Lesson 9 of 10Light & Geometric Optics

Unit 4 · Lesson 9 of 10

The Thin Lens Equation and Magnification

The thin lens equation and the magnification equation for lenses, the sign convention for converging and diverging lenses, and how to use them for cameras, projectors and magnifying glasses.

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Key things to remember

1Thin lens equation: $\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$
2Magnification: $M = \frac{h _{i}}{h _{o}} = - \frac{d _{i}}{d _{o}}$ ; negative $M$ means inverted.
3 $f$ is positive for converging lenses and negative for diverging lenses.
4Positive $d_{i}$ : real image on the far side of the lens. Negative $d_{i}$ : virtual image on the object's side.
5For a very distant object, the image forms at the focal point: $d_{i} \approx f$ .

A phone camera has to put a sharp image exactly on a sensor a few millimetres behind its lens, whether you are photographing a friend across the table or a mountain across a lake. A projector has to throw a slide's image several metres onto a screen. Designers do not get there by drawing ray diagrams; they use the thin lens equation. The good news: it is the same equation you used for mirrors, with the same sign rules.

\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}

The thin lens equation: f is the focal length, d_o the object distance and d_i the image distance, measured from the optical centre.

M = \frac{h _{i}}{h _{o}} = - \frac{d _{i}}{d _{o}}

Magnification for lenses, identical to the mirror version.

Sign convention for lenses

Quantity	Positive (+)	Negative (−)
Focal length $f$	Converging (convex) lens	Diverging (concave) lens
Object distance $d_{o}$	Real object (every case in this course)	Not used in this course
Image distance $d_{i}$	Real image, on the far side of the lens	Virtual image, on the object's side
Magnification $M$ and $h_{i}$	Upright image	Inverted image

Real is positive: the only change from mirrors is where a real image sits. For a lens it is on the opposite side from the object.

Fig 1The thin lens equation predicts this diagram: for an object beyond 2F, d_i comes out positive (real, on the far side, between F′ and 2F′) and M comes out negative with a size less than 1 (inverted and smaller).

Decide the sign of $f$ : converging is positive, diverging is negative.
Rearrange first: $\frac{1}{d _{i}} = \frac{1}{f} - \frac{1}{d _{o}}$ (or $\frac{1}{d _{o}} = \frac{1}{f} - \frac{1}{d _{i}}$ ).
Combine the fractions, then flip the result.
Interpret $d_{i}$ : positive is a real image on the far side; negative is a virtual image on the object's side.
Calculate $M = - \frac{d _{i}}{d _{o}}$ and $h_{i} = M h_{o}$ , then state SALT.
Check against a quick sketch: a diverging lens must give negative $d_{i}$ and $0 < M < 1$ .

Keep units consistent

The equation works in any length unit, as long as every distance uses the same one. If the focal length is in millimetres and the object distance is in metres, convert first. Magnification has no units.

Fig 2For a diverging lens f is negative, so the equation always gives a negative d_i (a virtual image on the object's side, between F and the lens) and a positive magnification less than 1 (upright and smaller).

Focusing a camera

For a very distant object,

\frac{1}{d _{o}}

is almost zero, so

d_{i} \approx f

: the image forms at the focal point. For a closer object,

\frac{1}{d _{o}}

is larger, so

\frac{1}{d _{i}}

is smaller and the image forms farther from the lens. That is why a camera focuses on nearby subjects by moving the lens slightly away from the sensor.

Loading interactive…

Drag the object toward a converging or diverging lens and check the thin lens equation and magnification against the drawn image.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

3.0 cm

tall object is

25.0 cm

from a converging lens with

f = 10.0 cm

. Find the image distance, magnification and image height.

Example 2

A magnifying glass has

f = 8.0 cm

and is held

6.0 cm

from an insect. Find the image distance and magnification.

Example 3

An object is

30.0 cm

from a diverging lens whose focal length has magnitude

20.0 cm

. Find the image distance and magnification.

Example 4

A projector lens (

f = 10.0 cm

) must form an image on a screen

5.00 m

away. How far from the lens should the slide be, and how tall is the image of a

24 mm

tall slide?

Where marks get lost

Using a positive focal length for a diverging lens. Diverging lenses have negative $f$ .
Forgetting to flip at the end and reporting $\frac{1}{d _{i}}$ as $d_{i}$ .
Mixing units, such as a focal length in mm with an object distance in m.
Placing a lens's real image on the object's side. For a lens, positive $d_{i}$ is on the far side.
Losing the minus sign in $M = - \frac{d _{i}}{d _{o}}$ and calling an inverted image upright.

Flashcards 7

Thin lens equation

Lens magnification

Sign of f for lenses

Positive d_i for a lens

+ 3 more

Practice

An object is

45.0 cm

from a converging lens with

f = 15.0 cm

. Find the image distance in cm, to 1 decimal place.

For the lens in the previous question (

d_{o} = 45.0 cm

d_{i} = 22.5 cm

), find the magnification, including its sign, to 2 decimal places.

A converging lens has

f = 20.0 cm

and an object is

12.0 cm

from it. Find the image distance in cm, including its sign, to 1 decimal place.

An object is

24.0 cm

from a diverging lens with

f = - 12.0 cm

. Find the image distance in cm, including its sign, to 1 decimal place.

A lens forms a real image

60.0 cm

from it when the object is

30.0 cm

away. Find the focal length in cm, to 1 decimal place.

1.5 cm

tall object is

5.0 cm

from a converging lens with

f = 4.0 cm

. Find the image height in cm, including its sign, to 1 decimal place.

A camera lens has

f = 50.0 mm

. A person stands

2.00 m

away. How far behind the lens must the sensor be, in mm, to 1 decimal place?

A lens calculation gives

f = - 15 cm

d_{i} = - 6.0 cm

and

M = + 0.60

. Which statement is correct?

Match each thin-lens result to what it tells you.

f = + 10 cm

f = - 10 cm

d_{i} = + 30 cm

d_{i} = - 30 cm

M = - 1.0

A very distant object is viewed through a converging lens with

f = 35 mm

. About how far behind the lens does the image form, in mm? (Type the number.)

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