Thin lenses | PhysicsRank

01 / Images at refracting surfaces and thin lenses

An image is a statement about where rays meet or appear to meet

Light from one object point spreads in many directions. An optical system forms an image of that point when the outgoing rays converge at another point, or when their backward extensions intersect consistently. An extended object is imaged by doing this for many points.

A real image occurs where the outgoing rays actually converge. A screen placed there can intercept the light pattern. A virtual image is located by backward extensions of diverging outgoing rays; those rays do not actually pass through that apparent image location.

Virtual does not mean invisible. An eye can intercept the rays and focus them onto its retina, so a virtual image can be seen. The distinction is whether the outgoing rays themselves converge at the stated image plane.

The simple formulas below use paraxial rays, close to the optical axis and at small angles. Spherical surfaces do not generally bring every wide-angle ray to one perfect point. Aberrations and diffraction set limits beyond this ideal construction.

STOP & THINK

A screen placed at the calculated image position receives a sharp pattern from the outgoing rays. What kind of image is this?

Real
Virtual
Neither, because only eyes can form images

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: An image is a statement about where rays meet or appear to meet
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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02 / Images at refracting surfaces and thin lenses

Fix one Cartesian axis before assigning any signs

For refraction and thin lenses in this chapter, light initially travels from left to right. Choose the positive x direction to the right and measure axial coordinates from the relevant surface vertex or thin-lens plane. A real object on the left therefore has u < 0.

The image coordinate v is positive on the right and negative on the left. For a lens in the same medium on both sides, f > 0 describes a converging lens and f < 0 a diverging lens. A radius R is positive when the surface’s centre of curvature lies to its right.

Transverse heights are signed upward. Lateral magnification m = h_image/h_object is negative for an inverted image and positive for an upright one. Its magnitude describes the size ratio.

Other valid conventions call a real-object distance positive and write different-looking formulas. The predictions agree when used consistently. Do not insert an unsigned object distance into a formula derived for the signed coordinate u.

Cartesian signs used throughout this chapter
Quantity	Positive meaning	Typical negative case
u or v	Point lies to the right of the chosen origin	A real object to the left has u < 0
R	Centre of curvature lies to the right	Centre lies to the left
f for a thin lens	Converging lens in the stated surrounding medium	Diverging lens
m	Upright relative to the object	Inverted relative to the object

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Fix one Cartesian axis before assigning any signs
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Images at refracting surfaces and thin lenses

A curved refracting surface has a different normal at each point

For a spherical boundary, the local normal passes through its centre of curvature. A paraxial ray striking at height y therefore meets a slightly tilted normal. This geometrical tilt gives the surface its focusing ability.

Let n1 be the incident-medium index and n2 the transmitted-medium index. With signed coordinates u, v and R as defined above, small-angle Snell refraction gives n2/v − n1/u = (n2 − n1)/R.

One way to see the signs is to write the incoming ray slope as −y/u, outgoing slope as −y/v and right-directed normal slope as −y/R. Apply n1 times the incoming angle relative to the normal equals n2 times the outgoing angle relative to the normal, then rearrange.

The refractive indices multiply different image and object terms. Replacing this relation by the thin-lens equation loses the effect of unequal media. The corresponding paraxial lateral magnification is m = n1v/(n2u), not generally v/u.

\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R},\qquad m=\frac{n_1v}{n_2u}

A single spherical refracting surface, paraxial rays, signed Cartesian coordinates increasing along incident light.

WORKED EXAMPLE

Solve a single surface before combining it into a lens

An object is 60 cm to the left of a spherical boundary. Light travels from index 1.00 to index 1.50. The centre of curvature lies 20 cm to the right of the vertex. Find image position and magnification.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: A curved refracting surface has a different normal at each point
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R},\qquad m=\frac{n_1v}{n_2u}
Meaning: A single spherical refracting surface, paraxial rays, signed Cartesian coordinates increasing along incident light.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Images at refracting surfaces and thin lenses

A plane boundary and an equal-index boundary check the model

Let the spherical radius grow without bound. The curvature term vanishes, so v = (n2/n1)u. A real object on the incident side then has a virtual image on that same side. This recovers the apparent-depth relation for near-normal viewing.

