OCR A Level Physics A: Electromagnetic spectrum, refraction and total internal reflection (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), section 4.4.2 Electromagnetic waves, statements (a) to (e). Where a value of c is needed, use 3.00 × 10⁸ m s⁻¹ and treat the refractive index of air as 1.00. Explanations are in the study guide and revision notes. See also the OCR A Level Physics hub, the printable checklist and all free 10-minute diagnostics.

Questions

1. State three properties that are common to all electromagnetic waves. [3]

2. Name the region of the electromagnetic spectrum that is most likely for each wavelength: (a) 2 × 10⁻² m, (b) 5 × 10⁻⁷ m, (c) 1 × 10⁻¹⁰ m. [3]

3. A wireless network uses radiation of frequency 5.0 GHz. An ultraviolet source emits radiation of wavelength 120 nm.

(a) Calculate the wavelength of the network radiation. [2] (b) Calculate the frequency of the ultraviolet radiation. [2] (c) Name the region of the spectrum for each radiation. [1]

4. A lamp is viewed through two polarising filters. One filter is fixed and the other is rotated slowly through 360° about the line of sight.

(a) Describe what is observed. [3] (b) Explain why this shows that light is a transverse wave. [2]

5. A microwave transmitter gives plane polarised waves with the electric field vertical. A receiver aligned with the field is placed in line with it. A metal grille of parallel horizontal rods is placed between them.

(a) Predict and explain the receiver signal. [2] (b) The grille is now turned through 90° in its own plane. State and explain the change. [2]

6. Light passes from air into glass of refractive index 1.58. The angle of incidence is 52.0°.

(a) Calculate the angle of refraction. [3] (b) Calculate the speed of light in the glass. [2]

7. A ray in air strikes the surface of a liquid with an angle of incidence of 65.0°. The angle of refraction is 36.0°. Calculate the refractive index of the liquid and the speed of light in it. [4]

8. A student investigates refraction through a rectangular acrylic block using a ray box. The readings are shown.

i / °	15.0	25.0	35.0	45.0	55.0	65.0
r / °	10.0	16.5	22.5	28.5	33.5	37.5

(a) Calculate sin i and sin r for the readings at 15.0° and 65.0°. [2] (b) Use these two points to find the gradient of a graph of sin i against sin r, and state the refractive index of the acrylic. [3] (c) Suggest why a best-fit line through all the points is preferable to using two readings. [1]

9. The refractive index of a glass is 1.62.

(a) Calculate the critical angle for a glass-air boundary. [2] (b) Calculate the critical angle for a boundary between this glass and water of refractive index 1.33. [3]

10. A student uses a ray box and a semi-circular glass block to find the critical angle.

(a) Explain why the ray is aimed at the centre of the flat face through the curved surface. [2] (b) Describe what is seen at the flat face as the angle of incidence is increased from 0° to beyond the critical angle. [3]

11. A small lamp lies at the bottom of a tank holding oil of refractive index 1.47 to a depth of 0.50 m.

(a) Calculate the critical angle for the oil-air surface. [2] (b) Calculate the radius of the circle on the surface through which light from the lamp can emerge. [2] (c) Explain what happens to light that reaches the surface outside this circle. [2] (d) The oil is replaced by water of refractive index 1.33 at the same depth. Calculate the new radius. [2]

12. Red light of wavelength 650 nm in air enters glass of refractive index 1.60.

(a) Calculate the frequency of the light. [2] (b) Calculate the speed of light in the glass. [2] (c) Calculate the wavelength in the glass. [2] (d) State what happens to the frequency of the light on entering the glass. [1]

Answers

1. Any three from: they are transverse waves [1]; they travel through a vacuum, needing no medium [1]; in a vacuum they move at 3.00 × 10⁸ m s⁻¹, or they transfer energy [1]. [3] Examiner insight: One property per mark; “they are waves” is too vague, and “travel at the same speed” needs “in a vacuum” to be secure.

2. (a) microwave [1]; (b) visible light [1]; (c) X-ray [1]. [3] Examiner insight: Match on the order of magnitude, using a reliable table; gamma rays overlap with X-rays at about 10⁻¹¹ m, so 10⁻¹⁰ m is the safer X-ray answer.

3. (a) λ = c/f = (3.00 × 10⁸) / (5.0 × 10⁹) [1] = 0.060 m [1] (b) f = c/λ = (3.00 × 10⁸) / (120 × 10⁻⁹) [1] = 2.5 × 10¹⁵ Hz [1] (c) microwave (network) and ultraviolet [1] [5] Examiner insight: Convert GHz and nm to base units before substituting; a power-of-ten slip here is the commonest way to lose both marks.

4. (a) The light is bright, then dims to darkness [1]; it brightens again [1]; two bright and two dark positions occur in 360° [1]. (b) Only a transverse wave can be plane polarised, with oscillation in one plane at right angles to its direction of travel [1]; the first filter transmits one plane, and the second blocks it when crossed, which longitudinal oscillations could not do [1]. [5] Examiner insight: Say that the filters are “crossed” at darkness; and an explanation of why it proves transverse needs the idea of oscillation perpendicular to the direction of travel.

