Measurement of Large Distances

MEASUREMENT OF LENGTH

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 10²m 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 \(\angle\)ASB in Fig. 2.2 represented by symbol \(\theta\) is called the parallax angle or parallactic angle.

As the planet is very far away,\(\frac{d}{D}\)<< 1, and therefore,\(\theta\) 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 = D \(\theta\) where \(\theta\) is in radians.

D=\(\frac{d}{D}\)

2.2 Parallax method                                             2.3

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 \(\alpha\) the angular size of the planet (the angle subtended by d at the earth), we have \(\alpha\)=d/D 

The angle \(\alpha\) 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) \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaafa % aaaa!36BD! 1'\)(minute of arc or arcmin) and (c) \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaaga % aaaa!36BE! 1''\)(second of arc or arc second) in radians. Use 360⁰=2\(\pi\) rad, 1⁰=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOnaiqaic % dagaqbaaaa!377C! 60'\)  and \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaafa % aaaa!36BD! 1'\)=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOnaiqaic % dagaGbaaaa!377D! 60''\)

Answer

(a) We have 360⁰=2\(\pi\) rad,1⁰ = (\(\pi\) /180) rad= 1.745×10^-2rad

(b) 1⁰ = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOnaiqaic % dagaqbaaaa!377C! 60'\) = 1.745×10^–2 rad

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaafa % aaaa!36BD! 1'\)=2.908\(\times\)10^-4 rad 2.91\(\times\)10^-4 rad

(c)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaafa % aaaa!36BD! 1'\)=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOnaiqaic % dagaGbaaaa!377D! 60''\)=2.908\(\times\)10^-4 rad

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaaga % aaaa!36BE! 1''\)=4.847\(\times\)10^-4 rad  ; 4.85\(\times\)10^-6rad.

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

 Answer

We have, parallax angle \(\theta\)= 400 From Fig. 2.3, AB = AC tan \(\theta\)

AC = AB/tan\(\theta\) = 100 m/tan 40⁰

EXAMPLE 3

The moon is observed from two diametrically opposite points A and B on Earth. The angle \(\theta\) subtended at the moon by the two directions of observation is 1^o 5\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGinayaafa % aaaa!36C0! 4'\). Given the diameter of the Earth to be about 1.276\(\times\)10⁷ m, compute the distance of the moon from the Earth.

Answer

We have\(\theta\) =1^o 5\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGinayaafa % aaaa!36C0! 4'\)=11\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGinayaafa % aaaa!36C0! 4'\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaqadaqaaiaaigdacaaIXaGaaGinaiabgEna0kaaiAdacaaIWaaa % caGLOaGaayzkaaWaaWbaaSqabeaakiadacUHYaIOaaGaey4LIq8aae % WaaeaacaaI0aGaaiOlaiaaiIdacaaI1aGaey41aqRaaGymaiaaicda % daahaaWcbeqaaiaaiAdaaaaakiaawIcacaGLPaaacaWGYbGaamyyai % aadsgaaeaacqGH9aqpcaaIZaGaaiOlaiaaiodacaaIYaGaey41aqRa % aGymaiaaicdadaahaaWcbeqaaiaaikdaaaGccaWGYbGaamyyaiaads % gaaaaa!5A6F! \begin{array}{l} = {\left( {114 \times 60} \right)^\prime } \times \left( {4.85 \times {{10}^6}} \right)rad\\ = 3.32 \times {10^2}rad \end{array}\)

since,\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabGymayaaga % Gaeyypa0JaaGinaiaac6cacaaI4aGaaGynaiabgEna0kaaigdacaaI % WaWaaWbaaSqabeaacqGHsislcaaI2aaaaOGaamOCaiaadggacaWGKb % aaaa!42EB! 1'' = 4.85 \times {10^{ - 6}}rad\)

Also b    AB   1 276\(\times\)10⁷

Hence from Eq. (2.1), we have the earth-moon distance

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGeb % Gaeyypa0JaamOyaiaac+cacaWGebaabaGaeyypa0ZaaSaaaeaacaaI % XaGaaiOlaiaaikdacaaI3aGaaGOnaiabgEna0kaaigdacaaIWaWaaW % baaSqabeaacaaI3aaaaaGcbaGaaG4maiaac6cacaaIZaGaaGOmaiab % gEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaaaaaO % qaaiabg2da9iaaiodacaGGUaGaaGioaiaaisdacqGHxdaTcaaIXaGa % aGimamaaCaaaleqabaGaaGioaaaakiaad2gaaaaa!5534! \begin{array}{l} D = b/D\\ = \frac{{1.276 \times {{10}^7}}}{{3.32 \times {{10}^{ - 2}}}}\\ = 3.84 \times {10^8}m \end{array}\)

EXAMPLE 4

The Sun’s angular diameter is measured to be 192\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiqaic % dagaGbaaaa!3779! 0''\).The distance D of the Sun from the Earth is 1.496 ×10¹¹ m.What is the diameter of the Sun ?

