Focal Length, Images, and Magnification

Lenses can produce images that are larger, smaller, upright, inverted, real, or virtual. The type of image formed depends on the type of lens and the position of the object relative to the lens's focal point.

Two important measurements help us describe lenses:

• focal length — describes how strongly a lens converges or diverges light
• magnification — compares the size of an image with the size of the original object

Ray diagrams allow us to predict an image's position, size, orientation, and type.

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The Principal Axis

The principal axis is an imaginary straight line passing through the centre of a lens.

It provides a reference line for drawing ray diagrams.

Important points such as:

• F
• 2F
• the optical centre

are positioned along the principal axis.

A good ray diagram should always begin by drawing the principal axis.

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Optical Centre

The optical centre is the central reference point of a thin lens.

In simple ray diagrams, a ray passing through the optical centre is treated as continuing approximately straight without changing direction.

This makes the central ray particularly useful when locating images.

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Focal Point

The focal point, F, is a point associated with the convergence or apparent divergence of rays that enter a lens parallel to the principal axis.

For a converging lens, parallel incoming rays are refracted so that they meet at the focal point on the opposite side of the lens.

For a diverging lens, parallel incoming rays spread apart after refraction. Their backward extensions appear to originate from a focal point on the same side as the incoming light.

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Focal Length

The focal length, f, is the distance between:

optical centre of lens → focal point

Focal length is usually measured in:

• metres
• centimetres
• millimetres

depending on the application.

For example, if the focal point is 20 cm from the optical centre:

f = 20 cm

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Focal Points on Both Sides

A thin lens has a focal point on each side.

These focal points are the same distance from the optical centre for an ideal symmetric thin lens in the same surrounding medium.

Therefore, if:

f = 10 cm

one focal point is 10 cm to the left and the other is 10 cm to the right.

This is why ray diagrams normally show:

F — lens — F

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The 2F Position

Ray diagrams for converging lenses often mark another position:

2F

This point is located twice the focal length from the lens.

If:

f = 15 cm

then:

2f = 30 cm

So 2F is 30 cm from the optical centre.

The object's position relative to F and 2F strongly affects the image produced by a converging lens.

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Converging Lenses

A converging lens is usually thicker in the centre than at the edges.

It is commonly called a:

convex lens

Parallel rays entering the lens are refracted toward the principal axis and meet at the focal point.

The focal point is therefore a real focus because the light rays actually pass through it.

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Diverging Lenses

A diverging lens is usually thinner in the centre than at the edges.

It is commonly called a:

concave lens

Parallel rays entering the lens spread apart after refraction.

The rays do not actually pass through the focal point.

Instead, their backward extensions appear to originate there.

The focal point is therefore a:

virtual focus

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Comparing the Focal Points

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Focal Length and Lens Strength

Focal length tells us something about the focusing strength of a lens.

A lens that bends light strongly brings rays to a focus over a shorter distance.

Therefore:

short focal length → stronger lens

long focal length → weaker lens

For example, a converging lens with:

f = 5 cm

focuses parallel rays more strongly than one with:

f = 50 cm

assuming comparable conditions.

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Real Images

A real image forms where actual light rays converge.

Because light really passes through the image position, a real image can usually be projected onto a screen.

For a single converging lens, an object placed beyond the focal point can form a real image.

Real images produced by a single converging lens are:

inverted

relative to the object.

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Virtual Images

A virtual image forms where light rays appear to originate.

The rays do not actually meet at the virtual image position.

Therefore, a virtual image cannot normally be projected directly onto a screen.

Examples include:

• an image viewed through a magnifying glass
• the image produced by a diverging lens
• an image in a plane mirror

Virtual images produced by a single lens for the usual cases studied here are upright.

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Identifying Real and Virtual Images from a Ray Diagram

Look at the rays after they pass through the lens.

If the actual refracted rays meet:

real image

If the actual rays spread apart but their backward extensions meet:

virtual image

Dashed lines are normally used for backward extensions because they do not represent actual paths travelled by the light.

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Principal Rays

Three standard rays are useful when constructing lens diagrams.

For a converging lens:

Parallel ray

Travels parallel to the principal axis before the lens and refracts through the far focal point.

Central ray

Passes through the optical centre and continues approximately straight.

Focal ray

Passes through the near focal point before reaching the lens and emerges parallel to the principal axis.

Only two correctly drawn rays are required to locate the image.

