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SCIENCE · INVESTIGATE AND EXPLAIN

Use a signed intermediate object in a two-lens system

Two lenses cannot generally be replaced by adding their separated focal lengths or image distances. First solve what the first lens alone would do. Here lens1 at global x=0 has f1=10 cm and object x=−30 cm, so its predicted image is at x=15 cm with magnification−1/2. A second lens at x=d then acts on the rays reaching that plane, not on the original object distance thirty centimetres away.

Learning outcome: Classify the second object as real, virtual or focal before solving each allowed separation. Retain both local distances and global positions. Explain why the d=10 case needs a negative second object distance even though the original object was real.

Starting recap

Use normal-based angles, signed conjugate distances and explicit optical-path/reflection phases. Separate geometric direction, amplitude and energy; preserve focal and critical boundaries.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

At d=10 cm, o2=−5 cm and i2=10/3 cm, so the final image lies at x=40/3 cm. m2=2/3 and total magnification−1/3. At d=20 cm, o2=5 cm gives i2=−10 cm and global x=10 cm: a virtual final image for lens2. At d=30 cm, i2=30 cm and the final image is at x=60 cm.

Open the experiment

Predict

At lens separation10 cm in the stated system, lens2 receives:

Experiment

Classify the second object as real, virtual or focal before solving each allowed separation. Retain both local distances and global positions. Explain why the d=10 case needs a negative second object distance even though the original object was real.

Treat an intermediate image as the next signed object

Worked teaching model · synthetic values · First thin lens at x=0 has f1=10 cm and object x=−30 cm; its image without the second lens would lie at x=15 cm. Second lens f2=10 cm lies at selected x=d. Both paraxial, no apertures/aberrations. d=10 gives a virtual object for lens2; d=25 gives its exact focal branch.

Reset keeps your prediction, captures and practice history.

Two lenses and signed conjugate coordinates

height (cm)global axis coordinate (cm)-30202-2

Spatial axes use the same pixels per physical unit, preserving geometry. 1: original object (solid).2: first lens (dashed).3: second lens (dotted).4: intermediate image if lens2 absent (dotted).5: final image location (dotted). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Two lenses and signed conjugate coordinates

original object

global axis coordinate (cm)height (cm)
-300
-301

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

first lens

global axis coordinate (cm)height (cm)
0-2
02

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

second lens

global axis coordinate (cm)height (cm)
20-2
202

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

intermediate image if lens2 absent

global axis coordinate (cm)height (cm)
150
15-0.5

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

final image location

global axis coordinate (cm)height (cm)
100
10-1

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Lens separation20cm
First intermediate image coordinate15cm
First magnification-0.5ratio
Second signed object distance5cm
Second virtual object branch0flag
Second focal-object branch0flag
Second image distance from second lens-10cm
Final global image coordinate10cm
Second magnification2ratio
Overall magnification-1ratio
Final image height-1cm

Use the first image as the second signed object, then translate the final coordinate back to the common axis.

Lens1 alone gives i1=10×30/(30−10)=15 cm and m1=−1/2. The second signed object distance is d−15, not d or15. Apply the same reciprocal relation with this signed object, multiply m1m2 and add lens2 position to its image distance. A predicted intermediate focus beyond an intercepting lens is not an actual completed first image; it defines the converging input rays for the next element.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Use the first image as the second signed object, then translate the final coordinate back to the common axis.

Lens1 alone gives i1=10×30/(30−10)=15 cm and m1=−1/2. The second signed object distance is d−15, not d or15. Apply the same reciprocal relation with this signed object, multiply m1m2 and add lens2 position to its image distance. A predicted intermediate focus beyond an intercepting lens is not an actual completed first image; it defines the converging input rays for the next element.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the optical model

Two lenses cannot generally be replaced by adding their separated focal lengths or image distances. First solve what the first lens alone would do. Here lens1 at global x=0 has f1=10 cm and object x=−30 cm, so its predicted image is at x=15 cm with magnification−1/2. A second lens at x=d then acts on the rays reaching that plane, not on the original object distance thirty centimetres away.

Build the signed account

The second signed object distance is o2=d−15 cm. When d exceeds15, the first rays have crossed and the second lens sees a real intermediate object to its left. When d is10, the second lens intercepts rays still converging toward x=15; that location is a virtual object for lens2 and o2=−5 cm. Apply the same reciprocal equation with this signed distance. The final global image coordinate is d+i2 and total magnification is m1m2.

Test the limits and evidence

This procedure also exposes a distinct focal boundary: with d=25 cm, o2=10 cm equals the second focal length and the final output is collimated, so no finite image location is assigned. An intermediate focus predicted beyond an intercepting lens is not an actual completed first image in the assembled system. Apertures, thick-lens principal planes and aberration are absent. The coordinate diagram labels image candidates rather than pretending every segment is a physical ray.

Worked original case

At d=10 cm, o2=−5 cm and i2=10/3 cm, so the final image lies at x=40/3 cm. m2=2/3 and total magnification−1/3. At d=20 cm, o2=5 cm gives i2=−10 cm and global x=10 cm: a virtual final image for lens2. At d=30 cm, i2=30 cm and the final image is at x=60 cm.

Predict and investigate

Classify the second object as real, virtual or focal before solving each allowed separation. Retain both local distances and global positions. Explain why the d=10 case needs a negative second object distance even though the original object was real.

Changed-case practice

Try the changed case before feedback

At separation30 cm with the stated focal lengths and original object distance, find the overall magnification.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original text and numerical cases checked 30 October 2026 against the primary explanation. All optical activity is virtual: no laser, sunlight viewing, eye exposure, physical lens alignment or device construction is requested. Public feedback is not independent mastery; retain first calculations, critique and changed-case evidence in the separate course workspace.

Apply

At lens separation10 cm in the stated system, lens2 receives:

Public practice checks a specific claim. It does not establish independent mastery.

Which independent observation would reject the approximation, sign convention or calibration premise rather than merely refit it?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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