Answer

Sun’s angular diameter \(\alpha\) 192\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiqaic % dagaGbaaaa!3779! 0''\).

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaIXa % GaaGyoaiaaikdacaaIWaGaey41aqRaaGinaiaac6cacaaI4aGaaGyn % aiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaI2aaaaOGaamOCai % aadggacaWGKbaabaGaaGyoaiaac6cacaaIZaGaaGymaiabgEna0kaa % igdacaaIWaWaaWbaaSqabeaacqGHsislcaaIZaaaaOGaamOCaiaadg % gacaWGKbaaaaa!5062! \begin{array}{l} 1920 \times 4.85 \times {10^6}rad\\ 9.31 \times {10^{ - 3}}rad \end{array}\)

Sun’s diameter

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpcaaI5aGaaiOlaiaaiodacaaIXaGaey41aqRaaGymaiaaicdadaah % aaWcbeqaaiaaiodaaaGccqGHxdaTcaaIXaGaaiOlaiaaisdacaaI5a % GaaGOnaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIXaGaaGym % aaaakiaad2gaaeaacqGH9aqpcaaIXaGaaiOlaiaaiodacaaI5aGaey % 41aqRaaGymaiaaicdadaahaaWcbeqaaiaaiMdaaaGccaWGTbaaaaa!53CB! \begin{array}{l} = 9.31 \times {10^3} \times 1.496 \times {10^{11}}m\\ = 1.39 \times {10^9}m \end{array}\)

Estimation of Very Small Distances: Size of a Molecule

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). 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 cm³ of oleic acid in alcohol to make a solution of 20 cm³.Then we take 1 cm³ of this solution and dilute it to 20 cm³, using alcohol.So, the concentration of the solution is equal to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % WcaaqaaiaaigdaaeaacaaIYaGaaGimaiabgEna0kaaikdacaaIWaaa % aaGaayjkaiaawMcaaiaadogacaWGTbWaaWbaaSqabeaacaaIZaaaaa % aa!4011! \left( {\frac{1}{{20 \times 20}}} \right)c{m^3}\) of acid/cm³ 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 cm³).

Volume of n drops of solution= nV cm³

Amount of oleic acid in this solution=-nv\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % WcaaqaaiaaigdaaeaacaaIYaGaaGimaiabgEna0kaaikdacaaIWaaa % aaGaayjkaiaawMcaaiaadogacaWGTbWaaWbaaSqabeaacaaIZaaaaa % aa!4011! \left( {\frac{1}{{20 \times 20}}} \right)c{m^3}\)

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 cm², then the thickness of the film.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG0b % Gaeyypa0ZaaSaaaeaacaWGwbGaam4BaiaadYgacaWG1bGaamyBaiaa % dwgacaWGVbGaamOzaiaadshacaWGObGaamyzaiaadAgacaWGPbGaam % iBaiaad2gaaeaacaWGbbGaamOCaiaadwgacaWGHbGaam4BaiaadAga % caWG0bGaamiAaiaadwgacaWGMbGaamyAaiaadYgacaWGTbaaaaqaai % aad+gacaWGYbGaaGPaVlaaykW7caWG0bGaeyypa0ZaaSaaaeaacaaI % XaaabaGaaGOmaiaaicdacqGHxdaTcaaIYaGaaGimaiaadgeaaaGaam % 4yaiaad2gaaaaa!6173! \begin{array}{l} t = \frac{{Volumeofthefilm}}{{Areaofthefilm}}\\ or\,\,t = \frac{1}{{20 \times 20A}}cm \end{array}\)

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^–5m to 10^–4m.

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 1010. 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.

Ranges of Lenghts

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 1026 m of the extent of the observable universe. Table 2.3 gives the range and order of lengths and sizes of some of these objects.

We also use certain special length units for short and large These are lengths.

1 fermi=1f=10^-18m

1 anstrong=1 Å =10^-10m

1 astronomical unit  = 1 AU (average distance of the Sun from the Earth)

= 1.496 \(\times\)1011 m

1 light year = 1 ly= 9.46 \(\times\) 10¹⁵ m (distance

that light travels with velocity of 3 \(\times\) 10⁸ m s^–1 in 1 year)

1 parsec= 3.08 \(\times\) 10¹⁶ m (Parsec is the distance at which average radius of earth’s orbit subtends an angle of 1 arc second)