In short

Lensmaker's equation. For a thin lens of material with refractive index n placed in air, with surface radii R_1 and R_2 (using the sign convention that a radius is positive if its centre of curvature lies to the right of the surface),

\boxed{\; \frac{1}{f} \;=\; (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \;}

A biconvex lens has R_1 > 0 and R_2 < 0, so both terms add up and f > 0. A biconcave lens flips both signs and gives f < 0.

Lens in a medium. If the lens sits in a medium of index n_m rather than air,

\frac{1}{f} \;=\; \left(\frac{n}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

A glass lens in water (n/n_m = 1.50/1.33 \approx 1.13) has a focal length about four times longer than in air — it is a much weaker lens.

Thin lenses in contact. Two thin lenses pressed together behave like a single lens of focal length f_{\text{eff}} given by

\boxed{\; \frac{1}{f_{\text{eff}}} \;=\; \frac{1}{f_1} + \frac{1}{f_2} \;} \quad\text{equivalently}\quad P_{\text{eff}} = P_1 + P_2

where the power in dioptres is P = 1/f_{\text{metres}}. Spectacle powers add directly — this is why Lenskart gives you a number like "-2.25 D" and not a focal length.

Silvered lens. A lens with one surface silvered behaves as an equivalent mirror of focal length F given by

\frac{1}{F} \;=\; \frac{2}{f_l} + \frac{1}{f_m}

where f_l is the focal length of the unsilvered lens and f_m is the focal length of the silvered surface treated as a mirror. The light passes through the lens twice, which is why f_l is doubled.

Hold a magnifying glass at arm's length and tilt it until sunlight catches the paper below. Slide it up and down until the bright spot shrinks to a pinpoint — that is the focal point, and the distance from the lens to the pinpoint is the focal length. Now swap the magnifying glass for a thinner one. The focal length is longer. Swap it for one made of a denser glass, and the focal length is shorter. Dip the whole thing in water and the focal length stretches so far that you may not find a focus at all.

Why? A lens is nothing but a piece of glass with two curved surfaces. Everything that governs how it bends light must be in those two numbers — the curvatures of the surfaces — and in the one number that describes how much the glass slows light, the refractive index. The lensmaker's equation is the single line that ties these three numbers to the focal length f. Once you have it, you can do more than use a lens: you can design one. You can say, "I want a lens of focal length +20 cm, and I'm going to grind it from BK7 glass whose n = 1.517" — and compute exactly how curved each surface must be. Every pair of spectacles, every phone camera, every telescope objective ever manufactured began as an application of this equation.

Where the equation comes from — two refractions in a row

A thin lens is just two refracting surfaces close together. Light entering the lens refracts once at the first surface; it travels through the glass (a negligible distance, for a thin lens); then refracts again at the second surface as it exits back into air. If you know how a single spherical surface refracts light — and this you learned in Refraction at Spherical Surfaces — the lens formula follows by applying that rule twice.

For one refraction from a medium of index n_1 into a medium of index n_2 across a surface of radius R, with object distance u and image distance v (all measured from the vertex, with the sign convention that distances measured in the direction of light propagation are positive):

\frac{n_2}{v} - \frac{n_1}{u} \;=\; \frac{n_2 - n_1}{R} \tag{1}

That is the single-surface formula. The lensmaker's derivation is to apply it twice: once at surface 1 (air \to glass) and once at surface 2 (glass \to air), and add.

Biconvex lens as two refracting surfacesA biconvex thin lens with left surface of radius R1 positive (centre of curvature to the right) and right surface of radius R2 negative (centre of curvature to the left). An object at O on the left refracts through surface 1 to form a virtual intermediate image I1 deep inside the glass; surface 2 then refracts that into the final image I2 on the right. The two vertices are essentially the same point for a thin lens. surface 1 (R₁ > 0) surface 2 (R₂ < 0) O object I₁ (intermediate) I₂ final image |u| v₁ (from surface 1) v (final)
A biconvex thin lens is two refracting surfaces sitting essentially at the same point. Surface 1 refracts light from the object O as if forming an intermediate image I₁ (dashed — it would form there if the glass extended to infinity). Surface 2 then refracts that intermediate image into the final image I₂. Adding the two surface formulas cancels the intermediate distance and gives the lensmaker's equation.

