## Related questions with answers

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

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

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

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. $\dfrac{1}{s_{01}}+\dfrac{1}{s_{i1}}=\dfrac{1}{f_1}$ from which we find that the image distance from the first lens is $s_{i1}=75 \, \, \text{cm}$ This image is now object for the second lens, and the object distance for the second lens is $s_{o2}=d-s_{i1}=-55 \, \, \text{cm}$ Now we can use that to compute final image of the system $\dfrac{1}{s_{o2}}+\dfrac{1}{s_{i2}}=\dfrac{1}{f_2}$ so that finally

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

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