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.