Refraction of Light by Lenses

Rays diagrams are normally used to illustrates the formation of images by lenses.

1. A ray parallel to the principal axis passes through or appears to diverge from the principal focus after refraction.
2. A ray of light passing through the principal focus of a lens is refracted parallel to the principal axis of the lens.
3. A ray of light through the optical center of the lens continues through undeviating (Not change direction)

The nature, position and size of the image formed by a lens depends on the position of the object in relation to the type of lens. For Example in converging lens when the object is between the lens and principal focus the image will be formed at the same side as the object but further from the lens. It is virtual, erect, and magnified. The image by concave lens is erect, virtual and reduced.

The nature, position and relative size of the image formed by convex lens for various positions of the object is summarized in the table below:

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus $F_2$	Highly diminished, point-sized	Real and inverted
Beyond $2F_1$	Between $F_2$ and $2F_2$	Diminished	Real and inverted
At $2F_1$	At $2F_2$	Same size	Real and inverted
Between $F_1$ and $2F_1$	Beyond $2F_2$	Enlarged	Real and inverted
At focus $F_1$	At infinity	Infinitely large or highly enlarged	Real and inverted
Between focus $F_1$ and optical centre O	On the same side of the lens as the object	Enlarged	Virtual and erect

The magnification is defined as the ratio of the height of the image and the height of the object. It is represented by the letter m. If h is the height of the object and h′ is the height of the image given by a lens, then the magnification produced by the lens is given by,

$m = \frac{\text{Height of the Image}}{\text{Height of the object}} = \frac{h'}{h}$

Magnification produced by a lens is also related to the object-distance u, and the image-distance ν. This relationship is given by

$m = \frac{h'}{h} = \frac{\nu}{u}$

Example

A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.

Solution

A concave lens always forms a virtual, erect image on the same side of the object.

Image-distance v = –10 cm;

Focal length f = –15 cm;

Object-distance u = ?

Since,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$

$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15}$

$\frac{1}{u} = \frac{-3+2}{30} = \frac{1}{-30}$

$u = -30 \text{ cm}$

Thus, the object-distance is 30 cm.

Magnification

$m = \frac{v}{u}$

$m = \frac{-10 \text{ cm}}{-30 \text{ cm}} = \frac{1}{3} = +0.33$

The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

The lens equation is given as

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

if sign convection is used for u, v and f the equation applies to both converging and diverging lenses for all cases of object and image.

Example 8

An object is placed 12 cm from converging lens of focal length 18 cm. Find the position of the image.

Solution

Since the lens is converging f = +18 cm.

$\frac{1}{v} = \frac{1}{18} - \frac{1}{12}, \quad v = -36$

The image is virtual.