5. Convex lens ray diagram for when the object is at a distance between F and the lens

• Ray diagram 5 (below): The formation of a virtual image by a convex lens when the object O is between F and the convex lens - here the convex lens is acting as a magnifying glass.

5a converging convex lens

• To construct the diagram 5a, as with the others, (i) from the top of the object you take the ray parallel to the principal axis down through the principal focus point F. That is what parallel rays with the object is standing on the axis.

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

• (ii) The second ray from the top of the object you take down through the centre of the lens without deviation.

• However, in this case these rays do not intersect to give you the position of the image - they diverge, but all is not lost to get to an image!

• Therefore you have to extrapolate back with the dotted line virtual rays until they intersect.

  • Virtual rays are where the rays from the object appear to come from, they do NOT exist in reality.

  • BUT, you can't construct the ray diagram to get to the characteristics of the image without using them!

• This then gives you the position and size of the virtual image - this time on the same side of the lens as the object.

• The image I is virtual, upright - right way up (erect, NOT inverted), bigger than the object (a 'magnifying glass' effect)

  • Unlike all the other situations for a convex lens, the (not real) virtual image is on the same side of the lens as the object and beyond the object.

  •  In this case the image is between the distances F and 2F to the left of the lens.

  • Along the line of the principal axis, the thicker (more curved) the convex lens, the shorter the focal length F and the greater the magnifying power of the lens.

• The thickness of the lens affects its magnifying power - the thicker the lens, the more powerful the magnifying glass.

• 

• 5b converging convex lens

• Above is quick sketch of how to do the ray diagram 5 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.

• This is the ray diagram for a convex lens acting as a magnifying glass.

  • You should know the magnification formula:

  • magnification = size (height) of image  ÷  size (height) of object

  • e.g. from diagram 5b, if the image was 20 mm high and the object was 4 mm high

    • magnification = 20 ÷ 10 = 2 (no units, but remember the two sizes must be in the same length units!)

    • 2nd example of calculation

    • Suppose the magnifying power of a lens is 3.0.

    • If an object is 2.0 cm high, calculate the size of the image.

      • Rearranging the magnification formula:

      • size of image = magnification x size of object

      • size of image = 3.0 x 2.0 = 6.0 cm

    • From the diagram you can also see the focal length of the lens is 2 cm.

  • This is a very simple example, but the method works for any given set of data.

  • All you need to be given is the (i) focal length of the lens, (ii) the size of the object and (iii) the distance from the lens to the object. From the graph diagram you can work out everything else e.g. the size of the image and the magnifying power of the lens.

  • Below is a more elaborate graph paper ray diagram 5c for a convex 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.

• 

  5c magnifying lens

  • (i) Draw a line from the top of the object to the lens and, after the lens, down through the focal point F to well beyond a distance of F - then extrapolate back with dotted lines

  • (ii) Draw a diagonal line from the top of the object down through the centre of the lens and beyond the intersection with ray (i).

    • In both cases, extrapolate back with dotted lines

    • The intersection of dotted lines from (i) and (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 down through F.

  • (iv) Draw a line from the bottom of the object diagonally down through the centre of the lens and beyond the intersection with ray (iii).

    • In both cases, extrapolate back with dotted lines.

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

  • The intersection of rays (iii) and (iv) gives you the position of the top of the upright virtual image.

  • From the graph, and measuring in 'squares' you can work out the ...

  • magnification = size of image / size of object = 16 / 5 = 3.2