Refraction by Lenses: The Lens Maker’s Formula and Thin Lens Equation

Two Surfaces, One Image: How a Lens Actually Works

In the previous topic, you learned what happens when light crosses a single curved boundary between two media. A real lens, though, has two curved surfaces. Light enters through one surface, travels through the lens material, and exits through the other. Each surface bends the light according to the single-surface refraction formula we already derived. The final image is the combined result of both refractions.

This topic builds directly on that foundation. By applying the single-surface formula twice (once at the front face and once at the back face) and using the thin lens approximation to stitch the two results together, you will arrive at two powerful equations: the lens maker’s formula, which connects the focal length of a lens to the curvature of its surfaces and the refractive index of the glass, and the thin lens equation, which relates the object distance, image distance, and focal length in any imaging situation.

Building the Lens Maker’s Formula Step by Step

The setup: a double convex lens

Fig 9.16: Image formation by a double convex lens viewed as refraction at two surfaces in sequence

Picture a double convex lens sitting in air. The lens material has a refractive index $n_2$ and the surrounding medium (air) has refractive index $n_1$. Let the two spherical surfaces have centres of curvature $C_1$ (for the front surface) and $C_2$ (for the back surface), with radii $R_1$ and $R_2$ respectively. An object O sits on the principal axis to the left of the lens.

The image formation happens in two stages:

• Stage 1: Light from the object hits the first surface (front face). This surface separates medium $n_1$ on the left from medium $n_2$ on the right. Applying the single-surface refraction formula here gives an intermediate image $I_1$.
• Stage 2: Before the rays actually reach $I_1$, they hit the second surface (back face). For this surface, the medium on the left is $n_2$ (the lens material) and the medium on the right is $n_1$ (air again). The intermediate image $I_1$ now serves as a virtual object for the second surface, which produces the final image I.

Step 1: Apply the single-surface formula at the first surface

The first surface has centre of curvature $C_1$ and separates medium $n_1$ (left) from medium $n_2$ (right). Using the single-surface refraction formula from the previous topic:

$\frac{n_1}{\text{OB}} + \frac{n_2}{\text{BI}_1} = \frac{n_2 - n_1}{\text{BC}_1} \qquad \text{(9.17)}$

Here, B is the point where the principal axis meets the first surface, OB is the object distance from B, $BI_1$ is the intermediate image distance from B, and $BC_1$ is the radius of curvature of the first surface.

Step 2: Apply the single-surface formula at the second surface

The second surface has centre of curvature $C_2$. Now the roles of the media flip: the medium on the left of this surface is $n_2$ (lens glass) and the medium on the right is $n_1$ (air). The intermediate image $I_1$ acts as the object for this surface, and D is the point where the axis meets the second surface.

Notice something important: $DI_1$ is measured against the direction of the incident light (because $I_1$ is on the same side as the incoming rays relative to this second surface), so it carries a negative sign. Applying the formula:

$-\frac{n_2}{\text{DI}_1} + \frac{n_1}{\text{DI}} = \frac{n_2 - n_1}{\text{DC}_2} \qquad \text{(9.18)}$

Step 3: Use the thin lens approximation to combine

Here is where the thin lens assumption enters. If the lens is thin enough that its thickness is negligible, then the two points B and D (where the axis meets the front and back surfaces) are essentially at the same location. This means:

$BI_1 = DI_1$

With this equality, we can simply add Equations (9.17) and (9.18). The $n_2/BI_1$ and $-n_2/DI_1$ terms cancel out, leaving:

$\frac{n_1}{\text{OB}} + \frac{n_1}{\text{DI}} = (n_2 - n_1)\left(\frac{1}{\text{BC}_1} + \frac{1}{\text{DC}_2}\right) \qquad \text{(9.19)}$

The intermediate image has disappeared from the equation. We now have a direct relationship between the original object distance, the final image distance, and the curvatures of the two surfaces.

Step 4: Find the focal length by placing the object at infinity

When an object is placed infinitely far away, the incoming rays are parallel to the principal axis, and the image forms at the focus of the lens. Setting $OB \to \infty$ (so $n_1/OB \to 0$) and $DI = f$ in Equation (9.19):

$\frac{n_1}{f} = (n_2 - n_1)\left(\frac{1}{\text{BC}_1} + \frac{1}{\text{DC}_2}\right) \qquad \text{(9.20)}$

The point where parallel rays converge (or appear to diverge from) is called the focus (focal point), and $f$ is the focal length of the lens. Every lens has two focal points, F and F’, one on each side. For a double convex or double concave lens, these two foci are equidistant from the optical centre (the central point of the lens).

Step 5: Apply the sign convention

Using the Cartesian sign convention (distances measured from the optical centre, positive in the direction of incident light):

• $BC_1 = +R_1$ (centre of curvature $C_1$ of the first surface is on the right, the positive direction)
• $DC_2 = -R_2$ (centre of curvature $C_2$ of the second surface is on the left, the negative direction)

Substituting into Equation (9.20) and writing $n_{21} = n_2/n_1$:

$\frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \qquad \text{(9.21)}$

This is the lens maker’s formula. It is the master equation that lens designers use: pick the glass (which fixes $n_{21}$), choose the surface curvatures $R_1$ and $R_2$, and the formula tells you the focal length you will get.

