CIE iGCSE Co-Ordinated Science P3.2.3 Thin converging lens Exam Style Questions Paper 4
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P3.1 — General properties of waves (Parts (a) & (b))
• Topic P3.2.3 — Thin converging lens (Part (c)(i) & (c)(ii))
• Topic P2.3.3 — Radiation / P2.3.4 — Consequences of thermal energy transfer (Part (d))
▶️ Answer/Explanation
(a) Amplitude (A) and Wavelength (W) on Fig. 11.1:
• Amplitude (A): Vertical double-headed arrow from the equilibrium position (middle line) to the crest or trough.
• Wavelength (W): Horizontal double-headed arrow between two consecutive crests (or troughs).
(b) Wave speed calculation:
\(v = f\lambda\)
\(v = 0.50 \times 0.078 = 0.039\text{ m/s}\)
(c)(i) Ray diagram for converging lens:
• Draw a ray from the top of the object parallel to the principal axis, which refracts through the focal point (F) on the other side.
• Draw a ray from the top of the object straight through the centre of the lens (undeviated).
• The image is formed where these two refracted rays meet.
• The image is inverted (upside down) and drawn with an arrow below the principal axis.
(c)(ii) Characteristics of the image when object is within focal length:
• Upright (same orientation as the object).
• Magnified (larger than the object).
• Virtual (cannot be projected on a screen; rays appear to diverge from the image).
(d) Daytime and nighttime temperature changes:
Daytime: The temperature of the Earth rises because the energy absorbed from the Sun (incoming infrared radiation) is greater than the energy emitted by the Earth back into space.
Nighttime: The temperature of the Earth falls because there is no incoming solar radiation, so the energy emitted by the Earth is greater than the energy absorbed.
Question
Mains potential difference is \(230\text{ V}\).
Calculate the electric current in the projector.
In another device, the object is placed between the principal focus and the lens.
On Fig. 12.2, draw rays to find the position of the image formed. Use an arrow to represent the image.
Topic codes:
• Topic P4.3.2 — Series and parallel circuits (Part (a) & (b))
• Topic P4.2.5 — Electrical energy and power (Part (c)(i))
• Topic P3.2.3 — Thin converging lens (Part (c)(ii) & (c)(iii))
▶️ Answer/Explanation
(a)(i) In a parallel circuit, the total current equals the sum of the currents in the branches.
Current in R = Total current − Current in \(10\Omega\) resistor
= \(0.32 – 0.18 = \mathbf{0.14\text{ A}}\)
(a)(ii) In a parallel circuit, the potential difference across each branch is the same as the cell voltage.
Potential difference across R = \(1.8\text{ V}\)
(b) For two resistors in parallel: \(\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_{\text{total}}} = \frac{1}{40} + \frac{1}{20} = \frac{1}{40} + \frac{2}{40} = \frac{3}{40}\)
\(R_{\text{total}} = \frac{40}{3} = \mathbf{13.3\ \Omega}\) (or \(13\ \Omega\) to 2 significant figures)
(c)(i) \(P = IV\), so \(I = P \div V\)
\(I = 750 \div 230 = \mathbf{3.3\text{ A}}\)
(c)(ii) When an object is placed between the principal focus and the lens, a virtual, upright and magnified image is formed on the same side as the object.
Rays to draw:
- A ray parallel to the principal axis refracts through the focal point on the other side
- A ray through the centre of the lens passes straight through
The rays diverge after passing through the lens, and are extended back to form a virtual image on the same side as the object.
(c)(iii) This arrangement is used as a magnifying glass.