OCR A-Level Physics B H557
Planning
Implementing
Analysis
Evaluation
Practical Skills
Use of Apparatus and Techniques
Fundamental Data Analysis
Imaging and Signalling
Sensing
Mechanical Properties of Materials
Waves and Quantum Behaviour
Space, Time and Motion
Creating Models
Out Into Space
Our Place in the Universe
Matter: Very Simple
Matter: Hot or Cold
Electromagnetism
Charge and Field
Probing Deep Into Matter
Ionising Radiation and Risk
Imaging and Signalling
The thin lens equation and the Cartesian convention
- 3 things to know
- Components 01, 02 and 03
Overview
Recall questions on the sign convention for object and image distances, calculating image position, focal length and power, the least possible object–image separation, and covering half a lens.
What you need to know
Answer each one before you reveal it.
In the Cartesian convention, which direction along the axis is taken as positive?
An object is 0.30 m in front of a converging lens of focal length 0.20 m. Calculate the image distance.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the thin lens equation as used in the Cartesian convention.
Model answer
1/v = 1/u + 1/f, with u, v and f all measured from the centre of the lens.
- 2
In the Cartesian convention, which direction along the axis is taken as positive?
Model answer
Distances measured from the lens in the direction the light is travelling are positive, and distances measured against it are negative.
- 3
What sign does the object distance take for a real object placed to the left of a lens, with light travelling from left to right?
Model answer
It is negative, because it is measured from the lens against the direction in which the light travels.
- 4
What does a positive value of the image distance tell you about the image?
Model answer
It tells you the image is real and forms on the far side of the lens from the object.
- 5
An object is 0.30 m in front of a converging lens of focal length 0.20 m. Calculate the image distance.
Model answer
1/v = −1/0.30 + 1/0.20 = 1.67 m⁻¹, so v = +0.60 m and a real image forms 0.60 m beyond the lens.
- 6
An object is 0.80 m in front of a converging lens of power +2.5 D. Calculate where the image forms.
Model answer
The focal length is 0.40 m, so 1/v = −1.25 + 2.50 = 1.25 m⁻¹ and v = +0.80 m, the same distance beyond the lens.
Other ways to work this
The same content, four other ways.
What is inside Levo
Six ways to work the same material.
Every question in Imaging and Signalling, on a schedule.
Attempt a question from memory, write the correct version, then attempt it again. Levo records the jump between your first attempt and your last, and brings each question back on the day you are due to forget it.
Or £46 a year, about £3.83 a month. Cancel anytime.