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IB HL Physics Option G Objectives
IB HL Physics Option G Objectives List and answers
12th Grade

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G.1.1: Outline the nature of electromagnetic (EM) waves.
- Oscillating electric charge produces varying electric and magnetic fields
- Electromagnetic waves are transverse, have the same speed
- Oscillating (SHM) charge produces an electric, which produces a magnetic field that is perpendicular.
G.1.2: Describe the different regions of the electromagnetic spectrum.
- Order of magnitude of the frequencies and wavelengths of different regions
- Identify a source for each region.
G.1.3: Describe what is meant by the dispersion of electromagnetic waves
Dispersion - separating of white light into its
component colors due to refraction.
G.1.4: Describe the dispersion of EM waves in terms of the dependence of refractive index on wavelength.
- As electromagnetic waves enter a new medium, velocity depends on the frequency of the wave
- Each frequency will have a different refractive index in the new medium.
G.1.5: Distinguish between transmission, absorption and scattering of radiation.
Transmission - electromagnetic radiation passing from one medium to another
Absorption - Photons are absorbed by material
Scattering - Deflection of electromagnetic radiation from original path due to collision with particles of medium.
G.1.6: Discuss examples of the transmission, absorption and scattering of electro magnetic radiation.
- Effect of the Earth’s atmosphere on incident electromagnetic radiation
Explanations for:
the blue colour of the sky
red sunsets or sunrises
the effect of the ozone layers
the effect of increased CO2 in the atmosphere.
G.1.7: Explain the terms monochromatic and coherent.
Coherent Waves - have a constant phase relationship.
Monochromatic Waves - single frequency (wavelength).
G.1.8: Identify laser light as a source of
coherent light.
Laser light is both coherent and monochromatic.
G.1.9: Outline the mechanism for the
production of laser light.
- Lower energy level electrons are pumped up to higher energy levels and stimulated to fall to energy levels corresponding to specific frequencies, producing laser light.
G.1.10: Outline an application of the use of a
- Medical applications
- Communications
- Technology (bar-code scanners, laser disks)
- Industry (surveying, welding and machining metals, drilling tiny holes in metals)
- Production of CDs reading and writing CDs, DVDs
G.2.1: Define the terms principal axis,
focal point, focal length and linear
magnification as applied to a
converging (convex) lens.
Principal axis - Line through the focal point of a lens and
the center of the lens
Focal point - Location on the principal axis where parallel
light rays converge after passing through the lens.
Focal length - Distance between the focal point and the
center of the lens.
G.2.2: Define the power of a convex lens and
the dioptre.
Power of a converging lens - The reciprocal of the focal length of the lens
Dioptre - The unit of power for a converging lens: 1 dioptre = 1 m-1
G.2.3: Define linear magnification.
Linear magnification - The ratio of the height of the image to the height of the object.
G.2.4: Construct ray diagrams to locate the
image formed by a convex lens.
- All rays incident on the lens from the object will be focused
- The image will be formed even if part of the lens is covered
G.2.5: Distinguish between a real image and
a virtual image.
Real image - formed when light rays actually converge on a location and can be projected onto a screen
Virtual image - formed by light rays that only appear to converge on a location and cannot be projected onto a screen.
G.2.6: Apply the convention “real is positive,
virtual is negative” to the thin lens
G.2.8: Define the terms far point and near point for the unaided eye.
For the normal eye, the far point may be assumed
to be at infinity and the near point is conventionally
taken as being a point 25 cm from the eye.
Near point - Distance between the eye and the nearest
object that can be brought comfortably into focus. “least distance of distinct

