AP Physics · topic 13
Geometric Optics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A mirror test with the wrong zero
A student reports the supplied pairs of incident and reflected ray angles. The incident angle is measured from the normal, but the reflected reading is measured from the mirror surface. Both rays lie in a plane perpendicular to the mirror. Angle-reading uncertainty is ±2°. Supplied data (Incident angle from normal / °; Reflected reading from surface / °): (20, 70), (35, 55), (50, 40), (65, 25). (a) Correct the reflected readings and decide whether the evidence supports the law of reflection. [3 marks] (b) Describe a ray diagram and an improved procedure. The mirror rotates by 5° with the incident ray fixed: predict the reflected-ray rotation. [3 marks]
Student's raw angle readings
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Reflected reading from surface / °
Horizontal axis (x): Incident angle from normal / °
Read the exact data as a table
| Incident angle from normal / ° | Reflected reading from surface / ° |
|---|---|
| 20 | 70 |
| 35 | 55 |
| 50 | 40 |
| 65 | 25 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Correct the reflected readings and decide whether the evidence supports the law of reflection.
Reveal the worked solution
- An angle from the surface is complementary to an angle from the normal, so r = 90°−raw reading.
- The four corrected angles match the incident angles. Plotting corrected r against i would give unit gradient and zero intercept within the stated uncertainty.
- The raw decreasing relationship is not a violation of reflection; it results from mixing reference directions.
- Draw the mirror, a normal at the incidence point, and incident and reflected rays on opposite sides of the normal with equal angles.
- Mark two well-separated points on each ray, draw thin lines, measure both angles from the same normal and repeat for a range of incident angles; use a narrow low-power beam without eye exposure.
- A mirror rotation of 5° rotates its normal by 5°. Since reflection preserves equality about the normal while the incoming ray is fixed, the outgoing ray rotates by 2×5° = 10°.
Answer (a) Corrected reflected angles are 20°, 35°, 50° and 65°, consistent with i = r. (b) The reflected ray rotates by 10°.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Testing a fibre from refraction data
A transparent material is surrounded by air of refractive index 1.00. Refraction measurements for light entering from air are supplied as sine values. Take c = 3.0×10⁸ m s⁻¹. A ray inside the material subsequently reaches a material–air boundary at 45° to the normal. Supplied data (sin(refracted angle); sin(incident angle)): (0, 0), (0.2, 0.3), (0.4, 0.6), (0.6, 0.9). (a) Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence. [3 marks] (b) Explain which wave quantities change on entry and evaluate whether all rays in a fibre necessarily remain trapped. [3 marks]
Air-to-material refraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): sin(incident angle)
Horizontal axis (x): sin(refracted angle)
Read the exact data as a table
| sin(refracted angle) | sin(incident angle) |
|---|---|
| 0 | 0 |
| 0.2 | 0.3 |
| 0.4 | 0.6 |
| 0.6 | 0.9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence.
Reveal the worked solution
- Snell's law gives nair sin i = nmaterial sin r. The graph gradient sin i/sin r is 1.50.
- Wave speed in the material is c/n = 2.0×10⁸ m s⁻¹. The critical angle obeys sin θc = 1/1.50, giving about 41.8°.
- The internal angle 45° exceeds θc and light travels from higher to lower index, meeting both conditions for total internal reflection.
- The boundary must transmit the same oscillation frequency. Since v = fλ and speed drops by a factor of 1.50, wavelength also drops by that factor.
- A ray arriving below the critical angle can refract into the surrounding lower-index material, so a transparent tube does not trap all entry directions.
- Real fibres have cladding, losses and bending limits. A tight bend can change incidence angles and allow previously guided light to escape.
Answer (a) n = 1.50; speed = 2.0×10⁸ m s⁻¹; critical angle ≈41.8°; total internal reflection occurs at 45°. (b) Frequency stays fixed; speed and wavelength fall on entry; only rays meeting the total-internal-reflection condition are trapped in the ideal model.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A camera that must refocus after the object moves
A converging thin lens forms a sharp real image of a 12 mm-high object placed 150 mm in front of it. The screen is 300 mm behind the lens. Use 1/f = 1/u + 1/v with positive distances for this real-object, real-image setup. The supplied calibration points use reciprocal distances in m⁻¹. Supplied data (1/u / m⁻¹; 1/v / m⁻¹): (2, 8), (4, 6), (6, 4), (8, 2). (a) Find focal length, image height and image orientation. Describe two principal rays that locate the image. [3 marks] (b) The object moves to 300 mm from the lens. Find the new screen distance and image height; explain what happens if it moves inside the focal length. [3 marks]
Thin-lens calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): 1/v / m⁻¹
Horizontal axis (x): 1/u / m⁻¹
Read the exact data as a table
| 1/u / m⁻¹ | 1/v / m⁻¹ |
|---|---|
| 2 | 8 |
| 4 | 6 |
| 6 | 4 |
| 8 | 2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find focal length, image height and image orientation. Describe two principal rays that locate the image.
Reveal the worked solution
- 1/f = 1/150+1/300 = 1/100 in mm⁻¹, so f = 100 mm, agreeing with the reciprocal-axis intercept 10 m⁻¹.
- Magnification magnitude is v/u = 300/150 = 2, giving image height 24 mm. A real image from a single converging lens is inverted.
- Draw a ray parallel to the principal axis refracting through the far focus, and a ray through the optical centre continuing straight. Their crossing determines the image position.
- 1/v = 1/100−1/300 = 1/150, so the screen must move to 150 mm behind the lens.
- Magnification magnitude is 150/300 = 0.50, giving 6 mm image height. This is a quantitative focusing requirement for a camera or projector design.
- With the object inside f, outgoing rays diverge and only their backward extensions meet. The virtual image can be viewed but not projected directly onto a screen.
Answer (a) f = 100 mm; image height magnitude = 24 mm; image is real and inverted. (b) New v = 150 mm and image height = 6 mm; inside f the image becomes virtual, upright and cannot be caught on a screen.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.