Assumptions. The lens is thin: the two surfaces are so close together that you can ignore the distance between their vertices. Rays travel close to the axis (paraxial approximation), so \sin\theta \approx \theta. The glass has refractive index n; it sits in air (n_{\text{air}} = 1). The sign convention is the one used throughout wiki-physics: distances measured along the direction of light propagation from the vertex are positive; against it, negative. A radius R is positive if the centre of curvature lies to the right of the vertex (light goes left to right), negative otherwise.

Step 1. Apply equation (1) at surface 1. Light goes from air (n_1 = 1) into glass (n_2 = n). The object is at distance u, and the image formed by surface 1 alone — as if the glass continued forever beyond — is at distance v_1.

\frac{n}{v_1} - \frac{1}{u} \;=\; \frac{n - 1}{R_1} \tag{2}

Why: equation (1) with n_1 = 1, n_2 = n, R = R_1. The intermediate image v_1 is virtual and lies inside what would be the continuation of the glass — we only need its position as a bookkeeping step, not as something you could actually see.

Step 2. The intermediate image I_1 now acts as the object for surface 2. Because the lens is thin, the distance from surface 1 to surface 2 is negligible, so the object distance for surface 2 is the same v_1 (measured from the common vertex). At surface 2, light goes from glass (n_1 = n) back into air (n_2 = 1), across a surface of radius R_2.

\frac{1}{v} - \frac{n}{v_1} \;=\; \frac{1 - n}{R_2} \tag{3}

Why: reapplying equation (1) with the media swapped. The final image distance v is measured from the same vertex because the lens is thin. The v_1 in equation (3) is exactly the v_1 of equation (2) — that is the whole point of "thin".

Step 3. Add equations (2) and (3) directly. The n/v_1 terms are opposite in sign — they cancel.

\left(\frac{n}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{n}{v_1}\right) \;=\; \frac{n - 1}{R_1} + \frac{1 - n}{R_2}
\frac{1}{v} - \frac{1}{u} \;=\; \frac{n - 1}{R_1} - \frac{n - 1}{R_2}

Why: the intermediate variable v_1 disappears because the intermediate image is a fiction — the real physics only cares about where the light ends up (the object at u, the image at v). When you add two correct sub-steps and the joining variable cancels, you know the derivation is clean.

Step 4. Factor out (n - 1) on the right.

\frac{1}{v} - \frac{1}{u} \;=\; (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \tag{4}

Step 5. Recognise the left side. For a thin lens, the relation between object distance, image distance, and focal length is the lens formula \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} (derived in Thin Lenses). Since this must hold for every object distance, the right-hand side of equation (4) must equal 1/f:

\boxed{\; \frac{1}{f} \;=\; (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \;} \tag{5}

Why: equation (4) has the same shape as the lens formula, so the right-hand side is 1/f. And the right-hand side depends only on the lens itself — its material (n) and its geometry (R_1, R_2) — not on where you put the object. That is the deep content: the focal length is a property of the lens, nothing else.

This is the lensmaker's equation. Three inputs, one output. Tell it the refractive index of your glass and the two surface radii, and it gives you the focal length.

Reading the equation — sign conventions and lens shapes

The signs in 1/R_1 - 1/R_2 look innocent, but everything depends on getting them right. Use this rule and never get confused again:

A radius is positive if the centre of curvature lies to the right of the surface (same direction as light propagation), and negative if it lies to the left.

Here is what that means for the four common shapes, with light travelling left to right in every case.

Four lens shapes with their R₁ and R₂ signsFour lens shapes arranged in a two-by-two grid. Top left: biconvex, R1 positive R2 negative, focal length positive (converging). Top right: biconcave, R1 negative R2 positive, focal length negative (diverging). Bottom left: plano-convex, R1 positive R2 infinite, focal length positive. Bottom right: meniscus (converging), both radii same sign; focal length depends on which radius is smaller in magnitude. biconvex R₁ > 0, R₂ < 0 → f > 0 biconcave R₁ < 0, R₂ > 0 → f < 0 plano-convex R₁ > 0, R₂ = ∞ → f > 0 converging meniscus R₁ > 0, R₂ > 0, |R₁| < |R₂| → f > 0
Four common lens shapes and what the lensmaker's equation says about each. The key rule: if light hits a surface that bulges *toward* the incoming ray, that surface has a positive $R$ (centre of curvature behind the surface, to the right). A flat surface has $R = \infty$, which makes $1/R = 0$ and drops out of the equation.

