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 \[ f=\frac{f_1f_2}{f_1+f_2}. \]

Hint:

For lenses in contact: \[ P=P_1+P_2 \] or \[ \frac1f=\frac1{f_1}+\frac1{f_2}. \] Always remember that powers add directly whereas focal lengths do not.

Answer 1

Concept: When two thin lenses are placed in contact, the image formed by the first lens acts as the object for the second lens. The net effect of the two lenses can be represented by a single equivalent lens whose focal length is called the equivalent focal length. The power of a lens is defined as \[ P=\frac{1}{f}. \] For lenses in contact, the powers add algebraically.

Step 1: Write the power of each lens. For the first lens, \[ P_1=\frac{1}{f_1}. \] For the second lens, \[ P_2=\frac{1}{f_2}. \]

Step 2: Use the law of addition of powers. When the lenses are placed in contact, \[ P=P_1+P_2. \] Substituting the expressions, \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}. \]

Step 3: Take the LCM and simplify. \[ \frac{1}{f} = \frac{f_2+f_1}{f_1f_2}. \] Therefore, \[ \frac{1}{f} = \frac{f_1+f_2}{f_1f_2}. \] Taking reciprocal on both sides, \[ f = \frac{f_1f_2}{f_1+f_2}. \]

Step 4: State the required result. Hence, the focal length of the combination of two thin lenses in contact is \[ \boxed{ f=\frac{f_1f_2}{f_1+f_2} }. \] Thus proved.