Exam questions - Lenses
1.
[2006]
A converging lens is used as a magnifying glass.
Draw a ray diagram to show how an erect image is formed by a magnifying glass.
2.
[2006]
A diverging lens cannot be used as a magnifying glass. Explain why.
3.
[2006]
A converging lens has a focal length of 8 cm.
Determine the two positions that an object can be placed to produce an image that is four times the size of
the object?
4.
[2008 OL]
A concave lens has a power of 0.1 cm−1. What is the focal length of the lens?
5.
[2002 OL]
A lens has a power of +50 m-1. What type of lens is it and what is its focal length?
6.
[2004]
Two converging lenses, each with a focal length of 10 cm, are placed in contact. What is the power of the
lens combination?
7.
[2002]
(i)
The power of a normal eye is +60 m-1. A short-sighted person’s eye has a power of +65 m-1.
Calculate the power of the contact lens required to correct the person’s short-sightedness.
(ii)
Calculate the focal length of the contact lens required to correct the person’s short-sightedness.
8.
[2006]
The power of an eye when looking at a distant object should be 60 m–1. A person with defective vision has a
minimum power of 64 m–1.
(i) Calculate the focal length of the lens required to correct this defect.
(ii) What type of lens is used?
(iii) Name the defect.
9.
[2002]
Draw a labelled diagram showing the optical structure of the eye.
10. [2002]
How does the eye bring objects at different distances into focus?
11. [2008]
(i) An eye contains a lens system and a retina, which is 2.0 cm from the lens system. The lens system consists
of the cornea, which acts as a fixed lens of power 38 m–1, and a variable internal lens just behind the cornea.
The maximum power of the eye is 64 m–1.
Calculate how near an object can be placed in front of the eye and still be in focus.
(ii) Calculate the maximum power of the internal lens.
(iii) Light is refracted as it enters the cornea from air as shown in the diagram.
Calculate the refractive index of the cornea.
(iv) Draw a diagram to show th path of a ray of light as it passes from water of refractive index 1.33 into the
cornea.
(v) A swimmer cannot see properly when she opens her eyes underwater.
When underwater why does the cornea not act as a lens?
(vi) What is the maximum power of the eye when underwater?
(vii) Why do objects appear blurred when underwater?
(viii)
Explain how wearing goggles allows objects to be seen clearly.
Exam Questions – Hookes Law and SHM
1. [2009][2007][2003]
State Hooke’s law.
2. [2002]
A mass at the end of a spring is an example of a system that obeys Hooke’s Law.
Give two other examples of systems that obey this law.
3. [2002]
(i) The equation F = – ks, where k is a constant, is an expression for a law that governs the motion of
a body.
Name this law and give a statement of it.
(ii) Give the name for this type of motion and describe the motion.
4. [2009]
(i) When a sphere of mass 500 g is attached to a spring of length 300 mm, the length of the spring
increases to 330 mm. Calculate the spring constant.
(ii) The sphere is then pulled down until the spring’s length has increased to 350 mm and is then
released.
Describe the motion of the sphere when it is released.
(iii) What is the maximum acceleration of the sphere?
5. [2007]
(i) When a small sphere of mass 300 g is attached to a spring of length 200 mm, its length increases
to 285 mm.
Calculate its spring constant.
(ii) The sphere is pulled down until the length of the spring is 310 mm.
The sphere is then released and oscillates about a fixed point.
Derive the relationship between the acceleration of the sphere and its displacement from the fixed
point.
(iii) Why does the sphere oscillate with simple harmonic motion?
(iv) Calculate the period of oscillation of the sphere.
(v) Calculate the maximum acceleration of the sphere.
(vi) Calculate the length of the spring when the acceleration of the sphere is zero.
6. [2002]
(i) The springs of a mountain bike are compressed vertically by 5 mm when a cyclist of mass 60 kg
sits on it.
When the cyclist rides the bike over a bump on a track, the frame of the bike and the cyclist
oscillate up and down.
Using the formula F = – ks, calculate the value of k, the constant for the springs of the bike.
(ii) The total mass of the frame of the bike and the cyclist is 80 kg.
Calculate the period of oscillation of the cyclist.
