Refraction at Spherical Surfaces

From Flat Boundaries to Curved Ones

So far, everything we have studied about refraction dealt with flat surfaces: light crossing a plane boundary, bending according to Snell’s law, and producing effects like apparent depth and lateral shift. But most practical optical devices (lenses, eyepieces, camera objectives) use curved surfaces, not flat ones. What happens when light meets a spherical boundary between two transparent media?

The good news is that the physics does not change. Any tiny patch on a spherical surface is essentially flat, so Snell’s law applies at every point of incidence, exactly as before. The normal at each point is simply the radius drawn to that point, passing through the centre of curvature (the centre of the sphere of which the surface is a part). The challenge is purely geometric: because the surface is curved, different rays hit it at different angles, and we need a single formula that relates the object position, image position, refractive indices, and the curvature of the boundary.

This topic derives that formula. Once you have it, applying it twice (once at each surface of a lens) will give you the lens maker’s formula and then the familiar lens equation, which we will explore in the topics that follow.

Setting Up the Geometry

Fig 9.15: Refraction at a spherical surface separating two media

Consider a spherical surface separating a medium of refractive index $n_1$ (on the left, where the object sits) from a medium of refractive index $n_2$ (on the right, where the refracted light goes). The surface has its centre of curvature at the point C and a radius of curvature $R$.

An object O is placed on the principal axis in medium $n_1$. A ray from O hits the spherical surface at a point N. At N, it refracts into medium $n_2$ and (after extending the refracted ray, if needed) meets the axis at the image point I.

Drop a perpendicular from N onto the principal axis. The foot of that perpendicular lands at a point M. The length MN is the height of the point of incidence above the axis.

We need one important simplification: the aperture (the lateral opening) of the spherical surface is small compared to the object distance, image distance, and radius of curvature. This is the same paraxial (close to the axis) condition we used for spherical mirrors. It means all angles involved are small enough that:

$\sin\theta \approx \tan\theta \approx \theta \quad (\text{in radians})$

Under this condition, the small segment MN is practically equal to the perpendicular distance from N to the axis.

Deriving the Refraction Formula

Step 1: Express the angles using geometry

With the small angle approximation, the tangent of each angle is simply the ratio MN divided by the distance along the axis from M to the relevant point:

$\tan(\angle NOM) = \frac{MN}{OM}, \quad \tan(\angle NCM) = \frac{MN}{MC}, \quad \tan(\angle NIM) = \frac{MN}{MI}$

Since all angles are small, replace each tangent by the angle itself (in radians):

$\angle NOM \approx \frac{MN}{OM}, \quad \angle NCM \approx \frac{MN}{MC}, \quad \angle NIM \approx \frac{MN}{MI}$

Step 2: Relate the angle of incidence to these angles

Look at triangle NOC (formed by the object O, the point of incidence N, and the centre of curvature C). The angle of incidence $i$ at vertex N is an exterior angle of this triangle. By the exterior angle theorem, an exterior angle equals the sum of the two non-adjacent interior angles:

$i = \angle NOM + \angle NCM$

Substituting the small-angle expressions:

$i = \frac{MN}{OM} + \frac{MN}{MC} \qquad \text{(9.13)}$

Step 3: Relate the angle of refraction to these angles

Now look at triangle NCI (formed by N, C, and the image point I). Here, $\angle NIM$ is the exterior angle at vertex I, and the two non-adjacent interior angles are $r$ (the angle of refraction at N) and $\angle NCI$. So:

$\angle NIM = r + \angle NCI$

Since $\angle NCI$ is the same as $\angle NCM$ in this configuration, rearranging gives:

$r = \angle NCM - \angle NIM$

Substituting the small-angle expressions:

$r = \frac{MN}{MC} - \frac{MN}{MI} \qquad \text{(9.14)}$

Step 4: Apply Snell’s law in the small-angle form

Snell’s law at the point of incidence states:

$n_1 \sin i = n_2 \sin r$

For small angles, $\sin i \approx i$ and $\sin r \approx r$, so this simplifies to:

$n_1\, i = n_2\, r$

Step 5: Substitute and simplify

Replace $i$ from Equation (9.13) and $r$ from Equation (9.14):

