What is an equivalent lens? Obtain an expression for focal length of a combination of two thin lenses placed in contact.

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> [!TLDR]
> For two thin lenses in contact, the combination behaves as a single 'equivalent lens' whose focal length satisfies $1/F=1/f_1+1/f_2$ -- i.e. powers simply add.
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> **Equivalent lens.**
> When two (or more) lenses are combined, the entire system can be replaced, for the purpose of image formation, by a single lens that produces exactly the same final image (same position and size) as the combination. This single lens is called the **equivalent lens**, and its focal length the **equivalent focal length**.
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> **Derivation for two thin lenses in contact.**
> Let two thin lenses $L_1$ (focal length $f_1$) and $L_2$ (focal length $f_2$) be placed coaxially in contact (negligible separation). An object O is placed on the axis at distance $u$ from the combination.
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> **Step 1 -- Lens $L_1$ forms an intermediate image $I_1$.**
> Using the lens formula for $L_1$ alone (image at $v_1$):
> $$\dfrac{1}{v_1} - \dfrac{1}{u} = \dfrac{1}{f_1} \qquad 	ext{...(i)}$$
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> **Step 2 -- This image $I_1$ acts as the object for lens $L_2$.**
> Since the lenses are in contact, the object distance for $L_2$ is the same $v_1$ (the image distance from $L_1$). Let the final image form at distance $v$ from the combination:
> $$\dfrac{1}{v} - \dfrac{1}{v_1} = \dfrac{1}{f_2} \qquad 	ext{...(ii)}$$
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> **Step 3 -- Add equations (i) and (ii).**
> $$\dfrac{1}{v_1}-\dfrac1u + \dfrac1v - \dfrac1{v_1} = \dfrac1{f_1}+\dfrac1{f_2}$$
> $$\dfrac1v - \dfrac1u = \dfrac1{f_1}+\dfrac1{f_2} \qquad	ext{...(iii)}$$
> …

Q.What is an equivalent lens? Obtain an expression for focal length of a combination of two thin lenses placed in contact.

Concept understanding — Power of a Lens and Combination of Lenses

Q.What is an equivalent lens? Obtain an expression for focal length of a combination of two thin lenses placed in contact.

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