For a plane boundary, the magnification n1v/(n2u) becomes one. The apparent axial distance changes, but the ideal paraxial transverse size is unchanged. Perspective changes can still alter the angle an observer sees.

If n1 = n2 at any spherical boundary, the right-hand side also vanishes and v = u. There is no refractive effect because the optical medium does not change. The apparent source remains the original point.

The sign of R alone does not decide whether a surface converges a given ray bundle. The index difference and the incoming bundle matter too. Solve the stated geometry and use the resulting ray directions to classify the image.

WORKED EXAMPLE

Recover apparent depth from the general surface equation

An object is 30 cm inside a medium of index 1.50, viewed nearly normally across a flat boundary from index 1.00. With light travelling toward the observer, take u = −30 cm. Find v and m.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: A plane boundary and an equal-index boundary check the model
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Images at refracting surfaces and thin lenses

Two nearby refracting surfaces become one thin-lens relation

A lens has two refracting surfaces. If its thickness is small compared with the relevant object, image and curvature scales, their vertices can be approximated by one plane. The first surface’s image becomes the second surface’s object.

For the same surrounding medium on both sides, adding the two paraxial surface equations cancels the intermediate image term. The result can be written 1/v − 1/u = 1/f, where the focal length contains the lens material and curvature information.

For a converging lens, an incoming ray parallel to the axis leaves toward the far focal point. A ray through the optical centre is approximately undeviated in the thin-lens model. Their intersection locates the image; backward intersections locate a virtual image.

The algebra and ray construction should agree. A calculated positive v for a real object means convergence on the outgoing side. A negative v means the outgoing rays diverge from an apparent point on the incident side.

\frac1v-\frac1u=\frac1f,\qquad m=\frac vu

Thin lens, paraxial rays, identical surrounding medium on both sides, Cartesian u and v. The index factors cancel from the magnification in this equal-medium case.

Explain this equation

Read it in words

For a thin lens, the signed image and object distances determine the focal length. Their signed ratio gives transverse magnification.

Symbols and units

$u$: Object coordinate measured from the lens along the chosen axis.Any length unit, used consistently.
$v$: Image coordinate measured from the same lens origin.Same length unit as u and f.
$f$: Signed focal length; positive for a converging lens in this convention.Same length unit as u and v.
$m$: Image height divided by object height, including orientation.Dimensionless.

Signs and operations

The positive axis follows incident light. A real object on the incident side has u < 0; a real image beyond the lens has v > 0. Thus m < 0 indicates inversion. The reciprocals are not the distances themselves.

When it applies

Use paraxial rays, a thin lens and the same surrounding medium on both sides, as in this chapter. Keep one sign convention throughout; formulas from a different convention may look different.

Make it concrete

For u = −30 cm and f = +10 cm, 1/v + 1/30 = 1/10 gives v = +15 cm. Magnification is −0.5: the real image is inverted and half as tall.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \frac1v-\frac1u=\frac1f,\qquad m=\frac vu
Read it as: For a thin lens, the signed image and object distances determine the focal length. Their signed ratio gives transverse magnification.
Symbols:
u: Object coordinate measured from the lens along the chosen axis.; Any length unit, used consistently.
v: Image coordinate measured from the same lens origin.; Same length unit as u and f.
f: Signed focal length; positive for a converging lens in this convention.; Same length unit as u and v.
m: Image height divided by object height, including orientation.; Dimensionless.
Signs and operations: The positive axis follows incident light. A real object on the incident side has u < 0; a real image beyond the lens has v > 0. Thus m < 0 indicates inversion. The reciprocals are not the distances themselves.
Conditions: Use paraxial rays, a thin lens and the same surrounding medium on both sides, as in this chapter. Keep one sign convention throughout; formulas from a different convention may look different.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Find an image with signs that agree with its ray geometry

A converging thin lens has f = +12 cm. An upright object is 36 cm to its left. Find image coordinate and magnification.

