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Physics
UTME / JAMB 2025
Refraction of Light
Solution
I. The colour of light depends on its frequency II. When white light is dispersed by a triangular prism, yellow is deviated more than green III. Rainbows are formed when rains fall heavily. Which of the above statements is/are correct about dispersion and colours?
A
I and III only
B
I only
C
II and III only
D
I and II only
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Answer: B. I only (I true; II false, yellow deviated less than green; III false, rainbows from light refraction in drops.)
Source: WAEC / JAMB / NECO Classroom Archive (Record #74359)
Physics
UTME / JAMB 2025
Refraction of Light
Solution
If the critical angle for a glass–air boundary is 45º, what is the refractive index of the glass?
A
\(\sqrt{2}\)
B
\(\sqrt{5}\)
C
\(\sqrt{3}\)
D
2\(\sqrt{2}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Answer: A
n = \(\frac{1}{\text{sinC}}\) = \(\frac{1}{sin45º}\) = \(\sqrt{2}\).
Source: WAEC / JAMB / NECO Classroom Archive (Record #74371)
Physics
UTME / JAMB 2025
Refraction of Light
Diagram
Solution
From the diagram above, the sine of the angle of refraction in glass is
Attached Figure / Diagram
A
0.4
B
0.3
C
0.2
D
0.1
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(_ag_g\) = \(\frac{\text{sini}}{\text{sinr}}\)
sinr = \(\frac{\text{sini}}{_ag_g}\)
sinr = \(\frac{0.5}{1.5}\)
sinr = 0.3
Source: WAEC / JAMB / NECO Classroom Archive (Record #74493)
Physics
UTME / JAMB 2025
Refraction of Light
Solution
What colour has the lowest frequency in the spectrum of white light?
A
Red
B
Yellow
C
Indigo
D
Blue
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
A. Red
Red light has the longest wavelength and the lowest frequency compared to the other colors listed.
ROYGBIV: arranged in increasing frequency or decreasing wavelength.
Source: WAEC / JAMB / NECO Classroom Archive (Record #74563)
Physics
UTME / JAMB 2025
Refraction of Light
Diagram
Solution
Given that SQ = 10cm and SR = 6cm, the refractive index of the block of glass shown in the above figure is
Attached Figure / Diagram
A
1.67
B
0.60
C
0.30
D
2.33
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
For a ray entering and exiting parallel faces of a glass slab, the refractive index is given by:
n = \(\frac{\text{real depth}}{\text{apparent depth}}\)
Where SR = 6cm is apparent depth, SQ = 10cm real depth.
n = \(\frac{10}{6}\) = 1.6666 ≈ 1.67
Source: WAEC / JAMB / NECO Classroom Archive (Record #74660)
Physics
WAEC / WASSCE 2025
Refraction of Light
Diagram
Solution
A ray of light is incident on mirror m\(_1\) and after reflection is incident on mirror m\(_2\) as shown in the diagram above. Calculate the angle of reflection of the ray at mirror m\(_2\).
Attached Figure / Diagram
A
60\(^0\)
B
90\(^0\)
C
30\(^0\)
D
120\(^0\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
i at m\(_1\) = 60^\circ → r = 60^\circ at m\(_1\) (law of reflection i ^\circ = r ^\circ) i at m\(_2\) = 30^\circ
i = r,
∴ r = 30^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #76365)
Physics
WAEC / WASSCE 2025
Refraction of Light
Diagram
Solution
A ray of light is incident on a body x as shown in the diagram above. What is the refractive index of the body?
Attached Figure / Diagram
A
1.33
B
1.63
C
1.50
D
1.49
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\lambda\) = \(\frac{\text{Sin i}}{\text{Sin r}}\) = \(\frac{Sin 60º}{Sin 32º}\) = 1.63
Source: WAEC / JAMB / NECO Classroom Archive (Record #76366)
Physics
WAEC / WASSCE 2025
Refraction of Light
Solution
The image of any real object formed by a diverging lens is always
A
magnified
B
inverted
C
erect
D
real
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
For all positions of the object formed by a concave (diverging) lens, the image is virtual, erect, and diminished.
Source: WAEC / JAMB / NECO Classroom Archive (Record #76368)
Physics
WAEC / WASSCE 2025
Refraction of Light
Diagram
Solution
2a.
(i) Fix a metre rule on the bench with the graduated face up.
(ii) Place the illuminated object at the zero end of the rule and the screen at the other end as illustrated in the diagram above.
(iii) Measure and record D, the distance between the object and the screen. Evaluate D\(^2\)
(iv) Place and move the converging lens between the illuminated object and the screen until a diminished, sharp image of the object is formed on the screen. Read and record the position, x\(_1\), of the lens. From this position, move the lens towards the object until another sharp image of the object is formed on the screen. Read and record the position, x\(_2\), of the lens.
(v) Evaluate and record L = (x\(_1\) - x\(_2\)), L\(^2\) and (D\(^2\) - L\(^2\)).
