Question

Two positive lenses with focal lengths of 0.30 m and 0.50 m are separated by a distance of 0.20 m. A small butterfly rests on the central axis 0.50 m in front of the first lens. Locate the resulting image with respect to the second lens.

Solution

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In this exercise we have a system of two lenses, so let's firstly write information we know

f1=30cmf2=50cmd=20cms01=50cm\begin{align*} f_1&=30 \, \, \text{cm}\\ f_2&=50 \, \, \text{cm}\\ d&=20 \, \, \text{cm}\\ s_{01}&=50 \, \, \text{cm} \end{align*}

One of the ways to solve this problem is to pretend as if the second lens does not exist and find the image for the first lens. 1s01+1si1=1f1\dfrac{1}{s_{01}}+\dfrac{1}{s_{i1}}=\dfrac{1}{f_1} from which we find that the image distance from the first lens is si1=75cms_{i1}=75 \, \, \text{cm} This image is now object for the second lens, and the object distance for the second lens is so2=dsi1=55cms_{o2}=d-s_{i1}=-55 \, \, \text{cm} Now we can use that to compute final image of the system 1so2+1si2=1f2\dfrac{1}{s_{o2}}+\dfrac{1}{s_{i2}}=\dfrac{1}{f_2} so that finally

si2=f2so2so2f2=26.19cm\boxed{s_{i2}=\dfrac{f_2 s_{o2}}{s_{o2}-f_2}=26.19 \, \, \text{cm}}

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