Новое решение
en/13.3.21.md
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| + | ### Statement | ||
| + | |||
| + | $13.3.21.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + | In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | ||
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| + | \[u=\frac{v-f}{vf}=\frac{-15-30}{-15\cdot30}=-10.\] | ||
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| + | Therefore, the object has to be closer than $10$ cm from the first lens. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.21.$ [Insert the problem statement] | |||
| ### Solution | |||
| In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | |||
| \[u=\frac{v-f}{vf}=\frac{-15-30}{-15\cdot30}=-10.\] | |||
| Therefore, the object has to be closer than $10$ cm from the first lens. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||