Combination of Thin Lenses in Contact

No single lens is perfect. A lone convex lens may converge light nicely, but it introduces colour fringes and imperfect focus. A single concave lens diverges light but cannot form a real image on its own. The real magic happens when you place lenses together. Every camera, every microscope eyepiece, and every telescope objective uses multiple lenses working as a team. So how do we figure out what a combination of lenses does? It turns out there is a beautifully simple rule.

Setting Up the Problem: Two Thin Lenses Touching

Fig 9.19: Image formation by a combination of two thin lenses in contact

Picture two thin lenses, A and B, with focal lengths $f_1$ and $f_2$, placed so they touch each other. An object O sits beyond the focus of the first lens A.

Since both lenses are thin and sit in contact, the gap between their optical centres is negligible. We treat both centres as sitting at the same point, call it P. This is the thin lens approximation at work: each lens has practically zero thickness, so stacking them does not introduce any meaningful separation.

Now think about what happens to the light step by step:

• Lens A alone would form an image at some point $I_1$.
• But light does not actually reach $I_1$ because it hits lens B first. Instead, lens B intercepts the converging rays and redirects them to produce the final image at a new point $I$.

An important subtlety: the image $I_1$ is a real image formed on the far side of both lenses. Since lens B sits between lens A and $I_1$, the rays heading toward $I_1$ are still converging when they hit B. From lens B’s perspective, these rays appear to come from a point beyond it. This makes $I_1$ a virtual object for lens B. We use the concept of $I_1$ only as a calculation tool to connect the two lens equations.

Deriving the Equivalent Focal Length

Let us write the thin lens equation for each lens separately. Object distance from P is $u$, intermediate image distance is $v_1$, and final image distance is $v$.

For lens A (object at $u$, image at $v_1$):

$\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \qquad \text{(9.27)}$

For lens B (object at $v_1$, image at $v$):

$\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \qquad \text{(9.28)}$

Now here is the clever step. Add these two equations together:

$\left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2}$

On the left side, $+\frac{1}{v_1}$ from the first equation and $-\frac{1}{v_1}$ from the second equation cancel each other perfectly:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \qquad \text{(9.29)}$

The intermediate image distance $v_1$ has vanished. We are left with a relationship that connects only the original object position $u$ and the final image position $v$ to the two focal lengths.

Now compare this with what a single lens of focal length $f$ would give:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Matching the two expressions:

$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \qquad \text{(9.30)}$

This is the equivalent focal length formula for two thin lenses in contact. The reciprocal of the combined focal length equals the sum of the reciprocals of the individual focal lengths. The two-lens system behaves exactly like a single lens whose focal length $f$ satisfies this equation.

Extending to Any Number of Lenses

The derivation above used two lenses, but the logic extends to any number. Each additional lens introduces one more thin lens equation, and each intermediate image distance cancels when the equations are added. For several thin lenses of focal lengths $f_1, f_2, f_3, ...$ all in contact:

$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + ... \qquad \text{(9.31)}$

Expressing the Result in Terms of Power

Recall that the power of a lens is $P = 1/f$ (with $f$ in metres). Since Eq. (9.31) already has reciprocals of focal lengths on both sides, converting to power is straightforward. Simply replace each $1/f$ with $P$:

$P = P_1 + P_2 + P_3 + ... \qquad \text{(9.32)}$

The net power of the combination is the algebraic sum of the individual powers. “Algebraic” is the key word here: convex lenses contribute positive values and concave lenses contribute negative values. Whether the combination as a whole converges or diverges light depends on which side wins.

For example, combining a $+5$ D convex lens with a $-3$ D concave lens gives $P = +5 + (-3) = +2$ D, a mildly converging system. But combining $+3$ D with $-5$ D gives $P = -2$ D, a net diverging system.

Total Magnification: A Product, Not a Sum

When several lenses sit in contact, the image formed by the first lens becomes the object for the second, and so on down the chain. Each lens introduces its own magnification. The overall magnification $m$ of the entire system is the product of the individual magnifications:

$m = m_1 \, m_2 \, m_3 \; ... \qquad \text{(9.33)}$

Notice that magnification multiplies rather than adds. This is because each lens scales the image produced by the previous lens. If the first lens doubles the image height and the second lens triples it, the combined system magnifies by a factor of $2 \times 3 = 6$, not $2 + 3 = 5$.

Why Combine Lenses at All?

Combining lenses gives optical designers two powerful advantages:

• Custom focal lengths and magnifications: By choosing suitable combinations of converging and diverging lenses, any desired effective focal length (and hence any magnification) can be achieved. This flexibility is essential for cameras, microscopes, and telescopes.
• Sharper images: A single lens always introduces some aberrations (distortions). A carefully chosen combination of lenses with different curvatures and glass types can cancel out these aberrations, producing much sharper and cleaner images.

Worked Example: A Three-Lens System (Example 9.8)

Fig 9.20: Three-lens system with focal lengths $+10$ cm, $-10$ cm, and $+30$ cm

Problem: An object O is placed 30 cm to the left of a three-lens system. The lenses have focal lengths $f_1 = +10$ cm, $f_2 = -10$ cm, and $f_3 = +30$ cm. The separation between the first and second lens is 5 cm, and between the second and third is 10 cm. Find the position of the final image.

Note: These lenses are not in contact (they are separated), so we cannot simply use $1/f = 1/f_1 + 1/f_2 + 1/f_3$. Instead, we trace the image step by step through each lens.

Step 1: Image formed by the first lens ($f_1 = +10$ cm)

The object is 30 cm to the left of the first lens, so $u_1 = -30$ cm. Apply the thin lens equation:

$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$

$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10}$

$\frac{1}{v_1} + \frac{1}{30} = \frac{1}{10}$

$\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$

$v_1 = +15 \text{ cm}$

The first lens forms a real image 15 cm to its right.

Step 2: Image formed by the second lens ($f_2 = -10$ cm)

The second lens sits 5 cm to the right of the first lens. The image from step 1 is at 15 cm from the first lens, which puts it at $15 - 5 = 10$ cm to the right of the second lens.

Although this image is real, the light rays are converging toward a point that lies on the far side of lens 2. So the image from lens 1 serves as a virtual object for lens 2, with $u_2 = +10$ cm (positive because it is on the same side as where outgoing light goes):

$\frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10}$

$\frac{1}{v_2} = \frac{1}{-10} + \frac{1}{10} = 0$

$v_2 = \infty$

The second lens produces an image at infinity. In practical terms, the light emerges from lens 2 as a parallel beam, neither converging nor diverging.

Step 3: Image formed by the third lens ($f_3 = +30$ cm)

The third lens is 10 cm to the right of the second lens. Since the light from lens 2 is a parallel beam, the object for lens 3 is effectively at infinity ($u_3 = \infty$):

$\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30}$

Since $1/\infty = 0$:

$\frac{1}{v_3} = \frac{1}{30}$

$v_3 = +30 \text{ cm}$

Result: The final image is formed 30 cm to the right of the third lens. A parallel beam entering a converging lens always comes to focus at the focal point, which is exactly what happened here.