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Lenses

The lens equation

  • 3 things to know
  • Unit AS 2

Overview

Lens equation and define each symbol and sign convention used with the lens equation.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

State the sign convention used with the lens equation.

Numbers

An object is 0.30 m from a converging lens of focal length 0.20 m. Calculate the image distance.

Numbers

An object is 0.20 m from a converging lens of focal length 0.12 m. Calculate the image distance.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the lens equation and define each symbol.

    Model answer

    1/f = 1/u + 1/v, where f is the focal length in metres, u is the distance from the object to the lens in metres and v is the distance from the lens to the image in metres.

  2. 2

    State the sign convention used with the lens equation.

    Model answer

    Real distances are taken as positive, so a real object and a real image have positive u and v, a converging lens has a positive focal length, and a diverging lens or a virtual image is given a negative value.

  3. 3

    What sign is given to the focal length of a converging lens and to that of a diverging lens?

    Model answer

    A converging lens is given a positive focal length and a diverging lens a negative focal length.

  4. 4

    What does a negative value of the image distance v tell you about the image?

    Model answer

    It tells you the image is virtual and lies on the same side of the lens as the object.

  5. 5

    An object is 0.30 m from a converging lens of focal length 0.20 m. Calculate the image distance.

    Model answer

    1/v = 1/0.20 − 1/0.30 = 5.00 − 3.33 = 1.67 m⁻¹, so v = 0.60 m and the image is real, and 1/0.30 + 1/0.60 = 5.00 m⁻¹ returns a focal length of 0.20 m as given.

  6. 6

    An object is 0.20 m from a converging lens of focal length 0.12 m. Calculate the image distance.

    Model answer

    1/v = 1/0.12 − 1/0.20 = 8.33 − 5.00 = 3.33 m⁻¹, so v = 0.30 m, and 1/0.20 + 1/0.30 = 8.33 m⁻¹ returns a focal length of 0.12 m as given.

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