A sanity check you should do once. For a symmetric biconvex lens made of glass (n = 1.50) with R_1 = +20 cm and R_2 = -20 cm:

\frac{1}{f} = (1.50 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.50 \cdot \frac{2}{20} = \frac{1}{20} \text{ cm}^{-1}

so f = +20 cm. Converging, as expected. Now flip to biconcave: R_1 = -20 cm, R_2 = +20 cm, same glass.

\frac{1}{f} = 0.50 \cdot \left(-\frac{1}{20} - \frac{1}{20}\right) = -\frac{1}{20}

so f = -20 cm. Diverging. The equation does the bookkeeping; your job is to put in the signs honestly.

Why water ruins a magnifying glass

Take your biconvex glass magnifying lens and dunk it into a bucket of water. Try to read something through it. The magnification collapses. The lens has not changed — but its focal length has quadrupled. Why?

When a lens sits in a medium of index n_m (not air), the derivation above gets repeated with n_1 = n_m instead of 1. The (n - 1) factors become (n - n_m) factors, and when you redo the algebra, you get

\frac{1}{f} \;=\; \left(\frac{n}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \tag{6}

For glass in air: n/n_m = 1.50/1 = 1.50, so the prefactor is 0.50. For glass in water: n/n_m = 1.50/1.33 = 1.128, so the prefactor is 0.128.

The focal length goes up by the ratio 0.50 / 0.128 \approx 3.9. A lens that focuses at 20 cm in air focuses at roughly 78 cm underwater — effectively, it is almost flat glass. This is exactly why divers wear masks: when water touches your cornea, the eye's refractive index (\approx 1.37) is too close to water's (1.33) to bend light meaningfully, and everything blurs. The mask puts an air gap between your eye and the water, restoring the corneal surface's full optical power.

Explore — focal length vs curvature

Interactive: focal length of a symmetric biconvex lens as a function of surface radius A plot of focal length f in centimetres versus surface radius R in centimetres for a symmetric biconvex crown-glass lens with R1 equals R and R2 equals negative R. The focal length equals R over 2n minus 2, which for n equals 1.5 is just R. A draggable vertical line lets you pick a value of R and read off f. surface radius R (cm), with R₁ = R, R₂ = −R focal length f (cm) 0 10 20 30 40 50 60 10 20 30 40 50 60 in air in water drag the red point along the axis
Focal length of a symmetric biconvex lens ($R_1 = R$, $R_2 = -R$) made of crown glass ($n = 1.50$), as a function of $R$. In air, $f = R$; in water, $f \approx 3.9\,R$. Drag the red point to change $R$ and read both focal lengths. Dunk a lens in water and its focal length quadruples — the lens is still curved, but the glass no longer slows light much more than the surrounding medium does.

Lenses in contact — why powers add

Put two thin lenses flat against each other, with focal lengths f_1 and f_2. Light from an object at distance u passes through the first lens, which would form an image at v_1 given by

\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \tag{7}

That image v_1 is the object for the second lens. Because the lenses are in contact, the distance between them is zero, so the object distance for lens 2 is v_1 (same measurement point):

\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \tag{8}

Add equations (7) and (8). The 1/v_1 terms cancel:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}

But the combination, treated as a single lens of focal length f_{\text{eff}}, must also obey \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_{\text{eff}}}. Comparing:

\boxed{\; \frac{1}{f_{\text{eff}}} \;=\; \frac{1}{f_1} + \frac{1}{f_2} \;} \tag{9}

Why: the intermediate image v_1 vanishes in exactly the same way the intermediate v_1 vanished in the lensmaker derivation — because two consecutive refractions "see" the intermediate as a handoff point, not as anything physical. Stack more lenses, and each new one just adds another 1/f_k to the sum.