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Constructing a Converging-Lens Ray Diagram

Follow this procedure:

1. Draw the principal axis.
2. Draw the lens.
3. Mark F on both sides.
4. Mark 2F on both sides.
5. Draw the object as an upright arrow.
6. Start a ray at the top of the object parallel to the principal axis.
7. After the lens, draw this ray through the far F.
8. Draw another ray from the top of the object through the optical centre.
9. Continue the central ray approximately straight.
10. Locate where the two refracted rays meet.
11. Draw the image.
12. Determine its position, size, orientation, and type.

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Object Beyond 2F

Consider an object positioned:

beyond 2F

The image forms:

between F and 2F

The image is:

• real
• inverted
• smaller than the object

So:

object beyond 2F → reduced image

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Object at 2F

If the object is positioned at:

2F

the image forms at:

2F on the opposite side

The image is:

• real
• inverted
• same size as the object

Therefore:

object at 2F → image at 2F

and:

magnification = 1

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Object Between F and 2F

If the object is:

between F and 2F

the image forms:

beyond 2F

The image is:

• real
• inverted
• larger than the object

Therefore:

object between F and 2F → enlarged real image

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Object at F

If the object is placed at the focal point, rays emerging from the lens are approximately parallel.

In the ideal thin-lens model, the image is effectively located infinitely far away.

This is a limiting case.

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Object Inside F

If the object is:

between F and the lens

the refracted rays diverge.

Extend them backward using dashed lines.

Their extensions meet on the same side of the lens as the object.

The image is:

• virtual
• upright
• enlarged

This is the arrangement used by a:

magnifying glass

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Converging-Lens Image Summary

This table is extremely useful for predicting images before drawing a ray diagram.

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Ray Diagrams for Diverging Lenses

For a diverging lens, two useful rays are:

Parallel ray

Travels parallel to the principal axis before the lens.

After refraction it diverges as though it came from the near focal point.

Central ray

Passes through the optical centre and continues approximately straight.

The refracted rays diverge.

Extend them backward with dashed lines.

Where the extensions meet is the virtual image position.

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Images from Diverging Lenses

For a normal real object in front of a single diverging lens, the image is always:

• virtual
• upright
• smaller
• between the lens and the near focal point

Changing the object distance changes the exact position and size, but these general characteristics remain the same.

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Image Size

The image formed by a lens may be:

• enlarged
• same size
• reduced

Instead of describing this only with words, we can calculate exactly how much larger or smaller the image is.

This quantity is called:

magnification

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Magnification

Magnification compares image size with object size.

Use:

magnification = image height / object height

or:

M = hᵢ/hₒ

where:

M = magnification

hᵢ = image height

hₒ = object height

Because magnification is a ratio of two lengths measured in the same units, it has:

no unit

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Interpreting Magnification

If:

M > 1

the image is larger than the object.

If:

M = 1

the image is the same size as the object.

If:

M < 1

the image is smaller than the object.

For example:

M = 3

means the image is:

3 times the object's size

M = 0.5

means the image is:

half the object's size

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Worked Example: Enlarged Image

An object is:

4 cm tall

Its image is:

12 cm tall

Calculate the magnification.

Use:

M = hᵢ/hₒ

M = 12/4

M = 3

The image is:

3 times larger in height than the object.

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Worked Example: Reduced Image

An object has a height of:

10 cm

Its image has a height of:

2.5 cm

M = hᵢ/hₒ

M = 2.5/10

M = 0.25

The image is one-quarter the object's height.

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Finding Image Height

The magnification equation can be rearranged.

Starting with:

M = hᵢ/hₒ

Multiply by hₒ:

hᵢ = M × hₒ

Suppose:

M = 2.5

hₒ = 6 cm

Then:

hᵢ = 2.5 × 6

hᵢ = 15 cm

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Finding Object Height

Starting with:

M = hᵢ/hₒ

Rearrange:

hₒ = hᵢ/M

Suppose:

hᵢ = 8 cm

M = 0.4

Then:

hₒ = 8/0.4

hₒ = 20 cm

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Units in Magnification Calculations

Image height and object height must use the same units.

For example:

hᵢ = 60 mm

hₒ = 3 cm

Convert first:

3 cm = 30 mm

Then:

M = 60/30

M = 2

Do not divide measurements expressed in different units without converting them first.

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Magnification and Image Orientation

At introductory level, magnification is often treated as a positive comparison of image and object sizes.

In more advanced lens calculations, a signed magnification may be used.

A common sign convention uses:

• positive magnification for upright images
• negative magnification for inverted images

For example:

M = −2

can indicate an image that is:

• twice the object's height
• inverted

Different courses may introduce sign conventions differently, so always use the convention required for the problem.

