Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 256–270 of 784 verified questions
Sort by:
Physics
WAEC / WASSCE 2008
Waves and Optics
In the day time, it is possible to see under shady areas such as under a tree because light has undergone
A
internal reflection
B
refraction
C
diffraction
D
diffuse reflection
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #35057)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
The distance between a concave mirror and an object placed in front of it is 1.0m. If the raduis of curvature of the mirror is 4.0m, the image formed will be
A
2.00m behind the mirror
B
2.00m in front of the mirror
C
1.25m in front of the mirror
D
1.25m behind the mirror
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The focal length of a concave mirror is negative
f = -2m
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{-2} = \frac{1}{v} - \frac{1}{1}\)
\(\frac{1}{v} = \frac{1}{-2} + \frac{1}{1}\)
\(\frac{1}{v} = \frac{1}{2}\)
v = 2m
The image distance is 2.0 m in front of the mirror
Source: Myschool Past Questions Archive (Record #35060)
Physics
WAEC / WASSCE 2008
Waves and Optics
The image of an object formed by a convex mirror is
A
real, errect and enlarged
B
real, inverted and enlarged
C
virtual, erect and diminished
D
virtual, erect and enlarge
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #35062)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
When viewed straight down, a fish in a pond is
A
father below than it appears to be
B
nearer than it appears to be
C
found at exactly the same point it is seen
D
seen beside its image
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
When viewed straight down, a fish in a pond appears nearer than it is. This is due to the bending of light as it travels from water to air, which causes the fish to appear closer to the surface than it is. This phenomenon is called refraction. As a result, the fish's actual position in the water is farther below than it appears to be when viewed from above.
Source: Myschool Past Questions Archive (Record #35064)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
A converging lens and a screen are placed 20 cm and 80 cm respectively from an object in a straight line so that a sharp image of the object is formed on the screen. If the object is 3cm high. Calculate the height of the image formed
A
1 cm
B
9cm
C
12cm
D
15cm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{v}{u} = \frac{hi}{ho} = \frac{60}{20} = \frac{hi}{3}\)
hi = 9cm
Source: Myschool Past Questions Archive (Record #35065)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
a sound wave is produced fron a source ans an echo is heard t seconds afterwards. If d is the distance of the reflecting surface from the source, v the speed, p the wavelength and T the period of the wave, then
A
d = \(\frac{2t}{\lambda}\)
B
d = \(\frac{\lambda t}{2T}\)
C
d = \(\frac{2T}{\lambda t}\)
D
d = \(\frac{\lambda T}{2t}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
V = \(\frac{2d}{t}\)
f\(\lambda = \frac{2d}{t} = \frac{\lambda}{T} = \frac{2d}{t}\)
Source: Myschool Past Questions Archive (Record #35069)
Physics
WAEC / WASSCE 2008
Waves and Optics
Both \(\gamma\)-rays and X-rays i. can be produced by thermionic emission ii. have low frequencies iii. have short wavelengths. Which of the statements above is/are correct?
A
i only
B
ii only
C
iii only
D
i and iii only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #35108)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
Explain plane polarized light
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Plane polarized light is a light wave vibrations that occur only in one plane perpendicular to the direction of propagation of the light. Light is a transverse wave, that can be plane polarized by reflection from surfaces.
Source: Myschool Past Questions Archive (Record #53538)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
(a) The equation y = \(\alpha\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.
(i) Given that x is in metres and t is in seconds, state the units of k and w
(ii) What physical quantity does \(\frac{W}{k}\) represent? Justify your answer
(iii) State whether the wave is travelling in the positive or negative x-direction.
(b)(i) What are beats?
(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency
(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.
(c) List two similarities between the human eye and the photographic camera.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) K is measured in m\(^{-1}\) and w rad s\(^{-1}\)
(ii) \(\frac{w}{k}\) represents speed of wave
w = 2\(\pi\)f; and
k = \(\frac{2 \pi}{\pi}\)
\(\frac{w}{k} \frac{2 \pi f}{2 \pi/\lambda} = f\lambda\)
But V = f\(\lambda\)
(iii) The direction of travel of the wave :s in the positive x - direction.
(b)(i) Beats are periodic rise and fall in amplitude of the sound produced when two notes of nearly equal frequencies are sounded together.
(ii) Beat frequency = \(f_1 - f_2\)
= 259 - 256 = 3Hz
(iii) f = \(\frac{1}{2L} \sqrt{T}{m}\)
259 = \(\frac{1}{2L} \sqrt{\frac{1200}{0.03}}\)
L = 0.386m
(c) 1. Both have converging or convex lenses.
2. Their interiors are black.
3. Both form real, inverted, and diminished image.
4. Images of objects are formed on the retina of the eye and the film of the camera.
Source: Myschool Past Questions Archive (Record #53560)
Physics
WAEC / WASSCE 2008
Waves and Optics
Solution
Using the above diagram as a guide, carry out the following instructions.
Fix a metre rule on the bench with its graduated side facing up.
