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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.

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Key term

State the thin lens equation as used in the Cartesian convention.

Key term

In the Cartesian convention, which direction along the axis is taken as positive?

Numbers

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. 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. 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. 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. 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. 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. 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.

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