iGCSE Physics (0625) 3.2.1 Reflection of light Paper 4 -Exam Style Questions- New Syllabus

Correct Answer: diffraction

Detailed solution: Diffraction is the spreading of waves around obstacles or through gaps. The sound waves from the violin bend around the edge of the doorway, allowing the teacher to hear the sound despite not having a direct line of sight.

Correct Answer: 450 Hz

Detailed solution: The relationship between wave speed, frequency, and wavelength is given by $v = f\lambda$. Using the values $v = 340\text{ m/s}$ and $\lambda = 0.75\text{ m}$, we calculate $f = \frac{v}{\lambda} = \frac{340}{0.75} \approx 453.3\text{ Hz}$, which rounds to $450\text{ Hz}$.

Correct Answer: large diffraction when gap size/doorway is similar to wavelength; high frequency/3500 Hz has (much) shorter wavelength AND there is less diffraction (with shorter wavelengths).

Detailed solution: Significant diffraction occurs when the wavelength is approximately the same size as the gap. The 450 Hz sound has a wavelength ($0.75\text{ m}$) similar to the doorway width, causing it to spread out widely. The 3500 Hz sound has a much shorter wavelength, resulting in minimal diffraction and poor spread.

Correct Answer: ray drawn from the violin to point X AND from point X to the teacher; mirror drawn at point X and labelled AND angle of incidence = angle of reflection; correct arrow on incident ray OR correct arrow on reflected ray.

Detailed solution: For the teacher to see the student via a plane mirror, light must travel from the violin to the mirror at X and then reflect to the teacher’s eye. The law of reflection states that the angle of incidence equals the angle of reflection, which must be shown by the drawn rays relative to the normal of the mirror placed at X.