In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.
The distance between the lenses that gives sharp image twice is 
and the ratio of the sizes of the image is 
.
Given:
The distance between the object and the screen is given as D, to find the images formed we need to find the relationship between magnification and sizes of both object and image.
Formula used:
Displacement Method: A method to determine the focal length of a converging lens by displacing the lens on the principal axis and gaining sharp and clear image at two position on the principal axis.
Magnification Law:
where
v is the length of the image formed, u is the size of the object placed on the principal axis, f is the focal length of the lens, M is the magnification, D is the distance between the object and the screen and L is the distance between the two position of lens which gives sharp image.
Explanation:
According to the displacement method, the focal length of the converging lens is
Therefore, the distance between the lenses that gives sharp image twice is
.
Case I: (
and
)
The magnification ratio of case I
Case II: (
and
)
The magnification ratio of case I
Hence, to find the ratio of the size of the images from the 2 points I and II is
.