The dioptre

Define the power of a lens as

P \;=\; \frac{1}{f}

with f in metres and P in dioptres (symbol D). A lens of focal length +50 cm has power +2 D. A lens of focal length -25 cm has power -4 D. Equation (9) then reads

P_{\text{eff}} \;=\; P_1 + P_2

— powers add. This is why spectacle prescriptions are written in dioptres: when the optician at Lenskart writes "-2.25 D" for your right eye, they are giving you a number that adds directly to whatever other correction you stack (for instance, if you also need a cylindrical correction for astigmatism, the two powers sum to give the total). A telescope's objective of power +1 D paired with an eyepiece of +50 D has a combined power of +51 D — though for instruments the lenses are not in contact, so equation (9) needs modification (you pick that up in Optical Instruments with the general two-lens formula 1/f = 1/f_1 + 1/f_2 - d/(f_1 f_2)).

Achromatic doublets — why spectacles and telescope objectives use two lenses

Crown glass (n \approx 1.517) and flint glass (n \approx 1.620) have different refractive indices — and more importantly, different dispersions, so n varies with wavelength differently in the two glasses. A single glass lens focuses blue light at a shorter focal length than red light (chromatic aberration). You can cancel this by cementing a weak diverging flint lens to a strong converging crown lens: the crown lens does the focusing, and the flint lens contributes just enough opposite-direction dispersion to cancel the colour spread. The total power is still positive (converging), but the image is colour-corrected.

Every refracting telescope you will ever look through — from the small Celestron at your college astronomy club to the 102 cm Dobsonian at the Indian Institute of Astrophysics's Kavalur observatory — has an achromatic doublet for its objective. The same principle goes into a good microscope objective, a phone camera's lens stack (five or six elements), and the high-index thinned lenses your optician grinds for strong prescriptions.

Worked examples

Example 1: Designing a reading glass

You want to grind a symmetric biconvex lens of focal length f = +25 cm from a piece of crown glass with n = 1.52. Both surfaces will have the same radius of curvature (in magnitude). Find R_1 and R_2.

Symmetric biconvex reading glassA symmetric biconvex lens drawn on the optic axis, with R1 positive on the left surface and R2 negative on the right surface, both equal in magnitude. The centres of curvature C1 is to the right of the lens and C2 is to the left. C₁ R₁ > 0 C₂ R₂ < 0 symmetric biconvex, |R₁| = |R₂| = R
The lens is symmetric: the left surface bulges left (centre of curvature C₁ to the right, so $R_1 > 0$), and the right surface bulges right (centre C₂ to the left, so $R_2 < 0$). Both radii have the same magnitude $R$.

Step 1. Write the symmetry condition. "Symmetric biconvex, equal radii in magnitude" means R_1 = +R and R_2 = -R for some positive R.

Why: the left surface has its centre of curvature to the right (so R_1 > 0); the right surface has its centre of curvature to the left (so R_2 < 0). Equal curvature in magnitude is the symmetry assumption.

Step 2. Plug into the lensmaker's equation.

\frac{1}{f} = (n - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n - 1) \cdot \frac{2}{R}

Why: \dfrac{1}{R_1} - \dfrac{1}{R_2} = \dfrac{1}{R} - \left(-\dfrac{1}{R}\right) = \dfrac{2}{R}. The two surfaces contribute equally and their effects add — this is what "symmetric" means for a lens.

Step 3. Solve for R.

R \;=\; 2(n - 1)\,f \;=\; 2(1.52 - 1)(25\text{ cm}) \;=\; 2(0.52)(25) \;=\; 26\text{ cm}

Why: rearranging 1/f = 2(n-1)/R gives R = 2(n-1)f directly. The arithmetic is clean because n = 1.52 was chosen to make it so.

Step 4. State the answer with signs.

R_1 = +26 \text{ cm}, \quad R_2 = -26 \text{ cm}

Why: the problem asked for both radii. The signs tell the grinder which way each surface bulges — without them, you could accidentally grind a biconcave lens.

Result: A symmetric biconvex crown-glass lens with R_1 = +26 cm and R_2 = -26 cm has focal length f = +25 cm.

Consistency check. Compute the power: P = 1/f = 1/0.25\text{ m} = +4 D. That's a reasonably strong reading glass — the kind of prescription someone aged 55 might need for close work. The lens in your grandmother's reading glasses is not dramatically different from this.

Example 2: A silvered biconvex lens — equivalent mirror

A biconvex lens has R_1 = +30 cm and R_2 = -20 cm, made of glass with n = 1.50. The second surface (the one on the right, with R_2 = -20 cm) is silvered so that it acts as a mirror. Find the focal length of the resulting equivalent mirror.