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Magnification from Image and Object Distances

For a thin lens, the magnitude of magnification can also be related to image and object distances:

|M| = |dᵢ/dₒ|

where:

dᵢ = image distance

dₒ = object distance

This means the geometry of a ray diagram connects:

• image size
• object size
• image distance
• object distance

For basic magnification problems, however, the height equation is often sufficient:

M = hᵢ/hₒ

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Worked Example Using Distances

An object is 15 cm from a lens.

Its real image forms 45 cm from the lens.

Magnitude of magnification:

|M| = 45/15

|M| = 3

The image is three times the object's height.

Because it is a real image formed by a single converging lens, it is also inverted.

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Predicting Size Without Calculating

A ray diagram can often tell you the relative image size without any calculation.

For a converging lens:

object beyond 2F → smaller

object at 2F → same size

object between F and 2F → larger

object inside F → larger

For a diverging lens:

image → smaller

for the ordinary real-object situations studied here.

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Predicting Orientation

For a single converging lens:

real images → inverted

virtual image inside F → upright

For a diverging lens with a real object:

virtual image → upright

This provides a quick way to check whether a ray diagram makes sense.

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Predicting Image Position

The object position provides clues about the image position.

For a converging lens:

object beyond 2F → image between F and 2F

object at 2F → image at 2F

object between F and 2F → image beyond 2F

object inside F → virtual image on object side

For a diverging lens:

image between lens and near F

for an ordinary real object.

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The Thin-Lens Equation

Image position can also be calculated using the thin-lens equation:

1/f = 1/dₒ + 1/dᵢ

where:

f = focal length

dₒ = object distance

dᵢ = image distance

This equation allows image position to be predicted mathematically.

Ray diagrams provide a visual prediction, while the lens equation provides a numerical result.

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Worked Example: Finding Image Distance

A converging lens has:

f = 10 cm

An object is:

dₒ = 30 cm

from the lens.

Use:

1/f = 1/dₒ + 1/dᵢ

1/10 = 1/30 + 1/dᵢ

Therefore:

1/dᵢ = 1/10 − 1/30

1/dᵢ = 3/30 − 1/30

1/dᵢ = 2/30

1/dᵢ = 1/15

Therefore:

dᵢ = 15 cm

The image forms 15 cm from the lens on the opposite side.

Since:

f = 10 cm

2f = 20 cm

the image is between F and 2F, exactly as the ray-diagram rules predict.

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Finding the Magnification

For the previous example:

dₒ = 30 cm

dᵢ = 15 cm

Magnitude of magnification:

|M| = 15/30

|M| = 0.5

Therefore, the image is:

half the object's height

This agrees with the prediction:

object beyond 2F → reduced image

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Ray Diagrams as Predictions

A well-constructed ray diagram should allow you to determine four important image characteristics.

A useful checklist is:

Position

Where is the image?

Size

Is it enlarged, reduced, or the same size?

Orientation

Is it upright or inverted?

Type

Is it real or virtual?

These can be remembered as the four main characteristics of an image.

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Checking a Ray Diagram

Suppose your ray diagram shows an object beyond 2F but produces an enlarged image beyond 2F.

Something is wrong.

The expected result is:

object beyond 2F → image between F and 2F, smaller

Known image rules therefore provide a useful way to check the accuracy of a diagram.

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Magnifying Glass

A magnifying glass is a converging lens.

To produce a magnified virtual image, the object must be:

inside the focal length

The image is:

• virtual
• upright
• enlarged
• on the same side of the lens as the object

The eye receives diverging rays and interprets them as coming from the enlarged virtual image.

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Cameras

A camera requires a real image because light must actually reach the image sensor.

A converging lens system focuses light onto the sensor.

For many ordinary photographs, the object is much farther away than the focal length.

The image on the sensor is typically:

• real
• inverted
• reduced

Digital processing then produces the displayed photograph in the expected orientation.

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Projectors

A projector also requires a real image.

The lens forms an enlarged image on a screen.

The object is positioned so that the lens produces:

• real
• enlarged
• focused image

A screen can display the image because actual light rays converge there.

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The Human Eye

The optical system of the human eye focuses light onto the retina.

The image formed on the retina is:

• real
• inverted
• reduced

The brain processes information from the retina so that we perceive the world normally.

The eye changes its focusing power to form clear images of objects at different distances.

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Microscopes

Microscopes use multiple lenses to produce large magnifications.

The objective lens produces an enlarged intermediate image.

The eyepiece then allows this image to be viewed at greater angular magnification.

The result allows extremely small structures to be observed.

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Telescopes

Telescopes use lenses or mirrors to collect light from distant objects.

In a refracting telescope, an objective lens forms an image of the distant object.

