Question

Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

Hint:

For lenses in contact: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \] Works like resistors in parallel — easy to remember!

Answer 1

Focal Length of Two Thin Lenses in Contact
When two thin lenses of focal lengths \( f_1 \) and \( f_2 \) are placed in contact coaxially, the focal length \( f \) of the combination can be derived using the lens formula.
Step 1: Lens Formula
The lens formula for a single lens is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \] where \( u \) is the object distance, \( v \) is the image distance, and \( f \) is the focal length of the lens.
Step 2: Consider Two Lenses in Contact
Let an object be placed at a distance \( u \) from the first lens (with focal length \( f_1 \)). Let the image formed by the first lens act as a virtual object for the second lens (with focal length \( f_2 \)). Since the lenses are in contact, the distance between the lenses is negligible.
- For the first lens:
\[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \implies \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u} \]
- For the second lens:
Let the final image be at distance \( v \) from the second lens. The object distance for the second lens is approximately \( u_2 = -v_1 \) (negative because it is on the opposite side of the lens). Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u_2} = \frac{1}{f_2} \implies \frac{1}{v} + \frac{1}{v_1} = \frac{1}{f_2} \]
Step 3: Combine the Two Lenses
From the first lens, \( \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u} \). Substitute into the second lens formula:
\[ \frac{1}{v} + \left(\frac{1}{f_1} + \frac{1}{u}\right) = \frac{1}{f_2} \implies \frac{1}{v} = \frac{1}{f_2} - \frac{1}{f_1} - \frac{1}{u} \]
Rewriting, we get the combined lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 4: Focal Length of Combination
By definition, the focal length \( f \) of the combination satisfies:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
Taking the reciprocal to get \( f \):
\[ f = \frac{f_1 f_2}{f_1 + f_2} \]
Conclusion:
Thus, the focal length of two thin lenses in contact is:
\[ \boxed{f = \frac{f_1 f_2}{f_1 + f_2}} \]
This shows that the combination acts as a single lens with a focal length determined by the harmonic sum of the individual focal lengths.

Answer 2

Step 1: Define optical power.
The power of a thin lens is \( P = \dfrac{1}{f} \), a measure of how strongly it converges (positive \( f \)) or diverges (negative \( f \)) parallel light falling on it.

Step 2: Powers of lenses placed in contact add.
Because the two lenses touch, the ray leaves the first lens and enters the second at essentially the same point, with no free space in between for the light to travel undeviated. The overall bending effect on the ray is therefore the combined converging (or diverging) effect of both lenses acting together at the same location: \[ P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 3: Write this as an equivalent single lens.
By definition, the combination itself behaves as one lens of power \( P = \dfrac{1}{f} \), so \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \implies f = \frac{f_1 f_2}{f_1 + f_2} \]

Step 4: Verify using a limiting case.
Take \( f_1 = f \) (a converging lens) and \( f_2 = -f \) (a diverging lens of equal magnitude) placed in contact. The formula gives \[ \frac{1}{f_{\text{eq}}} = \frac{1}{f} - \frac{1}{f} = 0 \implies f_{\text{eq}} \to \infty \] which correctly matches the physical expectation that equal and opposite lenses in contact cancel each other's focusing power completely, producing an afocal (zero-power) system, exactly as this power-addition rule should predict.

Conclusion:\[ f = \frac{f_1 f_2}{f_1 + f_2} \]