Question

Establish the formula for focal length of combination of two thin lenses placed in contact.

Hint:

This formula can be extended to any number of thin lenses in contact: \( \frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + ..... \). Remember to use the correct sign convention for the focal lengths of concave (negative) and convex (positive) lenses.

Answer 1

Step 1: Concept and Diagram: 
Consider two thin convex lenses L\(_1\) and L\(_2\) with focal lengths \(f_1\) and \(f_2\) respectively, placed in contact with each other. We want to find the focal length, F, of the combination, which behaves like a single equivalent lens. 

An object O is placed at a distance u from the optical center of the combination. The first lens L\(_1\) forms a real image I' at a distance v'. This image I' acts as a virtual object for the second lens L\(_2\), which forms the final real image I at a distance v. 
Step 2: Derivation using the Lens Formula: 
We apply the thin lens formula, \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), to each lens separately. 
For the first lens L\(_1\): 
Object distance = u 
Image distance = v' 
The lens formula is: \[ \frac{1}{v'} - \frac{1}{u} = \frac{1}{f_1} \quad .s (1) \] For the second lens L\(_2\): 
The image I' formed by L\(_1\) acts as the object for L\(_2\). Since the lenses are thin and in contact, the object distance for L\(_2\) is v'. 
Object distance = v' 
Final image distance = v 
The lens formula is: \[ \frac{1}{v} - \frac{1}{v'} = \frac{1}{f_2} \quad .s (2) \] Step 3: Combining the Equations: 
Now, we add equation (1) and equation (2): \[ \left( \frac{1}{v'} - \frac{1}{u} \right) + \left( \frac{1}{v} - \frac{1}{v'} \right) = \frac{1}{f_1} + \frac{1}{f_2} \] The terms \( \frac{1}{v'} \) and \( -\frac{1}{v'} \) cancel out: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \quad .s (3) \] Now, if we replace the combination of two lenses with a single equivalent lens of focal length F, it should form the same image I at distance v for the same object O at distance u. The lens formula for this equivalent lens would be: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad .s (4) \] Step 4: Final Formula: 
Comparing equation (3) and equation (4), we get the formula for the focal length of the combination: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] The power of the combination is the sum of the individual powers: \( P = P_1 + P_2 \).