IB DP Physics: Wavefronts, rays, reflection, refraction and diffraction (DP Physics) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover the first half of sub-topic C.3 (Wave phenomena) in the IB Physics guide, first assessment 2025: wavefronts and rays, reflection, refraction, Snell’s law, total internal reflection and diffraction. All of it is SL and HL. Where you need constants, use c = 3.00 × 10⁸ m s⁻¹ and take the refractive index of air as 1.00. Read the study guide first if the ideas are new. See also the course hub and the printable checklist.

Questions

1. Define a wavefront and a ray, and state how they are oriented to each other in a uniform medium. [2]

2. A ray of light meets a plane mirror at 28° to the surface of the mirror. Find the angle of incidence and the angle between the incident ray and the reflected ray. [3]

3. Water waves in a ripple tank cross from region 1, where their speed is 0.24 m s⁻¹ and their wavelength is 2.0 cm, into region 2, where their speed is 0.16 m s⁻¹. The rays in region 1 make 55.0° with the normal to the boundary.

(a) Find the frequency of the waves in region 2 and justify your answer. [2] (b) Find the wavelength in region 2. [1] (c) Find the angle between the rays in region 2 and the normal. [2]

4. A student measures the angle of incidence θ1 in air and the angle of refraction θ2 in a rectangular block. Two points on the best-fit line of a graph of sin θ1 (vertical axis) against sin θ2 (horizontal axis) are (0.171, 0.259) and (0.468, 0.707), written as (sin θ2, sin θ1).

(a) Find the refractive index of the block material. [3] (b) Find the speed of light in the block. [2]

5. Plane ripples have wavefronts 2.0 cm apart in deep water. They meet a boundary with shallow water, where the speed is 0.60 times the speed in deep water. The rays in deep water are at 50.0° to the normal. Describe the wavefront-ray diagram for the shallow region by finding the wavefront spacing, the angle of the rays to the normal, and the angle between the wavefronts and the boundary. [4]

6. A long glass rod of refractive index 1.48 is surrounded by a coating of refractive index 1.46.

(a) Calculate the critical angle at the rod-coating boundary. [2] (b) A ray in the rod meets the boundary at 84.0° to the normal and another at 78.0°. State, with reasons, what happens to each. [2]

7. A glass prism of refractive index 1.55 has angles of 45°, 45° and 90°. A ray enters one of the short faces along the normal and meets the long face at 45° to the normal.

(a) Show that the ray is totally internally reflected when the prism is in air. [3] (b) Determine whether the ray is still totally internally reflected when the prism is under water of refractive index 1.33. [2]

8. An AM radio transmitter sends waves of frequency 900 kHz. A hill about 400 m wide lies between the transmitter and a house, yet the house receives the signal. The lamp on the transmitter mast cannot be seen from the house. Explain the difference in terms of diffraction. [4]

9. Light of power 8.0 mW is incident on a glass surface. Of this, 0.32 mW is reflected and 0.28 mW is absorbed. Calculate the power transmitted, and state whether the frequency of the transmitted light differs from that of the incident light. [3]

10. A ray of light in air meets a flat glass layer of refractive index 1.58 at 70.0° to the normal. The glass layer lies on top of water of refractive index 1.33.

(a) Find the angle between the ray and the normal in the glass. [2] (b) Find the angle between the ray and the normal in the water. [2] (c) Show that, whatever the angle of incidence in air, total internal reflection cannot occur at the glass-water boundary. [4]

11. A semicircular block of refractive index 1.66 has a ray directed at the centre of its curved face along a radius, so that it hits the flat face at the centre.

(a) Explain why the ray is not deviated on entering the block, and calculate the critical angle at the flat face for glass to air. [3] (b) The block is turned so that the ray meets the flat face at 30.0° to the normal. Calculate the angle of the emerging ray to the normal. [2] (c) The block is turned until the angle at the flat face is 50.0°. Describe what happens to the ray. [2]

Answers

1. A wavefront is a line or surface joining points at the same phase of the wave [1]. A ray shows the direction of travel of the wave, and it is perpendicular to the wavefronts [1]. [2] Examiner insight: Answers that say “a ray is a line the wave follows” without the 90° relationship to wavefronts leave out half of what the question asks for.

2. The angle of incidence is measured from the normal: 90° − 28° = 62° [1]. The angle of reflection is also 62° [1]. The angle between the rays is 2 × 62° = 124° [1]. [3] Examiner insight: A common loss here is giving 28° as the angle of incidence. Write “from the normal” next to the 62°.

3. (a) f = v / λ = 0.24 / 0.020 = 12 Hz in region 1 [1]. The frequency is the same in region 2, 12 Hz, because the source sets it and the boundary cannot change it [1]. (b) λ2 = v2 / f = 0.16 / 12 = 0.0133 m, so 1.3 cm [1]. (c) sin θ2 / sin 55.0° = 0.16 / 0.24 [1], so sin θ2 = 0.6667 × 0.8192 = 0.5461 and θ2 = 33.1° [1]. [5] Examiner insight: A reason is needed alongside the value in part (a). Stating 12 Hz without saying why it is unchanged is likely to earn less than a full answer.

