Optics 021 · Where a lens puts the image

Thin Lens Imaging Bench

Where does a lens form the image of an object, and when can a screen catch it? Where the light from each point of the object meets again. With a converging lens and the object beyond the focal length, that point is behind the lens, and a screen there shows a sharp, upside-down picture. Inside the focal length, or with a diverging lens, the light only seems to come from the object’s side: a virtual image, seen through the lens but never on a screen.

Slide a lit F along an optical bench. The lens puts a sharp image on the screen, or a virtual one you only see through it; find where and how large.

Where the image forms, from the lens $\text{—}$

Magnification (negative: upside down) $\text{—}$

Blur patch on the screen $\text{—}$

Focal length $\text{—}$

Target distance in focal lengths $\text{—}$

Height of the F’s image $\text{—}$

Screen brightness against the default bench $\text{—}$

Screen pixels traced for this record $\text{—}$

The lens is an ideal thin lens and only rays close to the axis are traced, so there is no aberration, no dispersion and no diffraction. Its default power of 10 dioptres and 25.4 mm diameter match a Thorlabs LA1509 plano-convex lens of N-BK7 glass, focal length 100 mm; other powers are ideal lenses of the same size. The iris sits at the lens and limits the cone of light. The target is an F 10 mm tall cut out of a dark mask in front of a lamp, glowing evenly. For each point of the screen the model finds the disc of the target that sends it light through the iris and adds up the lit part of that disc exactly. The picture’s brightness is that lit fraction; the iris area and the screen distance scale every point alike and are given separately as the brightness relative to the default bench. Distances are measured from the lens: the image distance is negative when the image is virtual, on the target’s side, and the magnification is negative when the image is upside down. In the scene, heights and widths are drawn 4 times larger than distances along the bench and the lens is drawn thicker than it is; every number and the screen picture use true sizes.

Physics tutorial

Find where a lens puts the image

BackgroundOn an optical bench a lamp-lit target, a lens and a white screen stand on one rail. Slide the screen until the target appears on it sharp and upside down, and you have done what a projector does, and what a camera does when it focuses. The same three distances also tell why a magnifying glass works only close to the page.

Why it mattersKnowing where the image lands and how large it is tells you where the film or sensor goes in a camera, how far a projector must stand from the wall, and why some images can never be caught on a screen at all.

Start with the essentials

Focus question: Where does a lens form the image of an object, and when can a screen catch it?
One-sentence intuition: Each point of the target sends a cone of light through the lens. The lens turns every ray by an amount proportional to how far from the centre it crosses, so the whole cone meets again at one point, the image of that point, a distance $s'$ behind the lens. The target’s distance $s$ and the focal length $f$ fix it. When the target is inside the focal length, the cone still spreads after the lens; it only seems to come from a point on the target’s side, a virtual image that no screen can catch.

Core mathematical model

The thin-lens equation

\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}=P

The object distance $s$ and image distance $s'$ are measured from the lens; $f$ is positive for a converging lens and negative for a diverging one, and the power $P=1/f$ is in dioptres when $f$ is in metres. A negative $s'$ means a virtual image on the object’s side (OpenStax University Physics Vol. 3, Eq. 2.19). At $s=f$ the light leaves parallel and the image is at infinity.

Magnification

m=\frac{h'}{h}=-\frac{s'}{s}

The image height $h'$ over the object height $h$ (Eq. 2.22). A negative $m$ means upside down: every real image of a single lens is inverted, every virtual one upright. At $s=2f$ the image is the same size, $m=-1$ .

A screen in the wrong place shows a blur

b=D\left\lvert 1-\frac{L}{s'}\right\rvert,\qquad y_{L}=-\frac{L}{s}\,y

The cone through an iris of diameter $D$ that meets at $s'$ is $b$ wide where the screen stands, at $L$ . The patch for a target point at height $y$ is centred on the ray through the lens centre, at $y_{L}$ on the screen, so the picture lands at the same place and scale whether it is sharp or not. A smaller iris narrows every cone and keeps the picture sharper over a longer stretch of the bench.

