You are already familiar with some direct methods
for the measurement of length. For example, a
metre scale is used for lengths from $10^{–3}$ m to $10^2$ m. A vernier callipers is used for lengths to an
accuracy of $10^{–4}$ m. A screw gauge and a
spherometer can be used to measure lengths as
less as to $10^{–5}$ m. To measure lengths beyond these
ranges, we make use of some special indirect
methods.
Measurement of Large Distances Large distances such as the distance of a planet or a star from the earth cannot be measured directly with a metre scale. An important method in such cases is the parallax method. When you hold a pencil in front of you against some specific point on the background (a wall) and look at the pencil first through your left eye A (closing the right eye) and then look at the pencil through your right eye B (closing the left eye), you would notice that the position of the pencil seems to change with respect to the point on the wall. This is called parallax.
The
distance between the two points of observation
is called the basis. In this example, the basis is
the distance between the eyes.
To measure the distance D of a far away
planet S by the parallax method, we observe it
from two different positions (observatories) A and
B on the Earth, separated by distance AB = b
at the same time as shown in Fig. 2.2.
We
measure the angle between the two directions
along which the planet is viewed at these two
points. The ∠ASB in Fig. 2.2 represented by
symbol θ is called the parallax angle or
parallactic angle.
As the planet is very far away, $\frac{b}{D} << 1$ and therefore, θ is very small. Then we approximately take AB as an arc of length b of a circle with centre at S and the distance D as the radius AS = BS so that AB = b = Dθ where θ is in radians.
$D = \frac{b}{θ}$
Having determined D, we can employ a similar
method to determine the size or angular diameter
of the planet. If d is the diameter of the planet
and α the angular size of the planet (the angle
subtended by d at the earth), we have
α = d/D (2.2)
The angle α can be measured from the same
location on the earth. It is the angle between
the two directions when two diametrically
opposite points of the planet are viewed through
the telescope. Since D is known, the diameter d
of the planet can be determined using Eq. (2.2).
Example 1
Calculate the angle of (a) 10 (degree) (b) 1′ (minute of arc or arcmin) and (c) 1″(second of arc or arc second) in radians. Use $360^0$ = 2π rad, 10 = 60′ and 1′ = 60 ″
Calculate the angle of (a) 10 (degree) (b) 1′ (minute of arc or arcmin) and (c) 1″(second of arc or arc second) in radians. Use $360^0$ = 2π rad, 10 = 60′ and 1′ = 60 ″
Answer
(a) We have $360^0$ = 2π rad
10 = (π /180) rad = 1.745 × $10^{–2}$ rad
(b) 10 = 60′ = 1.745 × $10^{–2}$ rad
1′ = 2.908 × $10^{–4}$ rad = 2.91 × $10^{–4}$ rad
(c) 1′ = 60″ = 2.908 × $10^{–4}$ rad
1″ = 4.847 × $10^{–4}$ rad = 4.85 × $10^{–6}$ rad.
Example 2
A man wishes to estimate
the distance of a nearby tower from him.
He stands at a point A in front of the tower
C and spots a very distant object O in line
with AC. He then walks perpendicular to
AC up to B, a distance of 100 m, and looks
at O and C again. Since O is very distant,
the direction BO is practically the same as AO; but he finds the line of sight of C shifted
from the original line of sight by an angle θ
= $40^{0}$ (θ is known as ‘parallax’) estimate
the distance of the tower C from his original
position A.
 |
| Fig.2 |
Answer
We have, parallax angle θ = $40^{0}$ From Fig. 2, AB = AC tan θ
AC = AB/tanθ = 100 m/tan $40^0$ = 100 m/0.8391 = 119 m
Example 3
The moon is observed from
two diametrically opposite points A and B
on Earth. The angle θ subtended at the
moon by the two directions of observation
is 1o 54′. Given the diameter of the Earth to
be about 1.276 × $10^7$ m, compute the
distance of the moon from the Earth.
Answer
We have θ = 1°54′
= 114′
= (114 × 60)'' × (4.85 × $10^{-6}$ ) rad
= 3.32 × $10^{-2}$ rad,
since 1" = 4.85 × $10^{-6}$ rad
Also b = AB = 1.276 × $10^7$ m
Hence from Eq. (2.1), we have the earth-moon
distance,
D = $\frac{b}{\theta}$
= $\frac{1.276 x 10^7}{3.32 x 10^{-2}}$
D = 3.84 x $10^8$ m
Example 4
The Sun’s angular diameter
is measured to be 1920′′. The distance D of
the Sun from the Earth is 1.496 × $10^{11}$ m.
What is the diameter of the Sun ?
