Refraction at spherical surfaces

A pond looks shallower than it is, a fish in a curved glass bowl looks displaced, and the front surface of your eye bends light before any lens gets involved — all three are the same piece of physics: light crossing a curved boundary between two transparent media. Refraction at a spherical surface is the single geometric idea that makes this precise. Once you have one equation linking the object distance, the image distance, the radius of curvature and the two refractive indices, you can find where any image forms — real or virtual, in front of the surface or behind it — without redrawing a ray diagram every time. This relation is also not an isolated result: apply it twice, once at each face of a lens, and it hands you the lens maker's formula that governs every converging and diverging lens you will use for the rest of Class XII optics.

The core insight

A single spherical surface separating two media of refractive index $n_1$ and $n_2$ obeys $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ , with all distances measured from the pole using the Cartesian sign convention.

The core idea

Picture a spherical surface of radius $R$ separating a rarer medium ( $n_1$ ) from a denser one ( $n_2$ ), with a point object $O$ sitting on the principal axis in the rarer medium. One ray from $O$ travels straight along the axis and is undeviated; a second, nearby ray strikes the surface close to the pole $P$ and bends according to Snell's law. For a convex surface with light going from rarer to denser medium, these two rays meet at a real image point $I$ inside the denser medium. Distances are measured from the pole $P$ , along the direction of the incident light taken as positive — the same sign convention used for mirrors. Applying the small-angle (paraxial) approximation to Snell's law, $n_1\sin i = n_2\sin r \approx n_1 i = n_2 r$ , and working through the geometry of the ray triangles formed at $O$ , the centre of curvature $C$ , and the image point $I$ , gives the single relation that governs refraction at any spherical surface:

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

where: $n_1$ is the refractive index of the medium containing the incident ray (object side), $n_2$ is the refractive index of the medium containing the refracted ray (image side), $u$ is the object distance from the pole, $v$ is the image distance from the pole, and $R$ is the radius of curvature of the surface — all signed according to the Cartesian convention. This formula was derived for one specific picture — a convex surface, rarer-to-denser medium, forming a real image — but it is completely general. Feed in a concave surface, or a case that produces a virtual image, and the same equation returns the correct signed answer; you never need a separate derivation for each geometry, only careful bookkeeping of signs. A useful limit falls straight out of the formula: as $u\to-\infty$ (object at infinity), $v$ approaches the image-side focal length $f_2$ , giving $\dfrac{n_2}{f_2}=\dfrac{n_2-n_1}{R}$ , or $f_2=\dfrac{n_2R}{n_2-n_1}$ . This is exactly the idea that gets applied twice — once per lens face — to build the lens maker's formula in the next topic.

Figure

A convex spherical surface refracting a ray from an axial object O in air to a real image I inside denser glass

Derivation

Derivation ARefraction relation for a single spherical surface

The goal is to derive $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ for a convex surface separating a rarer medium from a denser one, using only the paraxial (small-angle) approximation applied to Snell's law and the geometry of one incident ray, so that the final relation connects only the two refractive indices and the three signed distances $u$ , $v$ , $R$ .

Step 1. Set up the ray. A ray leaves the axial object $O$ in the rarer medium ( $n_1$ ) and strikes the spherical surface at point $A$ , close to the pole $P$ , where the normal $NA$ passes through the centre of curvature $C$ . The angle of incidence is $i=\angle NAO$ and, after bending toward the normal on entering the denser medium, the angle of refraction is $r=\angle NAI$ , with $I$ the image point on the axis.

Step 2. Write Snell's law in paraxial form. For small angles, $\sin i\approx i$ and $\sin r\approx r$ , so Snell's law $n_1\sin i=n_2\sin r$ becomes simply $n_1 i=n_2 r$ .

Step 3. Express $i$ and $r$ using the exterior-angle property of triangles. In triangle $OAC$ , the exterior angle at $C$ equals the sum of the two opposite interior angles, giving $i=\angle AOP+\angle ACP$ . In triangle $ACI$ , similarly, $\angle ACP=r+\angle AIP$ , so $r=\angle ACP-\angle AIP$ . Because $A$ lies close to the pole, each of these angles is small and can be approximated by (height of $A$ )/(distance along axis), for instance $\angle AOP\approx h/(-u)$ , $\angle ACP\approx h/R$ , and $\angle AIP\approx h/v$ , where $h$ is the height of $A$ above the axis and the signs follow the Cartesian convention.

Step 4. Substitute into the paraxial Snell's law. Putting $i\approx h/(-u)+h/R$ and $r\approx h/R-h/v$ into $n_1 i=n_2 r$ gives $n_1\left(\dfrac{h}{-u}+\dfrac{h}{R}\right)=n_2\left(\dfrac{h}{R}-\dfrac{h}{v}\right)$ . The common factor $h$ cancels because it appears in every term — the result must hold for any nearby ray, not just this one.