(iii) Calculate the number of oscillations of the cyclist per second.
Exam questions - waves
Speed of sound = 340 m s-1
Speed of light = 3 × 108 m s-1
1.
[2007 OL][2010 OL]
The diagram shows the waveform of a musical note.
What is the name given to (i) the distance A, (ii) height B?
2.
[2007 OL][2010]
Explain what is meant by the frequency of a wave.
3.
[2005 OL]
What is meant by the amplitude of a wave?
4.
[2005][2006 OL]
Explain the difference between longitudinal and transverse waves.
5.
[2005 OL]
A wave motion has a frequency of 5 hz and a wavelength of 200 m. Calculate the speed of the wave.
6.
[2007 OL]
A tin-whistle produces a note of 256 Hz. Calculate the wavelength of this note.
7.
[2010 OL]
If the natural frequency of a string is 250 Hz calculate the wavelength of the sound wave produced.
8.
A mobile phone transmits at 1200 MHz from its antenna.
Calculate the length of its antenna, which is one quarter of the wavelength that it transmits.
(speed of light = 3×108 m s-1)
9. [2006 OL]
(i) A ship detects the seabed by reflecting a pulse of high frequency sound from the seabed. The
sound pulse is detected 0.4 s after it was sent out and the speed of sound in water is 1500 m s –1.
Calculate the time taken for the pulse to reach the seabed.
(ii) Calculate the depth of water under the ship.
(iii) Calculate the wavelength of the sound pulse when its frequency is 50 000 Hz.
10. [2004 OL][2002 OL][2005 OL][2009]
Explain the term diffraction.
11. [2004 OL][2005 OL]
Explain the term interference.
12. [2008]
Why does diffraction not occur when light passes through a window?
13. [2006]
A sound wave is diffracted as it passes through a doorway but a light wave is not. Explain why.
14. [2002]
Explain the term constructive interference.
15. [2002]
Explain the term coherent sources.
16. [2007]
What is the condition necessary for destructive interference to take place when waves from two coherent
sources meet?
Exam questions – Static Electricity
(permittivity of free space = 8.9 × 10-12 Fm-1; charge on the electron = 1.6 × 10-19 C)
1.
[2008 OL]
Give one effect of static electricity?
2.
[2004]
Identify two hazards caused by static electricity.
3.
[2003 OL]
The build-up of electric charge can lead to explosions. Give two examples where this could happen.
4.
[2003 OL]
How can the build-up of electric charge on an object be reduced?
5.
[2010 OL]
How would you detect the presence of an electric field?
6.
[2009 OL]
Name the instrument shown in the diagram.
7.
[2005 OL]
The diagram shows a gold leaf electroscope.
Name the parts labelled A and B.
8. [2005 OL]
(i) Explain why the gold leaf on the electroscope diverges when a positively charged rod is brought close to the
metal cap.
(ii) The positively charged rod is held close to the electroscope and the metal cap is then earthed.
Explain why the gold leaf collapses.
9. [2007 OL]
(i) The diagram shows a positively charged gold leaf electroscope.
Describe how an electroscope is given a positive charge.
(ii) What is observed when the cap of an electroscope is earthed?
(iii) Why does this happen?
(iv) How is the cap of the electroscope earthed?
10. [2008]
Describe how an electroscope can be charged by induction
11. [2003 OL]
Describe, with the aid of a labelled diagram, how you would charge a conductor by induction.
12. [2005 OL]
Give one use of an electroscope.
13. [2010 OL][2003 OL]
What is the unit of electric charge?
14. [2005][2003][2007 OL][2010 OL]
State Coulomb’s law of force between electric charges.
15. [2006][2005]
Why is Coulomb’s law an example of an inverse square law?
16. [2005]
Give two differences between the gravitational force and the electrostatic force between two electrons.
Exam solutions - Lenses
1. Object inside focal point
Two (appropriate) rays from object to lens
Two rays emerge correctly from lens
Rays produced back to form upright virtual image (on same side as
object)
2. The image is always diminished.
3. 1/u + 1/v = 1/f
Magnification = v /u = 4  v = 4u
For real image: 1/u + 1/4u = 1/8
 u = 10 cm
For virtual image: 1 /u - 1/4u = 1/8  u = 6 cm
4. P = 1/f
 f = 1/P = 10 cm.
5. P = 1/f
 f = 1/P
 f = .02 m. It is a converging (convex) lens.
6. P1 = 1/f1 = 10 m−1 , P2 = 1/f2 = 10 m−1
P = P1 + P2 = 20 m-1
7. (i) PTotal = P1 + P2
60 = 65 + P2
Power = - 5 m
(ii)P = 1/f  f = - 0.2 m
8. (i) P = P1 + P2
 60 = 64 + P2
P = 1/f

