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

Concept: For thin lenses in contact:

Image formed by first lens acts as object for second lens
Use lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

Step 1: Apply lens formula to first lens. Let object distance = \( u \), image formed by first lens = \( v_1 \). \[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \quad \cdots (1) \]
Step 2: Second lens. Since lenses are in contact, image of first lens becomes object for second lens. So object distance for second lens = \( v_1 \). Let final image distance = \( v \). \[ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \quad \cdots (2) \]
Step 3: Add equations (1) and (2). \[ \left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2} \] Cancel \( \frac{1}{v_1} \): \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 4: Equivalent focal length. For equivalent single lens: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Comparing: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 5: Final expression. \[ \frac{1}{f} = \frac{f_1 + f_2}{f_1 f_2} \] Taking reciprocal: \[ f = \frac{f_1 f_2}{f_1 + f_2} \]