Where the default lens’s 100 mm comes from

f=\frac{R}{n-1}=\frac{51.5\,\mathrm{mm}}{1.5168-1}=99.7\,\mathrm{mm}

A plano-convex lens with one surface of radius $R$ has this focal length. The Thorlabs LA1509 has $R=51.5\,\mathrm{mm}$ and is made of N-BK7, $n=1.5168$ at 587.6 nm, which gives 99.7 mm against the nominal 100 mm ± 1 %.

Common difficulties

Covering part of the lens does not cut off part of the image

Typical misconceptionClosing the iris, or covering half the lens, removes part of the picture.

Better mental modelEvery part of the lens receives light from every point of the target and sends it to the same image point. A smaller opening makes the whole image dimmer, and sharper when the screen is out of place, but the whole F is still there.

A real image exists without a screen

Typical misconceptionThe image is made by the screen.

Better mental modelThe light really meets at the image point and spreads again beyond it. A screen there only scatters it towards your eye; the image is there in the air, and you can see it by looking back towards the lens from beyond it.

A virtual image cannot be projected

Typical misconceptionMoving the screen far enough will always find the image.

Better mental modelFor a target inside the focal length, or with a diverging lens, the light leaving the lens spreads apart. No screen position collects it into a point: the image is where the light seems to come from, on the target’s side, and only an eye or a camera looking through the lens sees it.

Magnification is not a property of the lens alone

Typical misconceptionA lens magnifies by a fixed amount.

Better mental modelThe magnification depends on where the target is, not on the lens alone: the same 100 mm lens gives twice the size at 150 mm, the same size at 200 mm and a quarter at 500 mm.

Run the experiment

01
See a sharp image
Keep the default preset, Projector: larger and upside down.
What to observe: The screen shows the F sharp, upside down and twice its size, 20 mm tall. The image distance is 300 mm and the magnification −2.
02
Drag the target
Drag the lit F along the bench, away from the lens and back.
What to observe: Moving the target away brings the sharp image towards the lens and makes it smaller, so the screen blurs. The purple ghost in the scene shows where the image now is; move the screen there and it is sharp again.
03
Put it at twice the focal length
Load Twice the focal length: same size.
What to observe: With the F at 200 mm the image forms at 200 mm, the same size and upside down: magnification −1.
04
Make a camera
Load Camera: far target, small image.
What to observe: With the F 500 mm away the image is at 125 mm, just beyond the focal point, and a quarter of the size. Far objects all focus close to one focal length behind the lens.
05
Use it as a magnifying glass
Load Magnifying glass: virtual and upright.
What to observe: At 60 mm, inside the focal length, the rays still spread after the lens. The image is virtual, 150 mm on the target’s side, upright and 2.5 times larger. The screen shows only a blur, and the curve of blur against screen distance never reaches zero.
06
Swap in a diverging lens
Load Diverging lens: always virtual, smaller.
What to observe: A −100 mm lens spreads the light wherever the target is. The image is virtual, upright and 0.4 times the size, 60 mm in front of the lens.
07
Close the iris
Load Small iris: sharp without refocusing, then open the iris back to 25 mm.
What to observe: The screen is 60 mm beyond the sharp image. With a 3 mm iris the blur is 0.6 mm and the F is readable; at 25 mm it grows to 5 mm. The small iris lets through 1.4 % as much light as the open one.

Thin Lens Imaging Bench

Find where a lens puts the image

Core mathematical model

The thin-lens equation

Magnification

A screen in the wrong place shows a blur

Where the default lens’s 100 mm comes from

Common difficulties

Covering part of the lens does not cut off part of the image

A real image exists without a screen

A virtual image cannot be projected

Magnification is not a property of the lens alone

Run the experiment

See a sharp image

Drag the target

Put it at twice the focal length

Make a camera

Use it as a magnifying glass

Swap in a diverging lens

Close the iris

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