercury = 544.4 w/m2
(b) For Jupiter:
The net electromagnetic radiation of sun = 5.2AU
The net electromagnetic radiation of sun received by Jupiter = 1361 × 5.2 w/m2
The net electromagnetic radiation of sun received by Jupiter = 7077.2 w/m2

Chapter 6
Solution.1
The applications of the eyepiece are shown below: (Freudenrich )
 create and permit for variation in the telescope's exaggeration
 create a piercing image
 deliver contented eye aid (the space among your eye and the eyepiece while the image
is in concentration)
 regulate the telescope's Ground of opinion is realized edge-to-edge over the eyepiece
unaided (quantified on the eyepiece; factual or tangible - how much of the atmosphere
can be realized when that eyepiece is positioned in the telescope (accurate field =
ostensible field/magnification)
Aspects of the eyepiece for determining the magnification of the image:
It permits you to control the magnification of an eyepiece which subsidizes the
amalgamation with an assumed telescope. Magnification is resolute merely by separating
the focal length of the telescope through the focal length of the eyepiece. This means that
a minor number on an eyepiece contributes towards a higher magnification.
Solution.2
Magnifying power is a depraved standard for assessing telescopes. If a telescope has less
fortitude, it creates the subsequent image bigger which does not work properly, because it
will be due to the heavy lenses. And these heavy lenses are difficult to put in the right
place. Also if the glass will be thick it will resist the maximum light passing through it.

Solution.3
A spectrograph is a tool which divides the light waves by using its wavelengths and
archives these facts. A spectrograph characteristically has a multi-channel indicator
scheme or camera that notices and archives the range of light.
Numerous astronomers respect it as the greatest significant device that can be devoted to
a telescope due to the reason that they can disclose the chemical arrangement and
temperature.
Solution.4
X-ray telescope is an instrument which is used to observe and resolve X-rays by finding them
externally at the Earth's sphere. Due to atmospheric pre-occupation, X-ray telescopes must be
approved to higher elevations by using several objects such as: rockets, balloons or
by locating in the external orbit to the atmosphere.
The Ground's atmosphere strews or captivates high-energy radioactivity, defending us from
the destructive properties of X-rays, UV waves, and gamma waves. The atmosphere does a
decent work that telescopes intend to identify these rations of the electromagnetic range have
to be located externally from the atmosphere.
Solution.5
(a) Given:
Focal length of telescope = 1200mm
Mirror diameter = 203mm
The limiting magnitude of the telescope, Lim mv = 2 +5 D
Lim mv = 2 + (5× 203)
Lim mv = 1017
(b) Given:
Telescope’s Focal length = 1200mm
Eyepiece’s Focal length = 25mm
Magnification = (Telescope’s Focal length) ÷ (Eyepiece’s Focal length)

(Ashford and Tytell)
Magnification = 1200/25 = 48mm
(c) Given:
Telescope’s Focal length = 1200mm
Magnification = 150mm
Magnification = (Telescope’s Focal length) ÷ (Eyepiece’s Focal length)
Eyepiece’s Focal length = (Telescope’s Focal length) ÷ (Magnification)
Eyepiece’s Focal length = 1200 ÷150 = 8mm
References
Redd, Nola taylor " How fast does light travel? | the speed of light. " Space.com. Web 11
february 2020. 2018. < https://www.space.com/15830-light-speed.html. >.
Freudenrich, Craig " How telescopes work. " Howstuffworks. Web 11 february 2020. <
https://science.howstuffworks.com/telescope6.htm. >.
Ashford, Adrian r , David Tytell, " Telescope calculator: how does your telescope perform?. "
Sky and telescope. Web 11 february 2020. 2017. <
https://www.skyandtelescope.com/observing/skyandtelescope-coms-scope-
calculator/. >.