(vi) Repeat the procedure for D = 90, 80, 70, and 60 cm. In each case, evaluate L, L\(^2\), and (D\(^2\) - L\(^2\)). Tabulate your readings.
(vii) Plot a graph of D\(^2\) - L\(^2\) on the vertical axis against D on the horizontal axis.
(viii) Determine the slope, s, of the graph and evaluate K = \(\frac{\text{s}}{4}\). State two precautions taken to ensure accurate results.[21 marks]
bi. Distinction Between Real and Virtual Images [2 marks]
ii. Draw a ray diagram to show how a converging lens may be used to form a real diminished image of an object [2 marks]
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
2a. Table of observations
D(cm)
D\(^2\)(cm\^2\)
X\(_1\)(cm)
L = (X\(_1 - X_2\))
L\(^2\)(cm\(^2\))
100
10000
80.50
62.20
3868.8
90
8100
70.20
51.20
2621.4
80
6400
59.10
39.10
1528.8
70
4900
46.70
24.70
610.09
60
3600
29.00
21.40
457.96
S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{6250 - 3250}{100 - 60}\) = \(\frac{3000}{40}\) = 75
K = \(\frac{\text{s}}{4}\) = \(\frac{75}{4}\) = 18.75
Precautions:
(i) I carefully avoided parallax error by ensuring my eyes were directly in line with the markings on the metre rule while taking measurements to improve accuracy.
(ii) I adjusted the setup and focused precisely to ensure a sharp and well-defined image was formed on the screen, minimizing errors due to blurriness or misalignment.
(iii) I made sure that the experimental apparatus was stable and free from vibrations to avoid any disturbance during measurements.
(iv) I ensured that the metre rule was properly calibrated and positioned perpendicular to the setup to prevent measurement inaccuracies due to angle errors.
bi. A real image is formed when light rays converge and meet at a point, allowing it to be projected onto a screen. It is typically inverted relative to the object, as seen in concave mirrors or convex lenses when the object is beyond the focal point.
On the other hand, a virtual image is formed when light rays appear to diverge from a point but do not actually meet. This type of image cannot be projected onto a screen and is always upright, as seen in plane mirrors or convex mirrors.
bii. See diagram above.
Source: WAEC / JAMB / NECO Classroom Archive (Record #76580)
Physics
UTME / JAMB 2024
Refraction of Light
Solution
Calculate the depth of a swimming pool if the apparent depth is 10cm. ( Refractive index of water = 1.33 )
A
7.5cm
B
10.0cm
C
13.3cm
D
6.87cm
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Refractive index = \(\frac{\text{real depth}}{\text{apparent depth}}\)
Real depth = refractive index x apparent depth
Apparent depth = 10cm
Refractive index = 1.33
Real depth = 1.33 x 10 = 13.3cm.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70484)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
A ray of light travels from a less dense medium to a denser medium. How are the speed and frequency of the light affected? The speed
A
Increases and the frequency increases
B
decreases and the frequency decreases
C
increases and the frequency remains the same
D
decreases and the frequency remains the same
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
When light travels from a less dense medium (like air) to a denser medium (like glass or water), its speed decreases due to the higher optical density of the new medium.
The frequency of light remains constant when it passes from one medium to another. Although the speed changes, the frequency is determined by the source of the light and does not change with the medium.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72026)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
Chromatic aberration in lenses can be corrected by using a combination of
A
diverging and converging lenses.
B
an opaque material to cover the rims of lenses.
C
wide parallel beams on the lenses' surface.
D
a thin convex lens.
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Chromatic aberration occurs because different wavelengths of light are refracted by different amounts when passing through a lens.
Chromatic aberration can be corrected by using a combination of diverging (concave) and converging (convex) lenses, which help focus different colors of light to the same point.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72040)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
For a convex lens to form a real image, the distance between the object and the screen will be:
A
equal to the focal length of the lens
B
less than the focal length of the lens
C
twice the focal length of the lens
D
four times the focal length of the lens
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Twice the focal length of the lens: this is because, for a real image to be formed, the object must be placed outside the focal length of the lens, typically at a distance greater than twice the focal length.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72052)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
The inside portion of a hollow metal sphere of diameter 20 cm is polished. The portion will therefore form a
A
Concave mirror of focal length 5cm.
B
Concave mirror of focal length 10 cm.
C
Convex mirror of focal length 10 cm
D
Convex mirror of focal length 5cm
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Diameter = 20cm
The radius of curvature, R = 10cm
But, focal length, f = \(\frac{R}{2} = \frac{10}{2}\) = 5 cm.
Because the inside portion is polished, light reflects off the hollow, caved-in interior. A mirror that reflects light from its inner surface is concave.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72066)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
The phenomenon used for explaining the formation of a rainbow in the sky is
A
interference.
B
refraction.
C
reflection.
D
dispersion.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The formation of a rainbow is primarily due to dispersion, as it involves the separation of light into its constituent colors when it passes through water droplets in the atmosphere, it is refracted and dispersed due to varying wavelengths.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72070)