2. *Far point Distance between the eye and the furthest
object that can be brought into focus.
G.2.9: Define angular magnification.
Angular magnification - Ratio of the angle the image subtends at the eye to the angle the object subtends at the eye.
G.2.10: Derive an expression for the angular magnification of a simple magnifying glass for an image formed at the near point and at infinity
G.2.11: Construct a ray diagram for a compound microscope with final image formed close to the near point of the eye (normal adjustment).
Students should be familiar with the terms
objective lens and eyepiece lens
G.2.12: Construct a ray diagram for an
astronomical telescope with the final
image at infinity (normal adjustment).
G.2.13: State the equation relating angular
magnification to the focal lengths
of the lenses in an astronomical
telescope in normal adjustment
G.2.15: Explain the meaning of spherical
aberration and of chromatic
aberration as produced by a single
Chromatic aberration - Rays of different frequencies do not all converge at the same focal point due to dispersion by the lens.
Spherical aberration - Rays parallel to the principal axis do not all converge at the focal point.
G.2.16: Describe how spherical aberration in
a lens may be reduced.
- the lens can be ground to be slightly non‐spherical to adjust for the circle of least confusion
- using different combinations of lenses put together
- a stop (an opaque disc with a hole in it) is inserted before the lens so that the aperture size can be adjusted to allow only paraxial rays to enter (this reduces the light intensity and introduces diffraction of light)
G.2.17: Describe how chromatic aberration in
a lens may be reduced.
- using an achromatic doublet, made from a converging crown glass lens and a diverging flint glass lens that are adhered together
- Since the chromatic aberration of converging and diverging lenses is opposite, a combination of these two lenses will minimize this effect.
G.3.1: State the conditions necessary to
observe interference between two
-they must have the same phase
- the phase difference between them must remain constant
G.3.2: Explain, by means of the principle
of superposition, the interference
pattern produced by waves from two
coherent point sources.
- When two coherent point sources interfere they produce an interference pattern
- The dark lines are areas where the two waves interfere destructively, in between the dark points are areas of maximum amplitude where the waves interfere constructively
- When the two waves have both traveled an integer value of wavelengths they interfere constructively (assuming the sources are in phase)
- If the two waves have both traveled an integer value plus one half of a wavelength they interfere destructively.
G.3.3: Outline a double-slit experiment
for light and draw the intensity
distribution of the observed fringe
- Slit width must be small when compared to the slit separation so that diffraction effects of a single slit on the pattern are not considered
G.4.1: Describe the effect on the double-slit
intensity distribution of increasing the
number of slits.
when monochromatic light passes through a different number of slits we notice that as the number of slits increases:
- the number of observed fringes decreases
- the spacing between them increases
- the individual fringes become much sharper
G.4.2: Derive the diffraction grating formula
for normal incidence.
- The slits are very small so that they can be considered
to act as point sources
- They are also very close together such that d is small (10–6 m)
- Each slit becomes a source of circular wave fronts and the waves from each slit will interfere
- Consider the light that leaves the slit at an angle θ as shown, the path difference between wave 1 and wave 2 is dsinθ and if this is equal to an integral number of wavelengths then the two waves will interfere
constructively in this direction, similarly wave 2 will
interfere constructively with wave 3 at this angle, and wave 3 with 4 etc., across the whole grating. Hence if we look at the light through a telescope, that is bring it to a focus, then when the telescope makes an angle θ to the grating a bright fringe will be observed. The condition for observing a bright fringe is therefore:

dsinθ = nλ
G.4.3: Outline the use of a diffraction grating
to measure wavelengths.
- All elements have their own characteristic spectrum.
- An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through it when it is in a gaseous state
- If laser light is shone through a grating on to a screen, you will see just how sharp and spaced out are the maxima
- Measuring the line spacing and the distance of the screen from the laser, the wavelength of the laser can be measured.
G.5.1: Outline the experimental
arrangement for the production of
- A Coolidge tube is sufficient
- Intensity and hardness of the X-ray beam are controlled
- Electrons “boiled off” at cathode/filament get accelerated by high voltage to anode
- At anode, electrons hit metal, and produce X-rays
- X-rays proceed through walls of chamber
- Details of X-ray spectrum depend on metal, energy of electrons
- Very little of electron energy converted to X-rays
- Most of energy goes into heating anode
G.5.2: Draw and annotate a typical X-ray
Students should be able to identify the continuous
and characteristic features of the spectrum and the
minimum wavelength limit.
G.5.3: Explain the origins of the features of a
characteristic X-ray spectrum.
An electron which makes a transition from the M ‐shell to
a vacancy created in the K ‐shell gives rise to the Kβ line. If
an incident electron ionizes a target atom by ejecting an
L‐shell electron then electron transitions from the M‐shell
and N‐shell to fill the vacancy give rise to X‐ray lines called
the Lα and Lβ. respectively. It is apparent that the wavelength
of the lines in an X‐ray spectrum will be characteristic of a
particular element.
G.5.5: Explain how X-ray diffraction arises
from the scattering of X-rays in a
Each lattice ion (represented by a dot) acts as a source of
secondary waves. In general as these waves overlap they
will tend to interfere in a random manner. However, those
waves that are scattered at angles equal to the angle at which
the X‐rays are incident on the ion, will stand a chance of
reinforcing constructively with another scattered wave.Diffraction of x-rays is caused by the scattering of x-rays inside of a crystal.
G.5.6: Derive the Bragg scattering equation.
he direction of two such waves are shown in Figure 1843
by the ray labelled 1 that is scattered by the first layer and
by the ray labelled 2 that is scattered by the second layer.
Ray 2 travels an extra path difference AB + BC where AB =
BC = dsinθ, d being the spacing between adjacent crystal
layers and θ the angle between the incident X‐ray and the
crystal layer. The two waves will interfere constructively if
the path difference between them is an integral number of
wavelengths i.e.

