Derive an expression for an equivalent focal length of two thin lenses kept in contact.

To derive the expression for the equivalent focal length of two thin lenses kept in contact, let's break down the problem step by step.
Concept:
When two thin lenses are in contact, the focal lengths of the two lenses combine in such a way that their combined or equivalent focal length feqf_{\text{eq}} can be determined using the lens formula. This is based on the fact that both lenses contribute to the total optical power of the system.
Step 1: Lens Formula for a Single Lens
The lens formula for a single thin lens is:
1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
Where:
• ff is the focal length of the lens,
• vv is the image distance (distance from the lens to the image),
• uu is the object distance (distance from the lens to the object).
For thin lenses in contact, the object for the second lens is formed by the first lens, so the image formed by the first lens serves as the object for the second lens.
Step 2: Additive Nature of Optical Powers
When two lenses are kept in contact, the image formed by the first lens is used as the object for the second lens. The total or equivalent focal length feqf_{\text{eq}} is determined by the combined optical power of both lenses.
The optical power PP of a lens is given by:
P=1fP = \frac{1}{f}
For two lenses in contact, the total optical power PtotalP_{\text{total}} is the sum of the individual powers of the two lenses:
Ptotal=P1+P2P_{\text{total}} = P_1 + P_2
Where:
• P1=1f1P_1 = \frac{1}{f_1} is the power of the first lens with focal length f1f_1,
• P2=1f2P_2 = \frac{1}{f_2} is the power of the second lens with focal length f2f_2.
Thus:
Ptotal=1f1+1f2P_{\text{total}} = \frac{1}{f_1} + \frac{1}{f_2}
Step 3: Equivalent Focal Length
The total optical power of the system is also related to the equivalent focal length feqf_{\text{eq}}:
Ptotal=1feqP_{\text{total}} = \frac{1}{f_{\text{eq}}}
Equating the two expressions for the total power:
1feq=1f1+1f2\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2}
Step 4: Solve for feqf_{\text{eq}}
Rearranging the equation to solve for the equivalent focal length feqf_{\text{eq}}:
feq=11f1+1f2f_{\text{eq}} = \frac{1}{\frac{1}{f_1} + \frac{1}{f_2}}
Final Expression
The equivalent focal length feqf_{\text{eq}} of two thin lenses kept in contact is:
feq=f1f2f1+f2f_{\text{eq}} = \frac{f_1 f_2}{f_1 + f_2}
Conclusion:
Thus, the equivalent focal length of two thin lenses kept in contact is the harmonic mean of the individual focal lengths of the lenses.