Place the illuminated object at the 0cm end and the screen at the 100cm end of the rule such that the distance d between the illuminated object and screen is 100cm.
Record the distance d. Also, evaluate and record d\(^{2}\).
Place and move the Converging lens between the illuminated object and the Screen until a sharp diminished image of the object is formed on the screen.
Read and record the position Ih of the lens.
Now move the lens towards the object until another sharp image of the object is formed on the Screen.
Read and record the new position \(l_2\) of the lens.
Evaluate and record L= (\(l_1\) - \(I_2\)), also evaluate L\(^{2}\) and D = d\(^{2}\) - L\(^{2}\).
Repeat the procedure for four values of d= 85, 75, 65 and 55cm. Tabulate your readings.
Plot a graph or Don the vertical axis against d on the horizontal axis.
Determine the slope, s, of the graph.
Evaluate k = \(\frac{s}{4}\).
State two precautions taken to obtain accurate results.
(b)i. Explain the statement the focal length of a converging lens is 15cm.
ii. Distinguish between a real image and a virtual image.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
d(cm)
d\(^{2}\)(cm)
\(l_1\)
\(l_1\)
L=(\(l_1\)-\(l_2\))
L\(^{2}\)
D=(d\(^{2}\)-L\(^{2}\))
100
10,000
79.8
59.5
20.3
412.09
9587.90
85
7225
63.6
20.3
43.3
1874.89
5350.11
75
5626
51.6
23.6
28.0
784.00
4841.00
65
4225
38.3
21.6
16.7
278.89
3946.11
55
3025
33.5
26.5
07.0
049.0
2976.00
k = \(\frac{s}{4} = \frac{146.9}{4}\) = 36.733
Precautions:
Parallax error on reading metre rule was avoided.
The optical components are correctly aligned.
The vertical placement of the lens ensured.
Repeated readings (shown on the table)
(b)i. This means that the distance between the optical centre (or middle of lens) of the lens and its principal focus is 15cm.
Real image
Virtual image
(i) Can be received/formed on a screen
(ii) Formed by the intersection of actual rays
(i) Cannot be received/formed on a screen.
(ii) Formed by the intersection of imaginary rays.
Source: Myschool Past Questions Archive (Record #60035)
Physics
UTME / JAMB 2007
Waves and Optics
Solution
I. They should be identical
II. They should originate from the same source.
III. They should be coherent.
IV. They should be monochromatic.
from the statements above, the conditions for two waves to interfere are?
A
I,II and III only
B
I, III and IV only
C
II, III and IV only
D
I, II, III and IV
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Identical (I): The waves should have the same frequency and wavelength.
Coherent (III): The waves must maintain a constant phase relationship.
Monochromatic (IV): The waves should have the same frequency (which is closely related to being coherent).
Source: Myschool Past Questions Archive (Record #13935)
Physics
UTME / JAMB 2007
Waves and Optics
Solution
In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to
A
Increase the reverberation of sound in the hall
B
increase the amount of echo in the hall
C
reduce the hight of the ceiling from the floor
D
reduce the reverberation of sound in the hall
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
In a large auditorium perforated, absorbent materials are used to line the ceiling so as to reduce the effect of multiple reflection of echo called reverberation
Source: Myschool Past Questions Archive (Record #13939)
Physics
UTME / JAMB 2007
Waves and Optics
Solution
The phenomenon of light bending round an obstacle is?
A
diffraction
B
refraction
C
reflection
D
polarization
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The phenomenon of light bending around an obstacle is: diffraction
Source: Myschool Past Questions Archive (Record #13943)
Physics
UTME / JAMB 2007
Waves and Optics
Solution
The instrument used by designers to obtain different color pattern is called?
A
periscope
B
sextant
C
episcope
D
kaleidoscope
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Kaleidoscope is an optical instrument that creates patterns and designs by reflecting light through coloured materials.
Source: Myschool Past Questions Archive (Record #13945)
Physics
UTME / JAMB 2007
Waves and Optics
Solution
A glass plate 0.9cm thick has a refractive index of 1.50. How long does it take for a pulse of light to pass through the plate?
[c = 3.0 x 10\(^8\) ms-1]
A
4.5 x 10-11s
B
4.5 x 10-10s
C
4 .5 x 10-12 s
D
3.0 x 10-10s
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
N = 1.5 = \(\frac{(speed of light in glass)}{\text{speed of light in air}}\)
\(\frac{1.5}{1}\) = \(\frac{(3.0 \times 10^8)}{\text{V}}\)
V = \(\frac{(3.0 \times 10^8)}{1.5}\)
V = 2.0 x 10\(^8\) ms\(^{-1}\)
But x = Vt
t = \(\frac{\text{x}}{\text{Vt}}\)
t = (x)/v = \( \frac{9 x 10^{-3}}{2.0 x 10^8}\)
∴ Time = 4.5 x 10\(^{-11}\)S
Source: Myschool Past Questions Archive (Record #13957)