Step 5. Rearrange to the final form. Expanding and collecting terms gives $-\dfrac{n_1}{u}+\dfrac{n_1}{R}=\dfrac{n_2}{R}-\dfrac{n_2}{v}$ , and moving all the $R$ -terms to one side yields $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ , the required relation, valid for this convex, rarer-to-denser, real-image case and, by the generality of the sign convention, for every other case of spherical-surface refraction as well.

Worked example

A convex spherical surface of radius of curvature $20\ \text{cm}$ separates air ( $n_1=1$ ) from glass ( $n_2=1.5$ ). A point object sits $60\ \text{cm}$ from the pole, in air, on the axis. Find the position of the image formed inside the glass.

Step 1. Assign signs. The object is on the incidence side, so $u=-60\ \text{cm}$ . The surface bulges toward the incoming light with its centre of curvature on the outgoing (glass) side, so $R=+20\ \text{cm}$ .

Step 2. Substitute into the derived relation. $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ becomes $\dfrac{1.5}{v}-\dfrac{1}{-60}=\dfrac{1.5-1}{20}$ , that is, $\dfrac{1.5}{v}+\dfrac{1}{60}=0.025$ .

Step 3. Solve for $v$ . $\dfrac{1.5}{v}=0.025-\dfrac{1}{60}=0.025-0.01667=0.00833$ , so $v=\dfrac{1.5}{0.00833}\approx180\ \text{cm}$ .

Answer: the image forms $180\ \text{cm}$ from the pole, inside the glass — a real image, since $v$ is positive and the light is travelling into the glass.

Competitive stretch

Now let the same relation handle a flat surface. A small air bubble sits $6\ \text{cm}$ below the flat top of a large glass block ( $n=1.5$ ) and is viewed from directly above, in air. Treating the flat surface as spherical with $R\to\infty$ : light travels from the bubble ( $n_1=1.5$ ) into air ( $n_2=1$ ), so $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}=0$ , giving $\dfrac{n_2}{v}=\dfrac{n_1}{u}$ . With $u=-6\ \text{cm}$ , $\dfrac{1}{v}=\dfrac{1.5}{-6}=-0.25$ , so $v=-4\ \text{cm}$ . The bubble appears $4\ \text{cm}$ below the surface — exactly matching apparent depth = real depth $/n=6/1.5=4\ \text{cm}$ , confirming the single-surface formula reduces correctly to the apparent-depth relation when $R\to\infty$ .

Common traps

Setting $R=0$ instead of $R\to\infty$ for a flat surface. A flat surface is a spherical surface of infinite radius, so it makes the right-hand side of the formula zero, not undefined.
Swapping $n_1$ and $n_2$ . $n_1$ is always the medium of the incident ray (object side); $n_2$ is the medium of the refracted ray (image side). Reversing them flips every sign in the answer.
Misjudging the sign of $R$ . $R$ is positive when the centre of curvature lies on the same side as the outgoing light, negative when it lies on the incidence side — easy to get backwards for a surface curving away from the object.
Assuming the formula only gives real images. It is fully general: plug in the geometry and a negative $v$ correctly signals a virtual image. The "convex surface, real image" scenario used in the derivation is just one case the formula can produce, not a restriction on its use.

Ask the tutor: Want to see the two-surface derivation that turns this relation into the full lens maker's formula, or try a concave-surface practice problem?

Why does the same formula work for both convex and concave surfaces without changing the derivation?

Because the derivation never assumes a particular shape beyond a spherical surface — it only uses the paraxial approximation and the signed distances $u$ , $v$ , $R$ measured by the Cartesian convention. A concave surface, or a case producing a virtual image, simply plugs in different signs for the same three quantities. The geometry that was worked out for the convex, real-image case is general enough that the algebra automatically accounts for every other configuration.

How is this formula connected to the lens maker's formula I will meet next?

A thin lens has two spherical (or one spherical, one flat) surfaces. Applying this single-surface relation once at the first surface, using the image it forms as the object for the second surface, and applying the relation again at the second surface, then combining the two equations and cancelling the intermediate image distance, produces the lens maker's formula. This topic is therefore not separate from lenses — it is the derivation lenses are built from.

Do I need to memorise a different formula for when the object is in the denser medium instead of the rarer one?

No. The relation $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ already has $n_1$ and $n_2$ as separate, assignable labels — $n_1$ is simply whichever medium contains the incident ray, denser or rarer. Relabel $n_1$ and $n_2$ for the new situation, keep the sign convention consistent, and the same equation applies directly.