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:

When adding the equations for thin lenses in contact, the intermediate image term $\frac{1}{v_1}$ always cancels out nicely. This principle scales infinitely: for $n$ lenses in contact, the net focal length is simply $\frac{1}{F} = \sum \frac{1}{f_i}$.

Answer 1

Concept: When multiple thin lenses are placed coaxially in direct contact with one another, the total optical deviation produced by the combined system is equal to the algebraic sum of individual deviations produced by each single lens. In terms of image formation, the first lens creates an intermediate image of the object. This intermediate image then serves as a virtual object for the second lens, which subsequently produces the final image of the system. By applying the standard thin lens formula successively to each lens while neglecting the extremely small thickness of the lenses in contact, we can find the equivalent focal length ($F$) of the combined lens setup.

Step 1: Define the setup and analyze refraction through the first lens. Consider two thin converging lenses $L_1$ and $L_2$ having focal lengths $f_1$ and $f_2$ respectively, placed coaxially in contact with each other. Let $O$ be a point object placed on the common principal axis at a distance $u$ in front of the first lens $L_1$. In the absence of the second lens $L_2$, the lens $L_1$ would form a real image $I_1$ at a distance $v_1$ from its optical center. Applying the thin lens formula for the first lens $L_1$: \[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \quad \cdots (1) \]

Step 2: Analyze refraction through the second lens. In reality, the second lens $L_2$ intercepts the rays of light before they can converge at $I_1$. Therefore, the intermediate image $I_1$ acts as a virtual object for the second lens $L_2$. The second lens further deviates the light rays to form a final real image $I$ at a distance $v$ from the optical center of the combination. Since the lenses are thin and in close contact, their optical centers are treated as a single point. Applying the thin lens formula for the second lens $L_2$: \[ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \quad \cdots (2) \] Note that the object distance for the second lens is $+v_1$ (measured in the direction of light, as it is a virtual object), and the image distance is $v$.

Step 3: Combine equations to eliminate intermediate variables. To find the collective behavior of the system, add equation (1) and equation (2) together: \[ \left( \frac{1}{v_1} - \frac{1}{u} \right) + \left( \frac{1}{v} - \frac{1}{v_1} \right) = \frac{1}{f_1} + \frac{1}{f_2} \] Observe that the term $\frac{1}{v_1}$ appears with opposite signs and completely cancels out from the left-hand side: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \quad \cdots (3) \]

Step 4: Establish equivalent focal length equation. If we replace this pair of lenses with a single equivalent lens of focal length $F$ that creates a final image at the exact same position $v$ for an object placed at distance $u$, its standard expression via the lens formula must satisfy: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad \cdots (4) \] Comparing equation (3) and equation (4) directly, we equate the right-hand sides: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] We can also express this in terms of net optical power ($P = \frac{1}{f}$): \[ P = P_1 + P_2 \] Solving explicitly for $F$ by finding a common denominator: \[ \frac{1}{F} = \frac{f_2 + f_1}{f_1 f_2} \quad \Rightarrow \quad F = \frac{f_1 f_2}{f_1 + f_2} \] This completes the formal analytical derivation for the combination of two thin lenses in contact.