Everything Science Grade 11

CHAPTER 13. GEOMETRICAL OPTICS 13.2

The image of an object placed at a distance between 2 f and f from the lens is upside down or inverted.
This is because the rayswhich began at the topof the object, above the principal axis, afterpassing
through the lens end up below the principal axis. Theimage is called a real image because it is on the
opposite side of the lensto the object and you can trace all the light raysdirectly from the imageback
to the object.

The image is larger thanthe object and is locatedat a distance greater than 2 f away from the lens.

CASE 4:

Object placed at a distance less than f from the lens

ImageF 1 Object

F 2

O

R 1

R 3

R 2

f f f

Figure 13.10: An objectis placed at a distance less than f away from the converging lens. Three rays
are drawn to locate theimage, which is virtual,larger than the object and upright.

We can locate the position of the image by drawing our three rays. R 1 travels from the object to the
lens parallel to the principal axis and is bent bythe lens and then travels through the focal point. R 2
passes through the focal point before it entersthe lens and therefore must leave the lens parallel to
the principal axis. R 3 travels through the centre of the lens and doesnot change direction. The point
where R 1 , R 2 and R 3 intersect is the image ofthe point where they allstarted.

The image of an objectplaced at a distance lessthan f from the lens is upright. The image is called a
virtual image because it is on the same side of the lens as theobject and you cannot trace all the light
rays directly from the image back to the object.

The image is larger thanthe object and is locatedfurther away from the lens than the object.

Extension: The thin lens equation and magnification

The Thin Lens Equation

We can find the position of the image of a lensmathematically as thereis a mathematical

relation between the object distance, image distance, and focal length.The equation is:

1

f

=

1

do

+

1

di

where f is the focal length, dois the object distance and diis the image distance.

The object distance dois the distance from theobject to the lens. dois positive if the object

is on the same side of the lens as the light raysenter the lens. This should make sense: we