4. (a) The gradient of the line is the ratio sin θ1 / sin θ2 = n of the block [1]. Gradient = (0.707 − 0.259) / (0.468 − 0.171) = 0.448 / 0.297 [1]. n = 1.51 [1]. (b) v = c / n = 3.00 × 10⁸ / 1.51 [1] = 1.99 × 10⁸ m s⁻¹ [1]. [5] Examiner insight: Choose two points far apart on the line, not scattered data points, and show each subtraction before you divide.

5. The speed ratio equals the wavelength ratio, so the spacing is 0.60 × 2.0 = 1.2 cm [1]. sin θ2 = 0.60 × sin 50.0° = 0.60 × 0.7660 = 0.4596 [1], so θ2 = 27.4° to the normal [1]. The wavefronts are perpendicular to the rays, so they make 27.4° with the boundary [1]. [4] Examiner insight: Link the spacing, the bend and the wavefront angle. Saying only “the rays bend towards the normal” is unlikely to cover everything asked.

6. (a) sin C = n2 / n1 = 1.46 / 1.48 = 0.9865 [1], so C = 80.6° [1]. (b) At 84.0° the angle exceeds 80.6°, so the ray is totally internally reflected and stays in the rod [1]. At 78.0° the angle is below C, so part of the ray refracts into the coating and is lost [1]. [4] Examiner insight: For each ray, quote the comparison with C as well as the outcome. A bare “reflected” or “refracted” gives no reasoning to credit.

7. (a) For total internal reflection the angle must exceed C. sin C = 1.00 / 1.55 = 0.6452 [1]. C = 40.2° [1]. Since 45° > 40.2°, the ray is totally internally reflected [1]. (b) In water, sin C = 1.33 / 1.55 = 0.8581, so C = 59.1° [1]. Since 45° < 59.1°, there is no total internal reflection and part of the ray refracts out [1]. [5] Examiner insight: A “show that” question needs the critical angle value and the comparison written out. Quoting only the result of the comparison leaves a gap in the argument.

8. The wavelength of the radio wave is λ = c / f = 3.00 × 10⁸ / 9.00 × 10⁵ = 333 m [1]. This is comparable to the size of the hill, so the wave diffracts strongly round it and the wavefronts bend into the shadow [1]. Visible light has a wavelength of about 5 × 10⁻⁷ m, far smaller than the hill [1], so it diffracts negligibly and the hill casts a sharp shadow [1]. [4] Examiner insight: Compare the wavelength with the size of the obstacle using numbers. “Radio has a longer wavelength” alone does not say why the hill matters.

9. Transmitted power = 8.0 − 0.32 − 0.28 [1] = 7.4 mW [1]. The frequency is unchanged [1]. [3] Examiner insight: Energy conservation is the method. Show the subtraction and give the unit with the answer.

10. (a) 1.00 × sin 70.0° = 1.58 × sin θg [1]. sin θg = 0.9397 / 1.58 = 0.5947, so θg = 36.5° [1]. (b) 1.58 × sin θg = 1.33 × sin θw, which gives the same equation as 1.00 × sin 70.0° = 1.33 × sin θw [1], so sin θw = 0.7065 and θw = 45.0° [1]. (c) At the glass-water boundary, sin C = 1.33 / 1.58 = 0.8418, so C = 57.3° [1]. The largest possible angle in glass occurs when the angle in air is 90°: sin θg = 1.00 / 1.58, so θg = 39.3° [1]. Hence the angle in the glass can never exceed 39.3° [1], which is less than 57.3°, so total internal reflection never occurs [1]. [8] Examiner insight: The glass layer drops out of the final angle, which saves time. In part (c), state the largest angle that can arise before comparing it with C.

11. (a) The ray travels along a radius, so it meets the curved face at 90° to the surface, along the normal, and is not deviated [1]. sin C = 1.00 / 1.66 = 0.6024 [1], so C = 37.0° [1]. (b) 1.66 × sin 30.0° = 1.00 × sin θ [1], so sin θ = 0.830 and θ = 56.1° [1]. (c) 50.0° is greater than 37.0°, so there is total internal reflection [1]. The ray reflects at 50.0° to the normal and stays in the block, then leaves through the curved face along a radius [1]. [7] Examiner insight: Using Snell’s law at 50.0° gives a sine above 1, which signals that the ray cannot refract. Say so in words.

Where marks are usually lost

• Using the angle to the surface as if it were measured from the normal.
• Swapping the refractive indices, so that sin C comes out above 1.
• Leaving out the reason (frequency is set by the source) when asked about frequency at a boundary.
• Quoting a critical angle without comparing it with the angle of incidence.
• Describing a wavefront diagram without the changed spacing.
• Treating diffraction as a change in wavelength or speed.
• Giving “bigger wavelength” with no comparison to the size of the gap or obstacle.

Next steps

• Use the revision notes.
• Re-read the study guide.
• Visit the course hub and tick off the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization: sub-topic C.3 Wave phenomena (first half of the understandings).