2dsinθ = nλThis equation is known as the Bragg scattering equation
G.5.7: Outline how cubic crystals may be
used to measure the wavelength of
Students should be aware of the fact that the
structure of DNA was discovered by means of X-ray

By assuming a perfect cubic lattice array it is possible to
predict the resulting interference pattern produced. After measuring the lattice plane distances with x-rays of a specific wavelength, and by using cubic crystals of a known lattice plane distance, the wavelengths of x-rays can be measured.
G.5.8: Outline how X-rays may be used to
determine the structure of crystals.
Another consequence of Bragg’s work is the branch of
physics known as X-ray crystallography in which X‐rays
of known wavelengths are used to explore the crystal
structure of different elements and compounds. This is a
question of working back from the diffraction pattern to
determine the crystal structure that would produce such
a pattern. It was in this way that, in 1952, Francis Crick
and James Watson unraveled the structure of the DNA
G.6.1: Explain the production of interference
fringes by a thin air wedge.
Students should be familiar with the terms equal
inclination and equal thickness.

When the monochromatic light strikes the glass plate
some of it will be reflected down onto the wedge. Some
of the light reflected from the wedge will be transmitted
through the glass plate to the travelling microscope. A
system of equally spaced parallel fringes (fringes of equal
thickness) is observed.

The travelling microscope enables the fringe spacing to be
measured. The fringes can also be observed by the naked
G.6.2: Explain how wedge fringes can
be used to measure very small
Applications include measurement of the thickness
of the tear film on the eye and oil slicks.
G.6.3: Describe how thin-film interference is
used to test optical flats.
Wedge films can be used to test optical surfaces for flatness.
If a wedge is made of two surfaces one of which is perfectly
plane but the other has irregularities, the observed fringe
pattern will be irregular in shape. The irregular surface
can then be re‐polished until the fringes are all completely
parallel and of equal thickness.
G.6.5: State the condition for light to
undergo either a phase change of π,
or no phase change, on reflection
from an interface.
Hence if the path difference is an integral
number of half‐ wavelengths rays 1 and 2 will reinforce
i.e. ray 1 and 2 are in phase. However, rays 3, 5, 7 etc will
be out of phase with rays 2, 4, 6 etc but since ray 2 is more
intense than ray 3 and ray 4 more intense than ray 5, these
pairs will not cancel out so there will be a maximum of
G.6.6: Describe how a source of light gives
rise to an interference pattern when
the light is reflected at both surfaces
of a parallel film.
Consider light from an extended source incident on a thin
film. We also consider a wave from one point of the source
whose direction is represented by the ray shown. Some
of this light will be reflected at A and some transmitted
through the film where some will again be reflected at B
(some will also be transmitted into the air). Some of the
light reflected at B will then be transmitted at C and some
reflected and so on.
G.6.7: State the conditions for constructive
and destructive interference.
G.6.8: Explain the formation of coloured
fringes when white light is reflected
from thin films, such as oil and soap
If white light is shone onto the film then we can see why
we get multi‐coloured fringes since a series of maxima and
minima will be formed for each wavelength present in the
white light. However, when viewed at normal incidence,
it is possible that only light of one colour will under go
constructive interference and the film will take on this
G.6.9: the difference between
fringes formed by a parallel film and a
wedge film.
comparison between fringes formeD by
a parallel film anD tHose formeD by a
weDge film.

For a parallel film, the fringes are of equal inclination, that
is they form arcs of a circle whose centre is located at the
end of a perpendicular drawn from the eye to the surface
of the film.

For a wedge film, the fringes are parallel and of equal
G.6.10: Describe applications of parallel thin
Applications should include:

design of non-reflecting radar coatings for
military aircraft

measurement of thickness of oil slicks caused by

design of non-reflecting surfaces for lenses
(blooming), solar panels and solar cells.
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