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INTERACTIVE LAB

Move the object through the focal plane

Use a converging lens at object distances 30, 20, 10 and 5 cm, then switch to a diverging lens. Compare where the outgoing rays meet with the signed image coordinate.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Two nearby refracting surfaces become one thin-lens relation
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
\frac1v-\frac1u=\frac1f,\qquad m=\frac vu
Meaning: Thin lens, paraxial rays, identical surrounding medium on both sides, Cartesian u and v. The index factors cancel from the magnification in this equal-medium case.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Images at refracting surfaces and thin lenses

Moving the object across the focal point changes the image type

For a converging lens and a real object farther away than the focal distance, the image is real and inverted. As the object approaches the front focal point from outside, the real image moves farther away and its magnification magnitude increases.

At the front focal point, the outgoing paraxial rays are parallel. There is no finite image plane; saying the image is at infinity is shorthand for this limiting ray geometry. A finite magnification value is not defined there.

When the object moves inside the focal distance, outgoing rays diverge and their backward extensions meet on the object side. The image is virtual, upright and enlarged. This is the basic geometry of a magnifier.

A diverging thin lens in a uniform surrounding medium gives a real object a virtual, upright, reduced image. This familiar statement assumes a real incident object; a converging incoming bundle, represented by a virtual object, can produce other cases.

WORKED EXAMPLE

An object inside the focal distance gives a negative image coordinate

A converging lens has f = +10 cm and a real object at u = −6 cm. Find v and m.

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WORKED EXAMPLE

Test the usual diverging-lens case

A diverging lens has f = −15 cm and a real object at u = −30 cm. Find image coordinate and magnification.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Moving the object across the focal point changes the image type
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Images at refracting surfaces and thin lenses

Curvature and relative index determine the focal length

Keep the Cartesian axis directed left to right through a thin lens. Let R1 and R2 be the signed radii of the first and second surfaces encountered. For a biconvex lens in this orientation, R1 is positive and R2 negative.

Adding the surface equations gives 1/f = (n_lens/n_medium − 1)(1/R1 − 1/R2). The factor is the relative index, not simply n_lens − 1 unless the surrounding medium is approximated as index one.

A plane surface has infinite radius and contributes zero reciprocal curvature. A plano-convex lens therefore has one curvature term. Reversing a thin lens in the same surrounding medium swaps the surface roles and signs in a way that preserves its focal length.

Shape alone is insufficient to label a lens converging or diverging. If its material index equals the surrounding index, its optical power vanishes. If the index contrast reverses, the sign of its power reverses for unchanged geometry.

\frac1f=\left(\frac{n_\ell}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)

Thin-lens, paraxial lens-maker relation in the same surrounding medium on both sides. Signed radii share one axis.

WORKED EXAMPLE

Assign the signs of the two radii before calculating

A symmetric biconvex thin lens has material index 1.50 in air. Its first surface centre is 20 cm to the right of that surface and its second surface centre 20 cm to the left. Find f.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Curvature and relative index determine the focal length
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
\frac1f=\left(\frac{n_\ell}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)
Meaning: Thin-lens, paraxial lens-maker relation in the same surrounding medium on both sides. Signed radii share one axis.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Images at refracting surfaces and thin lenses

Changing the surrounding medium changes the same lens

Immersing a lens changes how much each surface bends light because refraction depends on an index ratio. The lens geometry can remain fixed while its focal length changes substantially.

For a lens with index 1.50 placed in a medium of index 1.25, the relative-index contrast is 1.50/1.25 − 1 = 0.20. In air it was 0.50. Its power therefore becomes 0.20/0.50 = 0.40 of its air power, and its focal length becomes 2.5 times as large.

If the surrounding index increases to 1.50, the contrast is zero and the surfaces no longer refract in this ideal model. An infinite focal length here means zero optical power, not a strongly focusing lens.

