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Optics · Thin Lenses

Geometric Optics and Image Formation

Trace rays through a convex lens to locate where the image forms, measure the image distance, and verify the thin-lens equation. Find the magnification and orientation for objects at different distances, plot 1/dᵢ against 1/dₒ to recover the focal length, and see why the aperture changes brightness but not image position.

Theory — Thin Lenses and Image Formation

Converging (Convex) Lenses

A convex lens bends parallel rays of light so they meet at a single point — the focal point — a distance f (the focal length) from the lens. When light from an object passes through the lens, the rays converge to form an image. Where that image forms, and how big it is, depends on how far the object sits from the lens.

Ray Tracing

Three principal rays locate the image of the tip of the object: a ray parallel to the axis bends to pass through the far focal point; a ray through the centre of the lens continues straight; and a ray through the near focal point emerges parallel to the axis. Where these rays cross is the image of the tip.

The Thin-Lens Equation

The object distance dₒ, the image distance dᵢ, and the focal length f are linked by the thin-lens equation.

Thin-Lens Equation 1/f = 1/dₒ + 1/dᵢ

f = focal length (cm), dₒ = object distance (cm), dᵢ = image distance (cm)
Rearranged: dᵢ = 1 / (1/f − 1/dₒ)

Magnification and Image Type

The magnification compares image size to object size. A real image (positive dᵢ) forms on the far side of the lens where the rays actually cross, and is inverted; a virtual image (the object inside the focal length) appears upright and on the same side.

Magnification M = dᵢ / dₒ  (size ratio)

M > 1 → image larger · M < 1 → image smaller
Real image: inverted · Object inside f: virtual, upright

Object at the Focal Point

If the object sits exactly at the focal point (dₒ = f), the rays leaving the lens are parallel — they never cross, so no image forms (the image is "at infinity"). This is the boundary between a real image (object beyond f) and a virtual image (object inside f).

Linearising — 1/dᵢ vs 1/dₒ

Rearranging the thin-lens equation gives 1/dᵢ = 1/f − 1/dₒ. So a graph of 1/dᵢ against 1/dₒ is a straight line whose intercept is 1/f — a neat way to recover the focal length from several measurements.

Object far away

Image forms close to the focal point, smaller and inverted.

Object near f

Image forms far away, much larger and inverted.

Larger aperture

More light collected (brighter image) but the image position is unchanged.

QuantityRelationshipNotes
Thin-lens equation1/f = 1/dₒ + 1/dᵢconvex lens, f > 0
Image distancedᵢ = 1/(1/f − 1/dₒ)positive ⇒ real image
MagnificationM = dᵢ/dₒ>1 larger, <1 smaller
Object at fdᵢ → ∞rays parallel, no image
Linear plot1/dᵢ vs 1/dₒintercept = 1/f

Instructions — Running the Virtual Experiment

The Ray Tracing tab lets you place the object and watch the rays converge to the image; the Thin-Lens tab records each trial and plots 1/dᵢ against 1/dₒ. Record every reading in your lab report with screenshots of the ray diagrams.

Part A — Verify the Thin-Lens Equation (Ray Tracing tab)
1
Open Simulation → Ray Tracing. The lens is convex with focal length f = 40 cm and diameter 120 cm. Place the object arrow at dₒ = 60 cm using the object slider.
2
Watch the marginal rays cross to the right of the lens. Read the image distance dᵢ from the readout (the ruler marks the axis). Confirm it matches 1/f = 1/dₒ + 1/dᵢ.
Part B — Magnification and Orientation (Ray Tracing tab)
1
Repeat for dₒ = 80 cm and dₒ = 50 cm. For each, record dᵢ, the magnification M = dᵢ/dₒ, and whether the image is real/virtual, upright/inverted, and larger/smaller.
2
Then slide the object to dₒ = 40 cm (exactly at the focal point) and observe that the rays emerge parallel — no image forms.
Part C — Aperture (Diameter) Effect (Ray Tracing tab)
1
Keep the object at dₒ = 60 cm and slide the diameter from 120 cm down to 60 cm. Notice the crossing point (image position) stays put while fewer rays reach it — the image gets dimmer, not relocated.
Part D — Recover the Focal Length (Thin-Lens tab)
1
Open Thin-Lens, record several object distances, and Plot & fit 1/dᵢ against 1/dₒ. The intercept gives 1/f, so f comes straight off the graph.

Simulation — Convex Lens

Geometric Optics Virtual LabPlace the object and find where the rays cross
object
image
rays
Convex lens · focal points marked F.

Lens & object

Image readout
Focal length f40 cm
Object distance dₒ60 cm
Image distance dᵢ120 cm
Magnification M2.00
Image typereal, inverted
Sizelarger
Rays cross where the image forms.
Each point is (1/dₒ, 1/dᵢ). The line's intercept on the 1/dᵢ axis is 1/f.

Record object distances (f = 40 cm)

Fit result
Intercept (1/f)— cm⁻¹
→ focal length f— cm
Expected f40 cm
dₒ (cm)dᵢ (cm)1/dₒ (cm⁻¹)1/dᵢ (cm⁻¹)M = dᵢ/dₒ
Click an object distance to record it.