An eyepiece then allows the observer to examine that image with increased angular magnification.

The focal lengths of the lenses are important in determining telescope performance.

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Focal Length and Cameras

Camera lenses with different focal lengths produce different fields of view and image scales.

In general:

shorter focal length → wider field of view

longer focal length → narrower field of view and larger image scale for distant objects

This is why different lenses are selected for:

• landscapes
• portraits
• wildlife
• sports
• astronomy

Real camera systems are more complex than a single thin lens, but the same fundamental optical principles apply.

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Measuring Focal Length Experimentally

A simple converging lens can have its focal length estimated experimentally.

One method is:

1. Aim the lens at a very distant object.
2. Place a screen behind the lens.
3. Move the screen until a sharp image forms.
4. Measure the distance between the lens and screen.

For a sufficiently distant object, incoming rays are approximately parallel.

Therefore, the sharp image forms approximately at the focal point.

The lens-to-screen distance is approximately:

f

Never use the Sun as the target for an unsupervised lens experiment. A converging lens can concentrate sunlight strongly enough to cause eye injury, burns, or fire.

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Lens Strength and Focal Length

A lens with a short focal length bends rays more strongly.

A lens with a long focal length bends rays less strongly.

Therefore:

stronger lens → shorter |f|

weaker lens → longer |f|

Lens power is related to focal length by:

P = 1/f

when f is measured in metres.

Lens power is measured in:

dioptres (D)

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Worked Example: Lens Power

A converging lens has:

f = 25 cm

Convert to metres:

f = 0.25 m

Use:

P = 1/f

P = 1/0.25

P = +4.0 D

A shorter focal length would correspond to a greater positive lens power.

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A Complete Lens Example

A converging lens has:

f = 12 cm

An object is placed:

36 cm from the lens

First determine the object's position relative to F and 2F.

F = 12 cm

2F = 24 cm

The object is beyond 2F.

Therefore, before doing any calculations, predict:

• image between F and 2F
• real
• inverted
• smaller

Now calculate the image distance:

1/f = 1/dₒ + 1/dᵢ

1/12 = 1/36 + 1/dᵢ

1/dᵢ = 1/12 − 1/36

1/dᵢ = 3/36 − 1/36

1/dᵢ = 2/36

1/dᵢ = 1/18

Therefore:

dᵢ = 18 cm

The image is between:

F = 12 cm

and:

2F = 24 cm

Now calculate the magnitude of magnification:

|M| = 18/36

|M| = 0.5

The image is half the object's height.

The mathematical result agrees with the ray-diagram prediction.

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Combining Ray Diagrams and Calculations

Ray diagrams and equations should support each other.

A ray diagram provides:

• visual position
• orientation
• approximate size
• real/virtual classification

Calculations provide:

• precise image distance
• precise magnification
• image height

If the two approaches strongly disagree, check the work.

For example, if a ray diagram predicts a reduced image but the calculation gives:

M = 4

there is probably an error in either the diagram or calculation.

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A Reliable Strategy for Lens Problems

When solving a lens problem:

1. Identify whether the lens is converging or diverging.
2. Draw the principal axis.
3. Mark the optical centre.
4. Mark F on both sides.
5. Mark 2F if useful.
6. Determine the object's position relative to F and 2F.
7. Predict the likely image before drawing.
8. Draw at least two principal rays.
9. Extend rays backward with dashed lines if necessary.
10. Locate the image.
11. Determine whether it is real or virtual.
12. Determine whether it is upright or inverted.
13. Determine whether it is enlarged, reduced, or the same size.
14. Calculate magnification if image and object sizes are known.
15. Check that the calculation agrees with the ray diagram.

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Common Mistakes

Confusing focal point and focal length

The focal point is a position.

Focal length is the distance from the lens to that position.

Thinking a lens has only one focal point

A thin lens has focal points on both sides.

Confusing F and 2F

2F is twice as far from the lens as F.

Thinking all converging-lens images are real

If the object is inside F, the image is virtual.

Thinking all virtual images are smaller

A converging lens can produce an enlarged virtual image.

Thinking a diverging lens forms a real image for an ordinary real object

It normally forms a virtual, upright, reduced image.

Drawing virtual rays as solid rays

Backward extensions should normally be dashed.

Thinking the rays actually travel to a virtual image

They only appear to originate from that position.

Using the wrong magnification ratio

Use:

M = image height/object height

not the reverse.

Giving magnification units

Magnification is a ratio and therefore has no unit.

Failing to convert units

Image height and object height must be expressed in the same units before calculating magnification.

