Question

Two thin converging lenses of focal length $f_1$ and $f_2$ are placed coaxially in contact. Derive expression for the focal length of the combination.

Hint:

This specific derivation heavily relies on the lenses being thin enough so that their optical centers effectively coincide.
If the lenses are slightly separated by a distance $d$, the formula uniquely changes to $1/F = 1/f_1 + 1/f_2 - d/(f_1 f_2)$.

Answer 1

Concept:
• When two thin lenses are placed perfectly in contact, they act as a single equivalent lens.

• The image intelligently formed by the first lens serves as a virtual object for the second lens.

• The total power of the optical combination is simply the algebraic sum of the individual powers.

Step 1: Analyze the First Lens
Consider two thin converging lenses $L_1$ and $L_2$ of focal lengths $f_1$ and $f_2$ respectively, placed coaxially in contact.
Let a point object $O$ be placed strictly on the principal axis at a distance $u$ from the optical center of the combination.
The first lens $L_1$ attempts to form an image at a point $I_1$ at a distance $v_1$.
Using the thin lens formula for the first lens alone:
\[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \]

Step 2: Analyze the Second Lens
This intermediate image $I_1$ behaves precisely as a virtual object for the second lens $L_2$.
The second lens intercepts the converging rays and forms the final real image $I$ at a distance $v$.
Applying the thin lens formula for the second lens, treating $v_1$ as the object distance:
\[ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \]

Step 3: Combine the Mathematical Equations
We algebraically add the two equations obtained from Step 1 and
Step 2:
\[ \left( \frac{1}{v_1} - \frac{1}{u} \right) + \left( \frac{1}{v} - \frac{1}{v_1} \right) = \frac{1}{f_1} + \frac{1}{f_2} \]
The term containing the intermediate image distance cleanly cancels out:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 4: Determine the Equivalent Focal Length
If this entire two-lens system is replaced by a single equivalent lens of focal length $F$ that forms the image $I$ at distance $v$ for an object at distance $u$, its formula would be:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \]
Comparing this generalized equation with our derived combination equation yields:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
This comprehensively proves the relation for lenses in contact.