$n_1 \left(\frac{MN}{OM} + \frac{MN}{MC}\right) = n_2 \left(\frac{MN}{MC} - \frac{MN}{MI}\right)$

The factor MN appears in every term, so divide through by MN:

$\frac{n_1}{OM} + \frac{n_1}{MC} = \frac{n_2}{MC} - \frac{n_2}{MI}$

Rearrange to collect the OM and MI terms on the left and the MC terms on the right:

$\frac{n_1}{OM} + \frac{n_2}{MI} = \frac{n_2}{MC} - \frac{n_1}{MC} = \frac{n_2 - n_1}{MC}$

This gives us:

$\frac{n_1}{OM} + \frac{n_2}{MI} = \frac{n_2 - n_1}{MC} \qquad \text{(9.15)}$

At this stage, OM, MI, and MC are magnitudes (positive lengths). We now bring in the sign convention to assign directions.

Step 6: Apply the Cartesian sign convention

Following the same convention used for mirrors:

• All distances are measured from the pole of the spherical surface (the point where the axis meets the surface).
• The direction of the incident light (left to right) is taken as positive.

For a real object on the left:

• The object distance OM is measured opposite to the light direction, so OM $= -u$ (meaning $u$ is negative for a real object).
• The image distance MI is measured in the light direction, so MI $= +v$.
• The radius MC is measured in the light direction (for the case shown, where C is on the transmission side), so MC $= +R$.

Substituting these into Equation (9.15):

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

Rearranging:

$\boxed{\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}} \qquad \text{(9.16)}$

What this formula tells you

Equation (9.16) connects four quantities: the object distance $u$, the image distance $v$, the refractive indices $n_1$ and $n_2$ of the two media, and the radius of curvature $R$ of the spherical boundary. Given any three, you can find the fourth.

A few important points:

• The formula holds for any curved spherical surface, whether convex or concave. The sign of $R$ automatically handles the geometry.
• It reduces to a simpler form for flat surfaces: as $R \to \infty$ (a plane surface), the right side goes to zero, and the equation becomes $n_2/v = n_1/u$, which gives the apparent depth relation.
• This single-surface formula is the building block for deriving the lens maker’s formula. A lens has two spherical surfaces; applying Equation (9.16) at each surface in turn and combining the results leads directly to the lens equation.

What Comes Next: From One Surface to a Full Lens

A thin lens is a transparent optical medium bounded by two surfaces, at least one of which is spherical. The idea is straightforward: light first refracts at the front surface (use Equation 9.16 once), then continues through the lens material and refracts again at the back surface (use Equation 9.16 a second time). Combining these two applications gives the lens maker’s formula, and from there the standard lens formula $1/v - 1/u = 1/f$. We will carry out that derivation in the next topic.

Worked Example: Image by a Glass Spherical Surface

Example 9.5: Light from a point source in air falls on a spherical glass surface ($n = 1.5$, radius of curvature $= 20$ cm). The light source is 100 cm from the glass surface. Where does the image form?

Identify the given quantities:

• The light is in air, so $n_1 = 1$ (refractive index of air).
• The glass has $n_2 = 1.5$.
• The object is to the left of the surface, so by sign convention, $u = -100$ cm.
• The centre of curvature is on the glass (transmission) side, so $R = +20$ cm.
• We need to find $v$ (the image distance).

Apply the formula:

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

Substitute the values:

$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$

The second term on the left simplifies ($-1/-100 = +1/100$):

$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$

Compute the right side:

$\frac{0.5}{20} = \frac{1}{40} = 0.025$

Isolate the term with $v$:

$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}$

Find a common denominator (200) for the right side:

$\frac{1}{40} = \frac{5}{200}, \quad \frac{1}{100} = \frac{2}{200}$

$\frac{1.5}{v} = \frac{5}{200} - \frac{2}{200} = \frac{3}{200}$

Solve for $v$:

$v = \frac{1.5 \times 200}{3} = \frac{300}{3} = 100 \text{ cm}$

Interpret the result:

$v = +100$ cm. The positive sign tells us the image forms on the same side as the refracted light (inside the glass), 100 cm from the spherical surface in the direction the light is travelling. Because the refracted rays actually converge at this point, the image is real.