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(A) The system consists of two lenses with focal lengths $f = 10 \, cm$ separated by $70 \, cm$. The object is placed $20 \, cm$ in front of the first

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$20 \, cm$ from the left lens

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(A) The system consists of two lenses with focal lengths $f = 10 \, cm$ separated by $70 \, cm$. The object is placed $20 \, cm$ in front of the first lens.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for the first lens:$u = -20 \, cm, f = 10 \, cm$$\frac{1}{v} - \frac{1}{-20} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20} \implies v = 20 \, cm$.This means the first lens forms an image $20 \, cm$ behind it. This image acts as the object for the second lens.The distance between the lenses is $70 \, cm$,so the distance of this image from the second lens is $70 - 20 = 50 \, cm$.To increase the light-collecting efficiency,we place a third lens of the same focal length $(f = 10 \, cm)$ at the position of the intermediate image ($20 \, cm$ from the first lens). This lens converges the diverging rays from the first lens,preventing them from missing the second lens,without altering the final image position.Thus,the third lens should be placed $20 \, cm$ from the first (left) lens.

In the given optical instrument,how can a third lens of the same focal length be inserted to increase the light-collecting efficiency without changing the positions of the object and the image? Calculate the position of that lens.

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