Because material indices can depend on wavelength, focal length can also depend on wavelength. This contributes to chromatic aberration: different spectral components can have different focal positions. Treating one f as universal across a broad spectrum is an additional approximation.

WORKED EXAMPLE

Reuse geometry while recalculating the relative index

The preceding biconvex lens has f = 20 cm in air and material index 1.50. Find its focal length when immersed in a medium of index 1.25.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Changing the surrounding medium changes the same lens
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

09 / Images at refracting surfaces and thin lenses

Optical power is an inverse focal length with a useful unit

Define the optical power of a thin lens as P = 1/f when f is expressed in metres. The unit m⁻¹ is called the dioptre, often written D. A converging lens has positive power and a diverging lens negative power in the convention used here.

Power describes how strongly a lens changes paraxial ray convergence. It is not electrical power and is not measured in watts. Keeping the unit attached helps prevent confusion with the same letter used elsewhere in physics.

For thin lenses in contact in the same surrounding medium, their powers add: Peq = P1 + P2. This follows by applying the lens equation successively at effectively the same plane and cancelling the intermediate image coordinate.

For lenses separated by a finite distance, simple contact addition is generally wrong. In air the effective paraxial system power includes a separation term, Peq = P1 + P2 − dP1P2, and the effective principal planes need not lie at either lens. For actual image locations, tracing each lens in sequence is usually clearer.

P=1/f\quad(f\text{ in metres}),\qquad P_{\mathrm{contact}}=P_1+P_2

The contact formula assumes thin lenses with negligible separation in a common surrounding medium.

WORKED EXAMPLE

Combine a converging and a diverging contact lens

Two thin lenses in contact in air have focal lengths +25 cm and −50 cm. Find equivalent power and focal length.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: Optical power is an inverse focal length with a useful unit
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
P=1/f\quad(f\text{ in metres}),\qquad P_{\mathrm{contact}}=P_1+P_2
Meaning: The contact formula assumes thin lenses with negligible separation in a common surrounding medium.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

10 / Images at refracting surfaces and thin lenses

An intermediate image becomes the next element’s object

For separated thin lenses, first solve the image formed by lens 1 as if lens 2 were absent. That intermediate point describes the incoming bundle at lens 2. Its coordinate relative to lens 2 determines whether the second element has a real or virtual object.

If lens 1 is at x = 0 and lens 2 at x = d, the second object coordinate is u2 = v1 − d. An intermediate point to the left of lens 2 gives u2 < 0; a point to its right gives u2 > 0 and represents a converging bundle arriving at the lens.

Apply the second lens equation using its own origin. The final global coordinate is x_final = d + v2. Forgetting to add d back mixes a local distance with a global position.

The total lateral magnification is the product m1m2 because the first image height is the second object height. Its sign tells whether the final image is upright or inverted relative to the original object.

WORKED EXAMPLE

Move the origin before applying the second lens equation

Lens 1 at x = 0 has f1 = +10 cm. An object is at x = −30 cm. Lens 2 at x = +25 cm has f2 = +5 cm. Find the final image coordinate and total magnification.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Images at refracting surfaces and thin lenses
Focus: An intermediate image becomes the next element’s object
Learning goals:
- Classify images from actual rays and backward extensions.
- Assign Cartesian object, image and radius signs consistently.
- Solve a single spherical surface including its index-dependent magnification.
- Use the thin-lens equation and explain focal-point limits.
- Derive lens power from signed curvature and relative index.
- Predict immersion and wavelength effects on focal length.
- Combine contact powers and trace separated lenses with moving origins.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

A real image is formed by converging outgoing rays; a virtual image by their backward extensions.
Here u is negative for a real object on the left, and a thin lens obeys 1/v − 1/u = 1/f.
Single-surface magnification includes n1/n2; equal-medium thin-lens magnification is v/u.
The front focal-point case gives parallel outgoing rays and no finite image plane.
Lens-maker curvature uses signed radii and the relative material index.
Contact powers add; separated lenses require position and origin tracking.
Total lateral magnification is the product of the stage magnifications.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A thin lens has f = +10 cm and object coordinate u = −20 cm. What is v?