Silvered lens as three optical elementsA biconvex lens with its right-hand surface silvered. Light enters through the unsilvered left surface (refraction by surface 1), reflects off the silvered right surface (reflection), and refracts again on the way back out through the left surface. The combined system is equivalent to a single concave mirror of effective focal length F. silvered incoming (through lens) outgoing (through lens again) refract 1 reflect refract 2
Light passes through the lens (refraction 1), reflects off the silvered back surface (reflection), and passes back through the lens (refraction 2). The combination behaves as a single equivalent mirror.

Step 1. Treat the system as three optical operations in sequence, and use the power-addition rule.

When a lens and a mirror act together in sequence, the effective power is

P_{\text{eff}} = P_l + P_m + P_l = 2P_l + P_m

because the light passes through the lens twice (once going in, once going out) and reflects off the mirror once.

Why: powers of optical elements in contact add just like the powers of thin lenses in contact did (equation 9), and the derivation is identical — each element imposes its own 1/f on the reciprocal of image distance, and successive contacts add these reciprocals. Light passes through the lens twice so P_l appears twice.

Step 2. Compute f_l, the focal length of the unsilvered lens, using the lensmaker's equation.

\frac{1}{f_l} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.50 - 1)\left(\frac{1}{30} - \frac{1}{-20}\right)
= 0.50 \cdot \left(\frac{1}{30} + \frac{1}{20}\right) = 0.50 \cdot \frac{2 + 3}{60} = 0.50 \cdot \frac{5}{60} = \frac{1}{24} \text{ cm}^{-1}

so f_l = 24 cm.

Why: this is the focal length the lens would have if neither surface were silvered. We need it first, because the silvered-lens formula treats the lens as a lens and the silvered surface as a mirror, separately.

Step 3. Compute f_m, the focal length of the silvered surface treated as a concave mirror of radius of curvature |R_2| = 20 cm.

For a concave mirror, f_m = R/2 = 20/2 = 10 cm.

Why: the silvered back surface is a spherical mirror. Its radius is |R_2| = 20 cm, and the mirror formula gives f_m = R/2. (See Spherical Mirrors for the derivation.) The sign is positive because it is a concave mirror — the reflective side curves toward the incoming light.

Step 4. Combine using the silvered-lens formula.

\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m} = \frac{2}{24} + \frac{1}{10} = \frac{1}{12} + \frac{1}{10}

Common denominator 60:

\frac{1}{F} = \frac{5}{60} + \frac{6}{60} = \frac{11}{60} \text{ cm}^{-1}

so F = 60/11 \approx 5.45 cm.

Why: the "2" on the lens term captures the double pass. The mirror term appears once because the light reflects off it once. Both terms are positive, so F > 0, and the equivalent mirror is converging — it focuses parallel light to a point about 5.45 cm in front of the silvered lens.

Step 5. Verify with power arithmetic.

P_l = 1/0.24\text{ m} = 4.17 D, so 2P_l = 8.33 D. And P_m = 1/0.10\text{ m} = 10.0 D. Total: P_{\text{eff}} = 8.33 + 10.0 = 18.33 D, giving F = 1/18.33 \approx 0.0546\text{ m} = 5.46 cm. ✓

Why: converting to dioptres and adding is an independent check. The agreement (5.45 cm by one method, 5.46 cm by the other — the tiny gap is rounding) tells you the power-addition picture is exactly equivalent to the focal-length formula.

Result: The silvered biconvex lens behaves as a converging mirror of focal length F \approx 5.45 cm.

Interpretation. A silvered lens is a common compact optical element: rear-view mirrors with a slight magnification, dental examination mirrors, the ophthalmoscope a doctor uses to look at your retina. Making a strongly converging mirror in a small footprint is easier with a silvered lens than with a deeply ground concave mirror.

Common confusions

If you want to use the lensmaker's equation, design a simple lens, and combine lenses in contact, you have what you need. What follows is the thick-lens treatment, the general two-surface derivation from scratch, and a deeper look at why the focal length is symmetric in the two surfaces.

