An object is 0.060 m from a converging lens of focal length 0.10 m. Where is the image?

1/v = 1/0.10 − 1/0.060 = −6.67 m⁻¹, so v = −0.15 m; the negative value shows a virtual image 0.15 m from the lens on the same side as the object.

An object is 0.30 m from a diverging lens of focal length −0.20 m. Where is the image and what is its nature?

1/v = −1/0.20 − 1/0.30 = −8.33 m⁻¹, so v = −0.12 m; the image is virtual and upright, 0.12 m from the lens on the same side as the object.

What are two equivalent expressions for the magnification produced by a lens?

Magnification is image height divided by object height, and it also equals v/u, the image distance divided by the object distance.

What is the magnification in the case where an object 0.30 m from a lens gives an image 0.60 m beyond it, and how tall is the image of a 20 mm object?

The magnification is 0.60/0.30 = 2.0, so the image of a 20 mm object is 40 mm tall.

As a distant object is brought slowly towards the principal focus of a converging lens, how does the real image change?

The image moves steadily further from the lens and grows larger, and as the object reaches the focus the image recedes towards infinity.

Edexcel A-Level Physics 9PH0

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Waves and Particle Nature of Light

Lens calculations and magnification

2 things to know
Papers 2 and 3

Overview

Recall questions on solving the thin lens equation for converging and diverging lenses, the two expressions for magnification, and what covering half a lens does.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Process

Where is the image formed by a diverging lens, and what happens when an object is placed at the principal focus of a converging lens?

Numbers

What does a magnification of less than 1 tell you about the image?

Questions from this subtopic

Each one with the model answer underneath.

1
An object is 0.060 m from a converging lens of focal length 0.10 m. Where is the image?
Model answer
1/v = 1/0.10 − 1/0.060 = −6.67 m⁻¹, so v = −0.15 m; the negative value shows a virtual image 0.15 m from the lens on the same side as the object.
2
An object is 0.30 m from a diverging lens of focal length −0.20 m. Where is the image and what is its nature?
Model answer
1/v = −1/0.20 − 1/0.30 = −8.33 m⁻¹, so v = −0.12 m; the image is virtual and upright, 0.12 m from the lens on the same side as the object.
3
What are two equivalent expressions for the magnification produced by a lens?
Model answer
Magnification is image height divided by object height, and it also equals v/u, the image distance divided by the object distance.
4
What is the magnification in the case where an object 0.30 m from a lens gives an image 0.60 m beyond it, and how tall is the image of a 20 mm object?
Model answer
The magnification is 0.60/0.30 = 2.0, so the image of a 20 mm object is 40 mm tall.
5
What does a magnification of less than 1 tell you about the image?
Model answer
The image is smaller than the object, so it is diminished.
6
As a distant object is brought slowly towards the principal focus of a converging lens, how does the real image change?
Model answer
The image moves steadily further from the lens and grows larger, and as the object reaches the focus the image recedes towards infinity.

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