What the lens maker’s formula reveals

• Convex lenses have positive focal lengths. For a double convex lens, $R_1 > 0$ and $R_2 < 0$, so $1/R_1 - 1/R_2$ is positive. Since $(n_{21} - 1) > 0$ for glass in air, $f$ comes out positive. A positive focal length means the lens converges light.
• Concave lenses have negative focal lengths. For a double concave lens, $R_1 < 0$ and $R_2 > 0$, making $1/R_1 - 1/R_2$ negative. The focal length is therefore negative, meaning the lens diverges light.
• The formula works for any lens shape, not just double convex. Plano-convex, plano-concave, meniscus: just plug in the correct signs for $R_1$ and $R_2$.

Arriving at the Thin Lens Equation

Starting from Equation (9.19) and using the focal length relation from Equation (9.20):

$\frac{n_1}{\text{OB}} + \frac{n_1}{\text{DI}} = \frac{n_1}{f} \qquad \text{(9.22)}$

Divide through by $n_1$:

$\frac{1}{\text{OB}} + \frac{1}{\text{DI}} = \frac{1}{f}$

In the thin lens approximation, B and D both sit at the optical centre. Applying the sign convention ($BO = -u$, so $OB = -u$ and $1/OB = -1/u$; $DI = +v$):

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \qquad \text{(9.23)}$

This is the thin lens formula. Despite being derived for a particular case (a convex lens forming a real image), it is universal: it works for convex and concave lenses alike, and for both real and virtual images. The sign convention handles every case automatically.

Comparing with the mirror equation

Notice how similar this looks to the mirror equation $1/v + 1/u = 1/f$. The only difference is the sign in front of $1/u$. For mirrors it is a plus; for lenses it is a minus. This difference comes from the geometry: in a mirror, the reflected light goes back the way it came, while in a lens, it continues forward.

Focal Points: First and Second

A thin lens has two focal points, both at the same distance $f$ from the optical centre but on opposite sides:

• First focal point (F): This is on the same side as the incoming light. For a convex lens, a ray passing through F emerges parallel to the axis. For a concave lens, a ray directed toward F (which is on the far side) emerges parallel.
• Second focal point (F’): This is on the far side. When parallel rays enter the lens, they converge at F’ (convex) or appear to diverge from F’ (concave).

Drawing Ray Diagrams: Three Simple Rules

Fig 9.17: Ray diagrams for (a) convex lens and (b) concave lens

To locate an image graphically, pick any two of these three standard rays from a point on the object:

1. Parallel ray: A ray travelling parallel to the principal axis refracts through the second focal point F’ (convex lens) or appears to diverge from the first focal point F (concave lens).

2. Central ray: A ray passing through the optical centre of the lens goes straight through without bending. This is the easiest ray to draw.

3. Focal ray:

   • Convex lens: A ray passing through the first focal point F emerges parallel to the principal axis after refraction.
   • Concave lens: A ray heading toward the second focal point F’ (on the far side) emerges parallel to the principal axis.

Where any two of these rays meet (or appear to meet, when extended backward) is the image point. You can use these rules to find images for any object position relative to the lens.

A reminder about real objects

Every point on a real, extended object sends out countless rays in all directions. We trace only two or three convenient ones, but all the rays from that point converge at (or appear to come from) the same image point. This is what makes the image sharp and well-defined.

Magnification by a Lens

Just as with mirrors, the magnification $m$ tells you how the image size compares to the object size:

$m = \frac{h'}{h} = \frac{v}{u} \qquad \text{(9.24)}$

Here, $h$ is the object height and $h'$ is the image height. The sign of $m$ carries important information:

• $m > 0$: The image is erect (upright). For a single thin lens, an erect image is always virtual.
• $m < 0$: The image is inverted (flipped upside down). An inverted image formed by a single thin lens is always real.
• $|m| > 1$: The image is magnified (larger than the object).
• $|m| < 1$: The image is diminished (smaller than the object).

Worked Example: The Vanishing Lens Trick

Example 9.6: During a magic show, a magician makes a glass lens with refractive index $n = 1.47$ disappear by dipping it in a trough of liquid. What must the refractive index of the liquid be? Could the liquid be water?

Think about what makes a lens visible. A lens bends light because its refractive index differs from the medium around it. If you could match those two refractive indices perfectly, the light would pass through without bending at all, and the lens would become invisible.

Apply the lens maker’s formula:

$\frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

For the lens to have no focusing effect, $1/f$ must equal zero. The curvatures $R_1$ and $R_2$ are fixed (the lens shape does not change), so the only way to make the right side zero is to set:

$n_{21} - 1 = 0 \implies n_{21} = 1$

This means $n_2/n_1 = 1$, so $n_2 = n_1$. The refractive index of the liquid must equal the refractive index of the glass.

Answer the question: The liquid must have $n = 1.47$. With $n_{21} = 1$, the focal length becomes infinite. The lens acts like a flat, transparent slab and effectively vanishes.

Could the liquid be water? No. Water has a refractive index of about 1.33, which is well below 1.47. The liquid could, however, be glycerine, which has a refractive index close to 1.47.