Table Of ContentFirst Edition, 2009
ISBN 978 93 80168 00 5
© All rights reserved.
Published by:
Global Media
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Chandni Chowk, Delhi-110 006
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Table of Contents
1. SI System of Units
2. Lenses
3. Refraction
4. Digital Storage
5. Frequency
6. Current and Electricity
7. Wave
8. Quantum Physics
9. Vectors
10. Kinematics
All units are assumed to be
The SI System of Units
SI units are used throughout science in many countries of the world. There are seven base
units, from which all other units are derived.
Base units
Every other unit is either a combination of two or more base units, or a reciprocal of a
base unit. With the exception of the kilogram, all of the base units are defined as
measurable natural phenomena. Also, notice that the kilogram is the only base unit with a
prefix. This is because the gram is too small for most practical applications.
Quantity Name Symbol
Length metre m
Mass kilogram kg
Time second s
Electric Current ampere A
Thermodynamic Temperature kelvin K
Amount of Substance mole mol
Luminous Intensity candela cd
Derived units
Most of the derived units are the base units divided or multiplied together. Some of them
have special names. You can see how each unit relates to any other unit, and knowing the
base units for a particular derived unit is useful when checking if your working is correct.
Note that "m/s", "m s-1", "m·s-1" and are all equivalent. The negative exponent form is
generally prefered, for example "kg·m-1·s-2" is easier to read than "kg/m/s2".
In
terms In terms of
Quantity Name Symbol
of base units
other
derived
units
square
Area m2
metre
cubic
Volume m3
metre
metre
Speed/Velocity per m s − 1
second
metre
per
Acceleration m s − 2
second
squared
kilogram
per
Density kg m − 3
cubic
metre
cubic
metre m3 kg −
Specific Volume
per 1
kilogram
ampere
per
Current Density A m − 2
square
metre
ampere
Magnetic Field Strength per A m − 1
metre
mole per
mol m −
Concentration cubic
3
metre
Frequency Hertz Hz
Force Newton N m kg s − 2
Pressure/Stress Pascal Pa m − 1 kg s − 2
Energy/Work/Quantity joule J N m m2 kg s − 2
of Heat
Power/Radiant Flux Watt W m2 kg s − 3
Electric Charge/Quantity
coulomb C s A
of Electricity
Electric
Potential/Potential
Volt V m2 kg s − 3 A - 1
Difference/Electromotive
Force
m − 2 kg − 1 s4
Capacitance Farad F
A2
Electric Resistance Ohm Ω m2 kg s − 3 A − 2
m − 2 kg − 1 s3
Electric Conductance siemens S
A2
Magnetic Flux Weber Wb V s m2 kg s − 2 A - 1
Magnetic Flux Density Tesla T kg s − 2 A - 1
Inductance Henry H m2 kg s − 2 A − 2
degree
Celsius Temperature °C K - 273.15
celsius
Luminous Flux Lumen lm cd sr
Illuminance Lux lx m − 2 cd sr
Activity of a
bequerel Bq s − 1
Radionuclide
Prefixes
The SI units can have prefixes to make larger or smaller numbers more manageable. For
example, visible light has a wavelength of roughly 0.0000005 m, but it is more
commonly written as 500 nm. If you must specify a quantity like this in metres, you
should write it in standard form. As given by the table below, 1nm = 1*10-9m. In standard
form, the first number must be between 1 and 10. So to put 500nm in standard form, you
would divide the 500 by 100 to get 5, then multiply the factor by 100 (so that it's still the
same number), getting 5*10-7m. The power of 10 in this answer, i.e. -7, is called the
exponent, or the order of magnitude of the quantity.
Prefix Symbol Factor Common Term
peta P 1015 quadrillions
tera T 1012 trillions
giga G 109 billions
mega M 106 millions
kilo k 103 thousands
hecto h 102 hundreds
deca da 101 tens
deci d 10 - 1 tenths
centi c 10 - 2 hundredths
milli m 10 - 3 thousandths
micro µ 10 - 6 millionths
nano n 10 - 9 billionths
pico p 10 - 12 trillionths
femto f 10 - 15 quadrillionths
Homogenous equations
Equations must always have the same units on both sides, and if they don't, you have
probably made a mistake. Once you have your answer, you can check that the units are
correct by doing the equation again with only the units.
For example, to find the velocity of a cyclist who moved 100 metres in 20 seconds, you
have to use the formula , so your answer would be 5m s −
1.
This question has the units , and should give an answer in m s − 1. Here, the
equation was correct, and makes sense.
Often, however, it isn't that simple. If a car of mass 500kg had an acceleration of 0.2 m s
− 2, you could calculate from F = ma that the force provided by the engines is 100N. At
first glance it would seem the equation is not homogenous, since the equation uses the
units , which should give an answer in kg m s − 2. If you look at the derived
units table above, you can see that a newton is in fact equal to kg m s − 2, and therfore the
equation is correct.
Using the same example as above, imagine that we are only given the mass of the car and
the force exerted by the engines, and have been asked to find the acceleration of the car.
Using F = ma again, we need to rearrange it for a, and we now have the formula:
. By inserting the numbers, we get the answer . You already know
that this is wrong from the example above, but by looking at the units, we can see why
this is the case: . The units are ms2, when we were looking for ms −
2. The problem is the fact that F = ma was rearranged incorrectly. The correct formula
was , and using it will give the correct answer of 0.2 m s − 2. The units for the
correct formula are .
, unless stated otherwise.
Lenses
Curvature of Wavefronts
Light can be viewed as beams travelling from point A to point B. However, from most
light sources, the light radiates outwards as a series of wavefronts. Hence, light from a
light source is bent - wavefronts of light have a property known as curvature.
Decreasing curvatures of wavefronts
As light travels further away from its source, its curvature decreases. At this point, it is
helpful to consider a sphere expanding gradually from a point. This represents a given
wavefront of light. As the sphere expands, the curvature of its surface decreases when we
look at any constant area of the surface. It should be noted at this point that light from a
source infinitely far away has 0 curvature. This is useful, as ambient light (light from a
source that is far away) can be assumed to have a curvature of 0, as the difference
between this and its actual curvature is negligible.
The curvature of a wavefront is given as:
.
In this formula, v equals the distance from the wavefront to the in-focus image depicted
by the light.
Power of lenses
Calculating the power of a lens
The function of a lens is to increase or decrease the curvature of a wavefront. Lenses
have a 'power'. This is the curvature which the lens adds to the wavefront. Power is
measured in dioptres, and is given by the formula:
,
where f equals the focal length of the lens. This is the distance between the lens and the
point where an image will be in focus, if the wavefronts entering the other side of the lens
are parallel.
The Lens Equation