f = 1/P  1/f = (-)4
(ii) Diverging / concave lens
(iii) Short sight / myopia


P2 = -4 ( m-1)
f = (-)¼ m
= (-)25 cm
9. See diagram.
10. It can change the shape of the lens which in turn changes the focal length of
the lens.
11.
(i) Pmax = 64 m-1 = 1/f
f = 0.0156 m = 1.56 cm
1/u = 1/v +1/f  1/u = 1/2 = 1/1.56  u = 7.14 cm
(ii) Pmax = P1 + P2
 64 = 38 + P2  P2 = 26 m-1
(iii)
 n = sin 37/sin 27
 n = 1.33
(iv) Both media have the same refractive index so there is no bending of light.
Draw a straight line passing from one medium to the other without bending.
(v) Because light does not refract at the cornea since there is no change in refractive index.
(vi) The maximum power of the eye is 64 m–1, but this includes the focusing power of the cornea (38
m-1) which doesn’t work underwater, so maximum power = 64 – 38 = 26 m–1.
(vii)
Because the internal lens by itself is not powerful enough to focus light on retina.
(viii) Because light which hits the cornea is coming from air and so there will be refraction here
(the cornea will now act as a lens).
Exam Solutions – Hookes Law and SHM
1. For a stretched string the restoring force is proportional to the extension.
A stretched spring obeys Hooke’s law.
2. Stretched elastic, pendulum, oscillating magnet, springs of car, vibrating tuning fork, object
bobbing in water waves, ball in saucer, etc.
3.
(i) Hooke’s Law states that when an object is stretched the restoring force is directly proportional to
the displacement, provided the elastic limit is not exceeded.
(ii) Simple harmonic motion; an object is said to be moving with Simple Harmonic Motion if its
acceleration is directly proportional to its distance from a fixed point in its path, and its
acceleration is directed towards that point.
4.
(i) When the mass of 500 g is attached the new force down = mg = (0.5)(g).
Because the spring is in equiblibrium this must be equal to the force up (which is the restoring
force).
Hooke’s law in symbols: F = k x
 (0.5)(g) = kx
 k = F/x = 0.5g/0.030
 k = 163.3 N m-1
(ii) It executes simple harmonic motion because the displacement is proportional to t he acceleration.
(iii) F = ma = kx
 a = kx/m = (163.3)(0.02)/(0.5) = 6.532 m s-2
OR
a = 2 x  2 = k/m = 163.3/0.5  a = 6.532 m s-2
5.
(i) F = ks
mg = ks
(0.30)(9.8) = (k)(0.085)
k = 34.6 N m-1
(ii) F = - ks

ma = - ks

a = - (k/m)s

a α -s 
a=-ks
(iii) Its acceleration is proportional to its displacement from a fixed point.
(iv) From above: ω2 = k/m  ω2 = 34.6 / 0.3 ω = 10.7 T = 2π/ω = 2π/10.7 = 0.58 ≈ 0.6 
T = 0.6 s
(v) This occurs when s is a maximum, i.e. when s = amplitude = 0.310 – 0.285 = 0.025 m.
a = -ω2s