Team Questions

Question 1. A convex lens has f = 40 cm. An object sits at dₒ = 60 cm. Use 1/f = 1/dₒ + 1/dᵢ to find the image distance. (Type just the number in cm — e.g. 120)
Question 2. For that trial (dₒ = 60, dᵢ = 120), what is the magnification M = dᵢ/dₒ? (Type just the number)
Question 3. With f = 40 cm and the object at dₒ = 80 cm, find the image distance. (Type just the number in cm)
Question 4. For a real image formed by a convex lens, is the image upright or inverted? (One word)
Question 5. What happens to the image when the object is placed exactly at the focal point (dₒ = f)? (Describe in a few words — e.g. "no image / rays parallel")
Question 6. On a plot of 1/dᵢ (y) vs 1/dₒ (x), what does the y-intercept equal? (A short expression — e.g. "1/f")
Question 7 — Challenge. A telescope uses a larger-diameter (aperture) lens to see faint stars. Does a larger diameter change where the star's image forms, or just how much light is collected? (Answer "position" or "light")

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include labelled screenshots of your ray diagrams.

Geometric Optics and Image Formation

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To use ray tracing through a convex lens to locate images and measure image distances, to verify the thin-lens equation 1/f = 1/dₒ + 1/dᵢ, to determine the magnification and orientation of the image at several object distances, to recover the focal length from a graph of 1/dᵢ against 1/dₒ, and to examine how the lens aperture affects the image.

Theory

For a thin convex lens, the object distance, image distance, and focal length obey 1/f = 1/dₒ + 1/dᵢ, and the magnification is M = dᵢ/dₒ. A real image (positive dᵢ) is inverted; an object placed at the focal point produces parallel emerging rays and no image. Rearranged, 1/dᵢ = 1/f − 1/dₒ, so a plot of 1/dᵢ versus 1/dₒ has intercept 1/f.

1/f = 1/dₒ + 1/dᵢ · M = dᵢ/dₒ
1/dᵢ = 1/f − 1/dₒ → intercept = 1/f (f = 40 cm)

Calculations — Sample: dₒ = 60 cm, f = 40 cm

Image distance: 1/dᵢ = 1/f − 1/dₒ = 1/40 − 1/60 = 0.025 − 0.0167 = 0.00833 cm⁻¹ → dᵢ = 120 cm

Magnification: M = dᵢ/dₒ = 120/60 = 2.0 (real, inverted, larger)

Results Table (f = 40 cm)

Trialdₒ (cm)dᵢ (cm)M = dᵢ/dₒOrientation / type
1601202.0real, inverted, larger
280801.0real, inverted, same size
3502004.0real, inverted, larger

A plot of 1/dᵢ against 1/dₒ was a straight line with intercept ≈ 0.025 cm⁻¹, giving f ≈ 40 cm.

Discussion

The measured image distances matched the thin-lens equation at every object position: dₒ = 60, 80, and 50 cm gave dᵢ = 120, 80, and 200 cm, with magnifications of 2.0, 1.0, and 4.0. All three images were real and inverted, as expected for an object beyond the focal length, and the image grew larger as the object approached the focal point. A graph of 1/dᵢ versus 1/dₒ was a straight line whose intercept, about 0.025 cm⁻¹, gave a focal length of 40 cm — matching the lens setting.

Reducing the lens diameter from 120 cm to 60 cm did not move the crossing point of the rays: the image stayed in the same place but fewer rays reached it, so it became dimmer. This is why a telescope uses a large aperture — to gather more light from faint stars, not to change where their images form. When the object was moved to exactly the focal point (dₒ = 40 cm), the rays emerged parallel and never crossed, so no image formed (the image is "at infinity"), marking the boundary between real and virtual images.

Conclusion

The thin-lens equation was verified for a convex lens of focal length 40 cm: measured image distances agreed with calculation, magnification followed M = dᵢ/dₒ, and the 1/dᵢ-vs-1/dₒ graph recovered f from its intercept. The aperture controlled image brightness but not position, and an object at the focal point produced no image.

Practice Questions

Show all work and include units. Use the thin-lens equation 1/f = 1/dₒ + 1/dᵢ and M = dᵢ/dₒ.

Question 1
A convex lens of focal length 15 cm forms an image of an object placed 25 cm away. Find the image distance and the magnification.
Hint: 1/dᵢ = 1/f − 1/dₒ; then M = dᵢ/dₒ.
Question 2
An object is placed 30 cm from a convex lens of focal length 20 cm. Where is the image, and is it real or virtual, upright or inverted?
Hint: positive dᵢ means a real, inverted image.
Question 3
A lens produces a real image at 60 cm when the object is at 40 cm. What is the focal length of the lens?
Hint: 1/f = 1/dₒ + 1/dᵢ = 1/40 + 1/60.
Question 4
An object is placed inside the focal length of a convex lens (dₒ < f). Describe the image: real or virtual, upright or inverted, larger or smaller. (This is how a magnifying glass works.)
Hint: dᵢ comes out negative — a virtual, upright, enlarged image.
Question 5
Explain why a graph of 1/dᵢ versus 1/dₒ is a straight line, and state what both its slope and its intercept represent.
Hint: 1/dᵢ = −1/dₒ + 1/f; slope −1, intercept 1/f.
Question 6 — Challenge
A camera lens of focal length 50 mm photographs a distant mountain (object at effectively infinity). Where does the image form, and what happens to the image position as you focus on a nearby subject 1.0 m away?
Hint: for dₒ → ∞, dᵢ → f; for dₒ = 1000 mm, solve 1/dᵢ = 1/50 − 1/1000 (the image moves slightly farther from the lens).