Assuming M = 0.5 means the image is 0.5 units tall

It means the image is half the object's height.

Judging image size only by how large the sketch looks

Use the ray geometry or magnification calculation. Ray diagrams may not always be perfectly to scale.

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Key Terms

Principal axis — reference line passing through the optical centre of a lens.

Optical centre — central reference point of a thin lens through which a ray is approximated as travelling straight.

Focal point (F) — point where parallel rays converge or from which diverging rays appear to originate.

Focal length (f) — distance between the optical centre and a focal point.

2F — position twice the focal length from the optical centre.

Converging lens — lens that brings parallel rays toward a focal point.

Diverging lens — lens that causes parallel rays to spread apart.

Real focus — focal point through which actual rays pass.

Virtual focus — focal point from which diverging rays appear to originate.

Real image — image formed where actual light rays converge.

Virtual image — image formed where light rays appear to originate.

Upright image — image having the same vertical orientation as the object.

Inverted image — image oriented upside down relative to the object.

Magnification (M) — ratio comparing image size with object size.

Image height (hᵢ) — vertical size of an image.

Object height (hₒ) — vertical size of the original object.

Object distance (dₒ) — distance between an object and the optical centre of a lens.

Image distance (dᵢ) — distance between an image and the optical centre of a lens.

Thin-lens equation — mathematical relationship connecting focal length, object distance, and image distance.

Lens power — measure of a lens's focusing strength.

Dioptre (D) — unit of lens power.

Ray diagram — diagram showing light-ray paths used to determine image characteristics.

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Key Takeaways

• The focal point is a position associated with parallel rays entering a lens.
• Focal length is the distance between the optical centre and focal point.
• Thin lenses have focal points on both sides.
• 2F is twice the focal length from the lens.
• Converging lenses bring parallel rays to a real focal point.
• Diverging lenses spread parallel rays so they appear to originate from a virtual focal point.
• Shorter focal lengths generally indicate stronger lenses.
• Real images form where actual light rays converge.
• Virtual images form where rays only appear to originate.
• Real images can normally be projected onto a screen.
• Ray diagrams can predict image position, size, orientation, and type.
• A converging lens can produce either real or virtual images.
• An object beyond 2F produces a smaller, inverted, real image between F and 2F.
• An object at 2F produces a same-size, inverted, real image at 2F.
• An object between F and 2F produces an enlarged, inverted, real image beyond 2F.
• An object inside F produces an enlarged, upright, virtual image.
• A diverging lens normally produces a reduced, upright, virtual image for a real object.
• Magnification compares image height with object height.
• M = hᵢ/hₒ
• M > 1 means an enlarged image.
• M = 1 means the same size.
• M < 1 means a reduced image.
• Magnification has no unit.
• Image distance, object distance, and focal length can be related using the thin-lens equation.
• Ray diagrams and calculations should produce consistent predictions.
• Focal length and magnification are important in cameras, magnifiers, microscopes, telescopes, projectors, corrective lenses, and the human eye.

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Check Your Understanding

1. Define focal point and focal length and explain the difference between them.
2. A converging lens has a focal length of 8 cm. How far from the optical centre are F and 2F?
3. Explain the difference between the focal point of a converging lens and the focal point of a diverging lens.
4. How can you determine from a ray diagram whether an image is real or virtual?
5. An object is placed beyond 2F in front of a converging lens. Predict the image's position, size, orientation, and type.
6. An object is placed between F and 2F. Predict the resulting image characteristics.
7. Explain what happens when an object is placed inside the focal length of a converging lens and relate this to a magnifying glass.
8. Describe the image normally produced by a diverging lens for a real object.
9. An object is 5 cm tall and its image is 15 cm tall. Calculate the magnification and interpret your answer.
10. An image is 4 cm tall and the magnification is 0.25. Calculate the height of the original object.
11. An object is 40 mm tall and its image is 6 cm tall. Calculate the magnification. Remember to convert the measurements to the same units.
12. A converging lens has f = 10 cm and an object is positioned 30 cm from the lens. Use 1/f = 1/dₒ + 1/dᵢ to calculate the image distance. Then use the result to determine the magnitude of the magnification.
13. Explain why a real image can normally be projected onto a screen but a virtual image cannot.
14. A student's ray diagram for an object beyond 2F shows an upright, enlarged image. Explain why the diagram must contain an error and state the correct image characteristics.
15. A converging lens has a focal length of 15 cm. An object is moved from 45 cm away, through 30 cm, then to 20 cm, and finally to 10 cm from the lens. Predict how the position, size, orientation, and type of image change as the object moves through these different regions.