5. (a) The signal is a maximum, or almost unaffected [1], as the rods are at right angles to the electric field so electrons are not driven along them and the wave passes [1]. (b) The signal falls to a minimum [1], since the rods are now parallel to the electric field; electrons are driven along them, so energy is absorbed or reflected [1]. [4] Examiner insight: Link the orientation of the rods to the electric field direction in each case; many answers state only “it blocks the waves”.

6. (a) 1.00 × sin 52.0° = 1.58 × sin r [1]; sin r = 0.7880 / 1.58 = 0.4987 [1]; r = 29.9° [1] (b) v = c/n = (3.00 × 10⁸) / 1.58 [1] = 1.90 × 10⁸ m s⁻¹ [1] [5] Examiner insight: Show the substituted Snell’s law line; give the angle to 3 significant figures, as the data are given to that precision.

7. n = sin 65.0° / sin 36.0° = 0.9063 / 0.5878 [1] = 1.54 [1]; v = c/n = (3.00 × 10⁸) / 1.54 [1] = 1.95 × 10⁸ m s⁻¹ [1] [4] Examiner insight: Air is medium 1 here, so n is sin i divided by sin r; inverting the ratio gives n below 1, which should alert you to a mistake.

8. (a) sin 15.0° = 0.259 and sin 10.0° = 0.174 [1]; sin 65.0° = 0.906 and sin 37.5° = 0.609 [1] (b) gradient = Δ(sin i) / Δ(sin r) = (0.906 − 0.259) / (0.609 − 0.174) [1] = 0.647 / 0.435 [1] = 1.49, so n = 1.49 [1] (c) A best-fit line averages out random errors in the readings. [1] [6] Examiner insight: A gradient needs both axes’ changes shown; remember the graph is sin i (y) against sin r (x), so the gradient is n directly.

9. (a) sin C = 1/n = 1/1.62 = 0.6173 [1]; C = 38.1° [1] (b) sin C = n₂/n₁ = 1.33/1.62 [1] = 0.8210 [1]; C = 55.2° [1] [5] Examiner insight: For a glass-water boundary the ratio is n₂/n₁ with the larger index on the bottom; using 1/n would give the glass-air value.

10. (a) The ray enters along a radius, so it meets the curved surface along the normal [1] and is not refracted there, so the only bending occurs at the flat face [1]. (b) At small angles a refracted ray leaves the flat face, bent away from the normal, with a weak reflected ray [1]. At the critical angle the refracted ray travels along the flat face [1]. Beyond it there is no refracted ray and all the light is reflected back into the glass: total internal reflection [1]. [5] Examiner insight: Name the normal when explaining why the ray is not deviated; the critical angle is an angle of incidence, not an angle of refraction.

11. (a) sin C = 1/1.47 = 0.6803 [1]; C = 42.9° [1] (b) radius = depth × tan C = 0.50 × tan 42.9° [1] = 0.46 m [1] (c) The angle of incidence exceeds the critical angle [1], and the light is totally internally reflected back into the oil [1]. (d) sin C = 1/1.33, so C = 48.8° [1]; radius = 0.50 × tan 48.8° = 0.57 m [1] [8] Examiner insight: Draw the right-angled triangle with the depth as the vertical side; and state both conditions for TIR in (c), not only that the angle is large.

12. (a) f = c/λ = (3.00 × 10⁸) / (650 × 10⁻⁹) [1] = 4.62 × 10¹⁴ Hz [1] (b) v = c/n = (3.00 × 10⁸) / 1.60 [1] = 1.88 × 10⁸ m s⁻¹ [1] (c) λ = v/f = (1.875 × 10⁸) / (4.615 × 10¹⁴) [1] = 4.06 × 10⁻⁷ m (406 nm) [1] (d) It is unchanged. [1] [7] Examiner insight: Use the unrounded values in a multi-step calculation; a quick check is that the wavelength in the glass equals 650 nm divided by 1.60.

Where marks are usually lost

• Measuring angles from the surface rather than the normal.
• Not converting GHz, MHz or nm before using c = fλ.
• Writing that the wave “bends towards the normal because it slows down” without saying the second medium has the higher refractive index.
• Describing polarisation as “waves vibrating up and down” and not as oscillation in a single plane.
• Reversing the effect of a metal grille on microwaves.
• Using sin C = 1/n for a boundary between two materials.
• Omitting one of the two conditions for total internal reflection.
• Saying the frequency changes when light enters glass.
• Plotting i against r and quoting a gradient as n.

Next steps

• Revise with the revision notes and re-read the study guide.
• Open the OCR A Level Physics hub and the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), section 4.4.2 Electromagnetic waves, for first assessment in 2017.