Mr Toogood's Physics

Refraction

All lenses employ refraction to bring light rays to a focus at a single point. When light enters a lens it is refracted. Refraction, at its most simple, is the change in direction that light experiences as it passes from one medium into another. When light travels into a medium with a higher refractive index than the one it is currently in, such as from air into glass, its direction changes, and it turns towards the normal line.

As the light leaves the glass into air, which has a lower refractive index, it bends away from the normal line. When we look refraction in a rectangular block, such as the one above, we can see that the emergent ray is parallel to the incident ray. If we had several rays, all parallel, incident on the block, the would all emerge parallel, so clearly, a simple rectangular block cannot be used to focus light rays.

Lenses are curved so that rays are refracted by different amounts depending in their distance from an axis through the centre of the lens. A ray travelling along this line would not be refracted at all, and the further the ray from this axis, the greater the angle through which it is refracted. Whereas with the parallel sided block, parallel incident rays emerge parallel, with a lens, parallel incident rays are brought to a focus at a single point.

As can be seen from the diagram above light is refracted as it passes into the lens, and again when it emerges from the lens. If the lens is fat the image it creates could be laterally translated, which causes distortions in the image. To prevent this lenses are manufactured to be as thin as possible. Light will still be refracted twice, but the effect is greatly reduced. To simplify diagrams and calculations we assume that the light is refracted once, at the centre of the lens.

The image made through a simple arrangement of lenses can be projected onto a screen, and is called a real image and will be inverted (upside down). Whether the image is magnified (larger) or diminished (smaller) will depend on how far the object is from the lens.

As you can see from the image above, the rays scattering from the toy Utahraptor (the object) spread out and do not reach the lens parallel to each other. This means that they will not be focused at the principal focus.

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Ray diagrams

The way that light behaves through lenses can be well understood by constructing ray diagrams, so it is important to learn and to be able to use these key terms associated with lenses and the diagrams that help describe how they work.

• Principal axis - This is the axis passing through the centre of the lens, and perpendicular to the lens axis.
• Principal focus - Otherwise called the focal point, this is the single point where parallel incident rays converge and are brought to a focus. When the incident rays are parallel to the principal axis, the focal point will also lie on the principle axis.
• Focal length - This is the distance from the lens axis of the focal point. Distances around the lens are often measured in integer multiples of the focal length, e.g. 1F, 2F.
• Focal plane - The focal plane is a plane perpendicular to the principal axis. If the incident rays are parallel but non-axial (not parallel to the principal axis) then the principal focus will not lie on the principle axis, but it will lie somewhere on the focal plane.

The lens also has two sides, the image side, from which incident rays approach the lens, and the objective side on which emergent rays form an image.

It is not always necessary to draw a realistic lens, the symbols below can be used. And in fact on a ray diagram ,the lens can be drawn any size, the rays would still refract at the lens axis.

When constructing a ray diagram it is useful to remember three standard rays,

1. Rays passing through the centre of a lens remain undeviated and can be drawn as one continuous line.





2. An incident ray parallel to the principal axis will be refracted and pass through the focal point.





3. An incident ray passing through the focal point on the objective side of the lens will emerge parallel to the principal axis.

You need to know how to draw ray diagrams for objects at different distances from a lens. When doing this, draw two of the standard rays above, emanating from one point on the object. Where they cross on the image side of the lens is where the image would appear.

• An object placed at a distance greater than 2F






Position	Between 1F and 2F
Nature	Real
Orientation	Inverted
Magnification	Diminished
Use	Camera lens



• An object placed at a distance equal to 2F






Position	At 2F
Nature	Real
Orientation	Inverted
Magnification	Same size
Use



• An object placed at a distance greater than 1F and less that 2F






Position	Greater than 2F
Nature	Real
Orientation	Inverted
Magnification	Magnified
Use	Projector



• An object placed at a distance equal to 1F






Position	At infinity
Nature	Real (at infinity)
Orientation	Inverted
Magnification	N/A
Use	Searchlight



• An object placed at a distance less than 1F






Position	On the objective side at 2F
Nature	Virtual
Orientation	Upright
Magnification	Magnified
Use	Magnifying glass.

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Magnification and the lens equation

In the examples above the distances are measured in terms of the focal length of the lens, but they can also be measured in metres, in which case the distance of the object from the lens is given the letter $u$ and the distance of the image from the lens is $y$.

The magnification of the image is a ratio of the size of the object to the size of the image.

As it is a ratio, magnification has no units. Looking at the diagram below we can see that the ray diagram can be understood as two similar triangles, and so, if the above magnification describes the ratio of the two opposite sides of the triangles it must also be true that the same ratio applies to the two adjacent sides of the triangles, which represent $u$ and $v$.

Therefore we can also describe the magnification as:

Looking at the ray diagrams geometrically allows us to derive an equation that links the object and image distances to the focal length of the lens.

And the angle θ (from the blue triangle) is:

We can now rearrange the term for $α$ to make $h_{i}$ the subject,

And substitute it into the equation for $θ$:

We can divide both sides by $h_{o}$ and then multiply out the brackets to give:

Which when given in terms of $\frac{1}{f}$ is called the lens equation:

Using this equation all positive values are real, and all negative values are virtual, so a negative value for $\frac{1}{v}$ would indicate that a virtual image had been created. The term $\frac{1}{f}$ is also known as the power of the lens, and has units of $\units{m^{-1}}$ or dioptres, symbol $\units{d}$.

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Worked example 1

Find the position, nature and magnification of the image when an object is placed $\quantity{20}{cm}$ from a convex lens with a focal length of $\quantity{15}{cm}$

The value of $f$ is $\quantity{15}{cm}$ and the value of $u$ is $\quantity{20}{cm}$, so using the lens equation we can find the position and nature of the image:

So,

As this value is positive it also tells us that the image is real. We should also know that as it is a convex lens making a real image it will be inverted. We cannot read this from the equation, we just have to know it!

The magnification is the ratio of the object position, $u$, and the image position, $v$:

So the image is magnified by a factor of three.

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Worked example 2

A converging lens forms a real image of a real object. If the image is twice the size of the object and $\quantity{90}{cm}$ from it, calculate the the focal length of the lens?

We can solve this either graphically, by drawing a ray diagram, or by using the lens equation. From the question we know the following:

• $M=2$
• $u+v=\quantity{90}{cm}$
• $f$ is unknown

The ray diagram of this situation is shown below, and as long as it is drawn to scale.

Firstly, the object and the image are drawn on the principle axis using an appropriate scale, such as $\quantity{1}{cm}=\quantity{10}{cm}$ . As we don’t know where the focal point is, we also do not know where the lens is either. We can, however, draw one standard ray from the top of the object to the top of the image. We can now draw the lens where the ray crosses the principle axis.

We can now fill in the rest of the ray diagram, including the lens, and another standard ray, which allows us to identify the position of the focal point. We can now measure this with a ruler and convert using our scale.

We can also use the lens equation to solve this problem. We know that the magnification is 2, which also means that $\frac{v}{u}=2$, and we know that $u+v=\quantity{90}{cm}$, therefore:

$v=2u$ and $v=90-u$

So

Therefore

We can now use the lens equation to find $f$

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