a = - (10.7)2 (0.025)

a = (-) 2.89 m s-2
(vi) This occurs at the fixed point when l = 0.285 m
6.
(i) F = – ks
 mg = – ks
m-1
(ii) k/m = ω2
ω = 38 s-1
T = 2π/ω = 0.16 s
(iii) f = 1/T ≈ 6
 60 × 9.8 = -k (.005)  588 = -k (.005)
 k = 1.2 × 105 N
Exam Solutions - waves
1. A = wavelength
B = amplitude
2. The frequency of a wave is the number of waves passing a fixed point per second.
3. Amplitude corresponds to the height of the wave.
4. A Transverse wave is a wave where the direction of vibration is perpendicular to the direction in
which the wave travels.
A Longitudinal Wave is a wave where the direction of vibration is parallel to the direction in
which the wave travels.
5. c = f λ

c = 5 × 200 = 1000 m s-1.
6. The speed of sound in air is 340 m s−1
c = fλ λ = c/f  λ = 340/256 = 1.33 m.
7. v = f λ
λ = v/f
λ = 340/250
= 1.36 m
8. λ = c/f
λ = (3 × 108)/(1.2 × 109)
λ = 0.25 m
Length of antenna = 0.25/4 = 0.0625 m.
9.
(i) 0.2 s.
(ii) v= s/t

s=v×t

s = 1500 × 0.2 = 300 m.
(iii) c = fλ

λ = c/f

λ = 1500/50000 = 0.03 m.
10. Diffraction is the spreading of waves around a slit or an obstacle.
11. Interference occurs when waves from two sources meet to produce a wave of different amplitude.
12. The window is too wide (relative to wavelength of light).
13. Diffraction only occurs when the width of the gap is approximately equal to the wavelength of the
wave. This is the case for a sound wave but the wavelength of a light-wave is very small
compared to the size of a doorway.
14. Constructive Interference occurs when waves from two coherent sources meet to produce a wave
of greater amplitude.
15. Two waves are said to be coherent if they have the same frequency and are in phase.
16. They must be out of phase by half a wavelength (this means that the crest of one wave will be
over the trough of the other.
1.
Exam solutions – static electricity
Lightning, static discharge, receive shock after walking across carpets, attracts objects, can damage
electronics.
2.
Electric shock / explosion in flour mills /explosion when fuelling aircraft/ damage to electronic devices /
electrical storm / static cling, etc.
3.
Dust e.g. flour mill explosions, inflammable vapours e.g. fuelling aircraft, lightning
4.
By earthing the object (i.e. using a conductor to connect the object to the earth which allows the charge to
flow to earth).
5.
Using an electroscope // electric field sensor // electric field meter
6.
A gold leaf electroscope
7.
A = insulation, B = metal/glass/plastic case
8.
(i) Some of the electrons at the bottom of the electroscope are attracted to the top due to the positive charge on
the rod and as a result there is an excess of positive charge on the bottom, including on the gold leaf.
Because similar charges repel the gold leaf moves away from the main section.
(ii) Some of the positive charges are repelled by the rod and so flow to the ground.
9.
(i) Bring a negatively charged rod close to the cap.
Earth the electroscope by touching it with your finger while still holding the rod close by.
Remove your finger, then remove the rod.
(ii) The leaves drop.
(iii) The positive charges move from the cap to the earth (or the negative charges move from the earth to the
cap).
(iv) By touching it with your finger.
10. See previous question.
11. Apparatus e.g. conductor (hanging from insulated thread or mounted on an insulating stand) and a
charged rod.
Bring a charged rod close to the conductor.
Earth the conductor by touching it with your finger while still holding the rod
close by.
Remove your finger, then remove the rod.
12. To detect charge
13. The coulomb
14. The force between two charges is proportional to the product of the charges and inversely proportional to
the square of the distance between them.
15. Force is inversely proportional to distance squared.
16. Gravitational force is much smaller than the electrostatic force.
Gravitational force is attractive, electrostatic force (between two electrons) is repulsive.