Answer
Sun’s angular diameter
α
= 1920"
= 1920 × 4.85 × $10^{-6}$ rad
= 9.31 × $10^{-3}$ rad
Sun’s diameter
$d = \alpha D$
$d = (9.31 \times 10^{-3})(1.496 \times 10^{11})$
$d = 1.39 \times 10^9$ m
Estimation of Very Small Distances:
Size of a Molecule
To measure a very small size, like that of a molecule ($$10^{–8} m to $10^{–10}$ m), we have to adopt special methods. We cannot use a screw gauge or similar instruments. Even a microscope has certain limitations. An optical microscope uses visible light to ‘look’ at the system under investigation. As light has wave like features, the resolution to which an optical microscope can be used is the wavelength of light (A detailed explanation can be found in the Class XII Physics textbook). For visible light the range of wavelengths is from about 4000 Å to 7000 Å (1 angstrom = 1 Å = 10-10 m).
To measure a very small size, like that of a molecule ($$10^{–8} m to $10^{–10}$ m), we have to adopt special methods. We cannot use a screw gauge or similar instruments. Even a microscope has certain limitations. An optical microscope uses visible light to ‘look’ at the system under investigation. As light has wave like features, the resolution to which an optical microscope can be used is the wavelength of light (A detailed explanation can be found in the Class XII Physics textbook). For visible light the range of wavelengths is from about 4000 Å to 7000 Å (1 angstrom = 1 Å = 10-10 m).
Hence an optical
microscope cannot resolve particles with sizes
smaller than this. Instead of visible light, we can
use an electron beam. Electron beams can be
focussed by properly designed electric and
magnetic fields. The resolution of such an
electron microscope is limited finally by the fact
that electrons can also behave as waves ! (You
will learn more about this in class XII). The
wavelength of an electron can be as small as a
fraction of an angstrom. Such electron
microscopes with a resolution of 0.6 Å have been
built. They can almost resolve atoms and
molecules in a material. In recent times,
tunnelling microscopy has been developed in
which again the limit of resolution is better than
an angstrom. It is possible to estimate the sizes
of molecules.
A simple method for estimating the molecular
size of oleic acid is given below. Oleic acid is a
soapy liquid with large molecular size of the
order of $10^{–9}$ m.
The idea is to first form mono-molecular layer
of oleic acid on water surface.
We dissolve 1 $cm^3$ of oleic acid in alcohol to
make a solution of 20 $cm^3$. Then we take 1 cm3 of this solution and dilute it to 20 $cm^3$, using
alcohol. So, the concentration of the solution is
equal to $\frac{cm^3}{20 \times 20}$ of oleic acid/$cm^3$ of
solution. Next we lightly sprinkle some
lycopodium powder on the surface of water in a
large trough and we put one drop of this solution
in the water. The oleic acid drop spreads into a
thin, large and roughly circular film of molecular
thickness on water surface.
Then, we quickly
measure the diameter of the thin film to get its
area A. Suppose we have dropped n drops in
the water. Initially, we determine the
approximate volume of each drop (V $cm^3$).
Volume of n drops of solution
= nV $cm^3$
Amount of oleic acid in this solution
= nV $\left(\frac{cm^3}{20 \times 20}\right)$
This solution of oleic acid spreads very fast
on the surface of water and forms a very thin
layer of thickness t. If this spreads to form a
film of area A $cm^2$, then the thickness of the
film
t = $\frac{Volume \ of \ the \ film}{Area \ of \ the \ film}$
t = $\frac{nV}{20 \times 20A}$ cm
If we assume that the film has mono-molecular
thickness, then this becomes the size or diameter
of a molecule of oleic acid. The value of this
thickness comes out to be of the order of $10^{–9}$ m.
Example 5
If the size of a nucleus (in
the range of $10^{–15}$ to $10^{–14}$ m) is scaled up
to the tip of a sharp pin, what roughly is
the size of an atom ? Assume tip of the pin
to be in the range $10^{–5}$ m to $10^{–4}$ m.
Answer
The size of a nucleus is in the range of $10^{–15}$ m and $10^{–14}$ m. The tip of a sharp pin is
taken to be in the range of $10^{–5}$ m and $10^{–4}$ m.
Thus we are scaling up by a factor of $10^{10}$. An
atom roughly of size $10^{–10}$ m will be scaled up to a
size of 1 m. Thus a nucleus in an atom is as small
in size as the tip of a sharp pin placed at the centre
of a sphere of radius about a metre long.
Range of Lengths
The sizes of the objects we come across in the universe vary over a very wide range. These may vary from the size of the order of $10^{–14}$ m of the tiny nucleus of an atom to the size of the order of $10^{26}$ m of the extent of the observable universe.
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| Table 1: Range and order of lengths |
Table 1gives the range and order of lengths and sizes of some of these objects. We also use certain special length units for short and large lengths. These are
1 fermi = 1 f = $10^{–15}$ m
1 angstrom = 1 Å = $10^{–10}$ m
1 astronomical unit = 1 AU (average distance
of the Sun from the Earth)
= 1.496 × $10^{11}$ m
1 light year = 1 ly = 9.46 × $10^{15}$ m (distance
that light travels with velocity of
3 × $10^8$ m/s in 1 year)
1 parsec = 3.08 × $10^{16}$ m (Parsec is the
distance at which average radius of earth’s orbit
subtends an angle of 1 arc second)
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