Answer in cm.

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SELF CHECK 2

A lens image has v = −12 cm for a real object at u = −8 cm in the same surrounding medium. What is lateral magnification?

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SELF CHECK 3

Two thin contact lenses in air have powers +6 D and −2 D. What is their equivalent focal length in centimetres?

Answer in cm.

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SELF CHECK 4

For a plane interface, an object at u = −40 cm in n1 = 1.60 is viewed through n2 = 1.20. What is the paraxial image coordinate v?

Answer in cm.

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SELF CHECK 5

A real object is at the front focal point of a converging thin lens. What happens to rays from an on-axis point?

They emerge parallel
They converge at twice the focal distance
They reverse direction

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SELF CHECK 6

A lens is surrounded by a medium with exactly the same refractive index as its material. In the ideal model its optical power is:

Unchanged
Zero
Infinite

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SELF CHECK 7

Lens 1 forms an intermediate image at global x = 18 cm. Lens 2 is at x = 30 cm. What object coordinate belongs in lens 2’s equation?

+18 cm
−12 cm
+48 cm

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

Turn a focal length into dioptresIntermediate · Free sample · Hints and guided reasoning Focus at a single surfaceAdvanced · Hints and guided reasoning Find the image without losing a sign.Intermediate · Hints and guided reasoning Combine positive and negative powerIntermediate · Hints and guided reasoning A virtual image has a signed positionIntermediate · Hints and guided reasoning A biconvex lens in a liquidIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

For a converging thin lens, what changes when the object moves from just outside to just inside the focal distance?

Compare your reasoning

Outside the focus the simple model gives a real inverted image; inside it gives a virtual upright image. At the focal position, outgoing paraxial rays are parallel and the image distance is unbounded.

Revisit the explanation →

A few days later: change the situation

Why can immersing a lens in a liquid change its focal length even though its surface curvatures stay fixed?

Compare your reasoning

Refraction depends on the lens index relative to the surrounding medium. Changing that contrast changes optical power. Geometry alone does not determine focal length.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑

What you will learn to do

Bring back one idea

An image is a statement about where rays meet or appear to meet

A screen placed at the calculated image position receives a sharp pattern from the outgoing rays. What kind of image is this?

Fix one Cartesian axis before assigning any signs

A curved refracting surface has a different normal at each point

Solve a single surface before combining it into a lens

A plane boundary and an equal-index boundary check the model

Recover apparent depth from the general surface equation

Two nearby refracting surfaces become one thin-lens relation

Read it in words

Symbols and units

Signs and operations

When it applies

Make it concrete

Find an image with signs that agree with its ray geometry

Move the object through the focal plane

Moving the object across the focal point changes the image type

An object inside the focal distance gives a negative image coordinate

Test the usual diverging-lens case

Curvature and relative index determine the focal length

Assign the signs of the two radii before calculating

Changing the surrounding medium changes the same lens

Reuse geometry while recalculating the relative index

Optical power is an inverse focal length with a useful unit

Combine a converging and a diverging contact lens

An intermediate image becomes the next element’s object

Move the origin before applying the second lens equation

What should stay with you

Try it without the examples

A thin lens has f = +10 cm and object coordinate u = −20 cm. What is v?

A lens image has v = −12 cm for a real object at u = −8 cm in the same surrounding medium. What is lateral magnification?

Two thin contact lenses in air have powers +6 D and −2 D. What is their equivalent focal length in centimetres?

For a plane interface, an object at u = −40 cm in n1 = 1.60 is viewed through n2 = 1.20. What is the paraxial image coordinate v?

A real object is at the front focal point of a converging thin lens. What happens to rays from an on-axis point?

A lens is surrounded by a medium with exactly the same refractive index as its material. In the ideal model its optical power is:

Lens 1 forms an intermediate image at global x = 18 cm. Lens 2 is at x = 30 cm. What object coordinate belongs in lens 2’s equation?

Take the ideas into a problem

What can you explain later?

Next study session

A few days later: change the situation