MYP Physics · topic MYP-11.05
Optical instruments and design
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 camera that must refocus after the object moves
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.