Physics · Ch 6 — Optics

Focal Length of Lenses in Contact

6.6.6

Focal Length of Lenses in Contact

For two thin lenses of focal length $f_1,f_2$ placed in contact at a common optical centre, writing the lens equation once for the first lens ( $\frac{1}{v'}-\frac{1}{u}=\frac{1}{f_1}$ ) and once for the second, using the first lens's image as the second lens's object ( $\frac{1}{v}-\frac{1}{v'}=\frac{1}{f_2}$ ), and adding the two equations together eliminates the intermediate image distance $v'$ , giving $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$ . Comparing this with the lens equation for a single equivalent lens of focal length $F$ , $\frac{1}{v}-\frac{1}{u}=\frac{1}{F}$ , gives $\boxed{\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}}$ , extendable to any number of lenses in contact as $\dfrac{1}{F}=\sum\dfrac{1}{f_i}$ , or equivalently as an algebraic sum of powers $P=P_1+P_2+\ldots$ (positive terms for convex lenses, negative …

Figure 6.37Lenses in contact

What this figure shows. Two thin lenses, labelled 1 and 2, share a common optical centre P and a common principal axis. An object at O forms an intermediate image at I' through the first lens alone, and this intermediate image then immediately acts as the object for the second lens, which forms the true final image at I -- writing the lens equation once for each lens using this intermediate image, then adding the two equations together, is exactly how the comb …

Misc Example 6.17Combined focal length and power of two lenses in contact

Worked out. A lens of focal length -70 cm (diverging) is placed in contact with a lens of focal length 150 cm (converging). Using 1/F = 1/f1 + 1/f2 = 1/(-70) + 1/150 = (-150+70)/10500 = -80/10500 gives F = -10500/80 = -131.25 cm. Since F is negative, the combination overall behaves as a diverging system of lenses, even though one of the two individual lenses was converging. The combined power is P = 1/F = 1/(-1.3125 m) = -0.76 dioptre, negative as expected for a net-diverging c …

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