8. Ray diagram showing the formation of a virtual image by a concave lens

• Ray diagram 7b

• The formation of an image O by a concave lens.

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  7b diverging concave lens

  • (i) The ray from the top of the object is taken down through the centre of the lens undeviated.

    • Apart from the axis line, this is essentially a 2 ray diagram for an object 'standing' on the axis line.

    • The object is standing on the principal axis of the concave lens.

  • (ii) You then take the ray parallel to the principal focus and diverge it in line with the principal focus F - that's how parallel rays behave. You extrapolate ray (i) back down to F with a dotted line.

  • Where the two lines intersect gives you the top of the image and hence its size too.

  • A concave lens always produces a virtual image, the right way up, smaller than the object and is situated somewhere between the object and the lens and on same side as object).

    • These image characteristics are independent of the position of the object - it can be anywhere!

    • Reminder: A virtual image cannot be projected onto a screen.

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  • 7c concave lens

  • Above is quick sketch 7c of how to do the ray diagram 7b on graph paper. If done very carefully to scale, you can then calculate the height of the image I and the distance from the lens to the image I.

• Below is a more elaborate graph paper ray diagram 7d for a concave lens where the object is placed at a distance less than F from the lens, BUT, above the central axis of the lens - four rays are marked (i) to (iv), each intersecting pairs of lines ('rays') gives you the top and the bottom of the image.

  • 7d concave lens

  • (i) Draw a line from the top of the object O to the lens and, after the lens, diverging - you then extrapolate back to the principal focus F - this is how rays parallel to the principal axis behave.

  • (ii) Draw a diagonal line from the top of the object down through the centre of the lens.

    • The intersection of dotted line (i) and line (ii) gives you the position of the top of the upright virtual image.

  • (iii) Draw a line from the bottom of the object parallel to the principal axis and, after the lens, diagonally diverged and extrapolated with a dotted line back to  F.

  • (iv) Draw a line from the bottom of the object diagonally down through the centre of the lens.

    • The intersection of dotted ray (iii) and ray (iv) gives you the position of the bottom of the upright virtual image.

  • Note the image is again smaller than the object - it always is for a concave lens.