Lenses work by refracting light at a glass-air boundary. Although refraction occurs at the boundary, we will treat all lenses as bending the rays at the lens axis. You may want to go back to Unit 2 to revise refraction.
The lens in the eye is a convex or converging lens. This means that the lens makes rays of light come together, or converge.
The rays parallel to the principal axis are converged onto the principal focus. The focal length is the distance between the lens axis and the principal focus (strictly speaking, the focal plane).
Thicker lenses bend light more, and are therefore described as more powerful. Powerful lenses have short focal lengths. The power of a lens is measured in dioptres (D) and is given by the formula:
Power = 1
focal length (m)
|The optician's prescription for a lens is +0.2 D What is the focal length in metres?|
The principal focus of a convex lens is called real. The images made by convex lenses are in most cases real. This means that the image can be projected onto a screen. We will see later how images are made with ray diagrams.
The concave lens splits rays parallel to the principal axis, which is why it is called a diverging lens.
The principal focus is virtual because the rays do not pass through it , but diverge as if they had come from it. Images in concave lenses are always virtual because they cannot be projected onto a screen.
The power of a concave lens is always negative, for example -1.5 D.
We can determine where an image lies in relation to the objects by using a ray diagram. We can do this by using two simple rules:
Draw a ray from the top of the image parallel to the principal axis. This ray bends at the lens axis and goes through the principal focus.
Draw a ray from the top of the lens through the centre of the lens.
Where the two rays meet, that is where the image is found. The diagrams shows how we do a ray diagram step-by-step:
Step 1 Draw the ray parallel to the principal axis.
Step 2 Draw the refracted ray so that it passes through the principal focus.
Step 3 Draw a ray from the top of the object through the middle of the lens. This ray is undeviated.
Step 4 Where the rays meet, that is where the image is.
It is a good idea to draw your ray diagrams on graph paper as the following ray diagrams are. Be careful with your drawing; a small change in the angle of the undeviated ray can lead to quite a big change in the final position of the image. And PLEASE... Be a good chap and use a sharp pencil.
Click here to look at a ray diagram done on graph paper
This diagram shows where an object is at a distance of greater than twice the focal length. The image is inverted (upside down), real, and diminished (smaller).
What is the image like if the object is at 2F?
|Question 3||What is the image like if the object is between 2F and F?|
|Question 4||What is the image like if the object is at F?|
|Question 5||What is the image like if the object is less than F?|
For a concave lens, the process is similar, except that we extend the refracted parallel ray back to the virtual principal focus.
|What is the image like for a concave lens?|
The Lens Formula
Lens diagrams have the main disadvantage that there is uncertainty in precisely where the image is. Therefore the use of the lens formula is better. The lens formula is:
[f - focal length; u - object distance; v - image distance]
|An object of height 1.6 cm is placed 50 cm from a converging (convex) lens of focal length 10 cm. What is the position of the image?|
|v = 1/0.08 = 12.5 cm|
It does not matter if you work in cm, as long as you are consistent. However if you are going to use dioptres you must work in metres.
The magnification is worked out using this simple formula:
Since v is in metres, and u is in metres, M has no units.
|What is the magnification in the example above? What is the size of the image?|
|M = 12.5 ÷ 50 = 0.25|
|Image 1.6 × 0.25 = 0.4 cm = 4 mm|
The convention for the equation is that real is positive. For a concave lens, the focal length is negative, because the principal focus is virtual. If the image position gives a negative value, then the image is virtual.
|Question 7||Find the position and size of a pound coin, 2.2 cm in diameter placed 20 cm from a converging lens of focal length 40 cm|
|Question 8||The same coin is now placed 20 cm in front of a diverging lens of focal length 40 cm. What is the position and the size of the image now?|
Defects of Vision
The optician uses lenses to correct defects of vision. There are two defects that we will think about here:
Myopia (short sight), where the cornea is curved too much, or the lens is too powerful. The power of the eye is too great and the image of far objects is formed in front of the retina. Far objects appear blurred.
Hypermetropia (long sight) is where the eye is not powerful enough, leading to the image of near objects is behind the retina. Near objects appear blurred.
Short sight is corrected by using a concave (diverging lens) to make the parallel rays move apart. The optician will find the far point, which is the furthest distance that the unaided eye can focus a clear image.
|The far point of a patient is 60 cm. If the distance from the lens to the retina is 0.02 cm, what is the power of the eye. What is the power of the eye with a corrective lens? What is the power of the corrective lens? What kind of lens is it and what is its focal length?|
|P = 1/f = 1.67 + 50 = 51.67 D|
|For the corrected eye
|Power of the corrective
lens = Power of the corrected eye - power of the uncorrected eye
Power = 50 - 51.67 = -1.67 D
|Since the power is negative, the lens must be concave (diverging).|
|Its focal length is 0.6 m|
The diagram shows how short sight can be corrected using a diverging lens.
Spectacle lenses are not biconcave as shown, as they look odd. Instead the front is slightly convex and the back is much more concave. It's a matter of taste, you know...
The near point is the point that the eye can focus onto a close up object. The eye is made more powerful by the lens being made fatter. Generally the eye can change power by about 5 dioptres.
|What is the change in focal length represented by 5 dioptres?|
|What would the near point be of the patient in the example above, assuming the eye has a power change of 5 D?|
|At its most powerful, the eye would have a power of 56.67 D|
|Power = 57.67 D = 1/u + 1/0.02 = 1/u + 50|
|1/u = 7.67|
|u = 0.13 m|
|What is the near point of the patient's eye in the examples above with corrective lenses on?|
With long sight, a converging lens is used to make the eye more powerful. The corrective lens is shown in the diagram below:
A patient has a near point of 3.0 metres. Assuming that the distance from the front of the eye to the retina is 2 cm,
(a) what is the power of her eye unaided?
(b) What power of eye will she need to read a book at 25 cm?
(c) What power should the lens be and what kind of lens is it?
(d) What is the focal length?
Astigmatism is a condition where the cornea is not spherically curved. It has different curves in different directions, so that images are sharper in one direction than another. Lenses are made to compensate for the differences. Astigmatism can also be combined with short or long sight.
Contact lenses can be placed directly on the eye. They correct in exactly the same way as the lenses for spectacles. Monocles are single lenses to correct the defect in one eye only. They are associated with the stereotype of an upper-class twit, and are, nowadays, a prop for period drama.
Lenses work by bending light
Convex lenses converge rays together onto a principal focus
Concave lenses diverge light rays.
Concave lenses correct short sight
Convex lenses correct long sight.
Specially shaped lenses are needed to correct astigmatism