The general refraction-at-a-spherical-surface formula

Go back to equation (1): n_2/v - n_1/u = (n_2 - n_1)/R. Where does this come from? Consider a single spherical refracting surface of radius R, with object O on the axis at distance u to the left, and image I formed at distance v to the right. Take a ray from O that strikes the surface at height h above the axis. Apply Snell's law n_1 \sin\theta_1 = n_2 \sin\theta_2. In the paraxial limit, \sin\theta \approx \theta, and the geometry gives

n_1 \left(\frac{h}{|u|} + \frac{h}{|R|}\right) \;=\; n_2 \left(\frac{h}{v} - \frac{h}{|R|}\right)

(up to sign handling). Rearranging and attaching signs to match our convention:

\frac{n_2}{v} - \frac{n_1}{u} \;=\; \frac{n_2 - n_1}{R}

This is the single-surface formula. The lensmaker's equation is this formula applied twice, with the intermediate image distance cancelling out. Nothing more.

The thick lens — why thickness matters

If the two surfaces are separated by a distance t (the thickness of the lens), the intermediate image distance v_1 from surface 1 becomes the object distance u_2 = v_1 - t for surface 2 (subtracting the thickness because the object for surface 2 is measured from surface 2's vertex). Redoing the derivation with this correction gives

\frac{1}{f} \;=\; (n - 1)\left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)\,t}{n\,R_1 R_2}\right]

For a 5-mm-thick lens with R_1 = R_2 = 50 mm and n = 1.5, the third term contributes about (0.5)(5)/(1.5 \cdot 50 \cdot 50) = 6.7 \times 10^{-4} mm⁻¹, compared to the main terms of order 2 \times 10^{-2} mm⁻¹ — a correction of order 3\%. For ordinary spectacles and simple lab lenses this is negligible; for a camera lens it is the difference between a sharp image and a blurry one, and every design program computes it exactly.

Why the focal length is symmetric under R_1 \leftrightarrow R_2 and sign-flip

A curious feature of the lensmaker's equation: flipping the lens around end-to-end (so the ray hits what used to be surface 2 first) leaves f unchanged. The algebraic statement is: swap R_1 \leftrightarrow -R_2 and R_2 \leftrightarrow -R_1 (the negatives come from the convention — the centre of curvature that was on the right is now on the left). Substituting:

\frac{1}{f'} = (n-1)\left(\frac{1}{-R_2} - \frac{1}{-R_1}\right) = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{f}

So f' = f: the lens has the same focal length no matter which way you flip it. This is not a coincidence — it is a statement of the time-reversal symmetry of geometric optics. Light rays are reversible; if a ray from O reaches I after one trip through the lens, a ray from I will reach O after the reverse trip. The focal length, which is a property of the lens itself rather than of the direction of travel, cannot care about the direction.

Equivalent lens of a separated pair

For two thin lenses separated by distance d (not in contact), the effective focal length is

\frac{1}{f} \;=\; \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}

The equivalent single lens has focal length f and sits at a position depending on d. Two limiting cases: (i) if d = 0 (in contact), we recover equation (9); (ii) if d = f_1 + f_2 (a Keplerian telescope in normal adjustment), then 1/f = 0, so f = \infty. A telescope of this arrangement has infinite effective focal length — meaning parallel rays in come out parallel again but at a different angle, which is precisely what an angular magnifier does.

The lensmaker's equation in water, revisited

Equation (6) was stated but not derived in the main text. The derivation is the same as before, but with the surroundings n_1 = n_m instead of 1. At surface 1, going from medium (index n_m) into glass (index n):

\frac{n}{v_1} - \frac{n_m}{u} \;=\; \frac{n - n_m}{R_1}

At surface 2, going from glass back into medium:

\frac{n_m}{v} - \frac{n}{v_1} \;=\; \frac{n_m - n}{R_2}

Add, and divide through by n_m:

\frac{1}{v} - \frac{1}{u} \;=\; \left(\frac{n}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \;=\; \frac{1}{f}

So the rule is: everywhere you had n in the air-derivation, use n/n_m in the medium-derivation. The formula is the same shape — only the strength of the refraction (the prefactor) changes. This explains why the lens shape doesn't determine whether it focuses or diverges underwater; the relative index n/n_m does. A lens of plastic with n = 1.34 placed in water with n_m = 1.33 has an almost zero prefactor — it is essentially invisible, the same way a glass rod in a beaker of matched-index oil disappears from view.

Where this leads next