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CIE iGCSE Co-Ordinated Science P3.1 General properties of waves Exam Style Questions Paper 4

Question

(a) Microwaves are transverse electromagnetic waves.
(i) State a use of microwaves.
(ii) Describe a transverse wave.
(b) Fig. 10.1 shows a ray of visible light travelling in air incident on the boundary between air and glass at an angle of 48°. The refractive index of the glass is 1.25.
(i) Calculate the angle of refraction of the light ray.
(ii) Draw the refracted ray on Fig. 10.1.
(c) Fig. 10.2 shows a ray of visible light travelling in glass.
The light is incident on the boundary between glass and air at an angle greater than the critical angle.
Continue the path of the light ray.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P3.3 — Electromagnetic spectrum (Part (a)(i))
• Topic P3.1 — General properties of waves (Part (a)(ii))
• Topic P3.2.2 — Refraction of light (Part (b)(i), (b)(ii))
• Topic P3.2.2 — Refraction of light / Total internal reflection (Part (c))

▶️ Answer/Explanation

(a)(i) Any one from:

  • Satellite television transmissions.
  • Mobile phones (cell phones).
  • Microwave ovens.

Microwaves are a region of the electromagnetic spectrum with wavelengths ranging from about 1 mm to 1 m. They have several important applications: satellite television uses microwaves for communication as they can pass through the Earth’s atmosphere; mobile phones use microwaves for wireless communication; microwave ovens use microwaves to heat food by causing water molecules to vibrate.

(a)(ii) In a transverse wave, the vibrations / oscillations are perpendicular to the direction of propagation / direction of travel / direction of energy transfer.

Waves can be classified as transverse or longitudinal based on the direction of particle vibration relative to the direction of wave propagation. In a transverse wave, the particles of the medium vibrate at right angles (perpendicular) to the direction in which the wave travels. Examples include electromagnetic waves (including light, microwaves, X-rays), water waves, and seismic S-waves.

(b)(i) Angle of refraction = 36°.

The refractive index (\( n \)) is defined as the ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)):

\( n = \frac{\sin i}{\sin r} \)

Given: \( n = 1.25 \), \( i = 48^\circ \)
Rearranging: \( \sin r = \frac{\sin 48^\circ}{1.25} \)
\( \sin 48^\circ \approx 0.7431 \)
\( \sin r = \frac{0.7431}{1.25} = 0.5945 \)
\( r = \sin^{-1}(0.5945) \approx 36.5^\circ \approx 36^\circ \)

(b)(ii) The ray should be drawn continuing as a straight line into the glass block in the top right region, refracted towards the normal.

When light travels from air (less dense medium) into glass (more dense medium), it slows down and bends towards the normal. The refracted ray should be drawn with a smaller angle to the normal than the incident ray (i.e., closer to the normal line), continuing in a straight line into the glass block.

(c) The ray should be drawn continuing as a straight line in the bottom right region (back into the glass), with the angle of incidence equal to the angle of reflection (total internal reflection).

When light travels from a denser medium (glass) to a less dense medium (air) at an angle of incidence greater than the critical angle, total internal reflection occurs. In total internal reflection:

  • The light ray is reflected back into the glass (the denser medium).
  • The angle of incidence equals the angle of reflection.
  • No light is transmitted into the air.

The critical angle for the glass-air boundary can be calculated using \( \sin c = \frac{1}{n} \), where \( n \) is the refractive index of glass (approximately 1.5), giving a critical angle of about 42°. Since the incident angle in the question is greater than the critical angle, total internal reflection occurs, and the ray should be drawn reflecting back into the glass at an equal angle to the normal.

Question

(a) (i) Seismic P-waves are longitudinal waves. Describe a longitudinal wave.
(ii) P-waves travel at 6200 m/s in rock. The frequency of a P-wave is 12 Hz.
Calculate the wavelength of the P-wave.
(b) Waves spread out when they pass through a narrow gap.
(i) State the name of this effect.
(ii) Explain whether sound waves with wavelength of 1.2 m will spread out when passing through a 1.0 m wide doorway.
(iii) The wavelength of red light is 700 nm. Explain why red light travels in a straight line through the doorway in (b)(ii).
(c) Light waves travelling in air refract when incident on a boundary with a transparent material.
A light ray incident on the boundary at an angle of 57° is refracted at an angle of 44°, as shown in Fig. 12.1.
(i) Define refractive index.
(ii) Calculate the refractive index of the transparent material.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P3.1 — General properties of waves
• Topic P3.2.2 — Refraction of light

▶️ Answer/Explanation

(a)(i) vibrations / oscillations are parallel ; (vibrations are parallel) to the direction of propagation / direction of travel / direction of energy transfer
In longitudinal waves, the particles of the medium vibrate parallel to the direction of energy transfer. This creates regions of compression and rarefaction.

(a)(ii) 520 m

v = fλ
6200 = 12 × λ
λ = 6200 ÷ 12 = 517 m ≈ 520 m

The wavelength is the distance between successive points of the same phase in the wave.

(b)(i) diffraction
Diffraction is the spreading out of waves when they pass through a gap or around an obstacle.

(b)(ii) yes (sound waves spread out) ; wavelength is similar to width of gap
Significant diffraction occurs when the gap size is similar to the wavelength of the wave. Since 1.2 m is close to 1.0 m, sound waves will spread out through the doorway.

(b)(iii) wavelength is much less than width of gap
Diffraction is negligible when the wavelength is much smaller than the gap. Light waves have very short wavelengths (700 nm) compared to the 1.0 m doorway, so they travel in straight lines.

(c)(i) ratio of the speeds of a wave in two different regions
The refractive index is a measure of how much a wave slows down when entering a medium. It can be calculated using Snell’s law.

(c)(ii) 1.2

n = sin i ÷ sin r
n = sin 57° ÷ sin 44°
n = 0.839 ÷ 0.695 = 1.21 ≈ 1.2

The refractive index indicates how much light is bent when entering the material. A higher value means greater bending and slower light speed in the medium.

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