Edits to “Statement”, “Solution”, “Answer”
en/13.3.13.md
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| ### Statement | |||
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| − | $13.3.13.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
| $13.3.13.$ Is a person who sees normally in water nearsighted or farsighted? | |||
| ### Solution | |||
| In a person with normal vision, the eye focuses light exactly on the retina thanks to the cornea and the lens. The refractive index of the cornea ($n_{\text{cornea}} \approx 1.376$) and the lens ($n_{\text{lens}} \approx 1.386 - 1.406$) provide the necessary refractive power ($\approx 60 \text{ D}$), most of which ($\approx 40 \text{ D}$) is produced by the cornea due to the significant difference in the refractive indices of air ($n_{\text{air}} \approx 1$) and the cornea. | |||
| Water has a refractive index of $n_{\text{water}} \approx 1.33$, which is very close to that of the cornea. Because of this, the cornea almost completely stops refracting light in water (since the difference $n_{\text{cornea}} - n_{\text{water}}$ is very small). The main focusing is now provided solely by the lens, but its optical power ($\approx 20 \text{ D}$) is insufficient to focus light on the retina. As a result, a normal eye in water becomes hypermetropic (farsighted) — the image is focused behind the retina. | |||
| − | If a person sees clearly in water, it means their eye somehow compensates for this lack of corneal refraction. This is possible if, in the air, their eye excessively refracts light (due to an overly strong cornea or an elongated eyeball), meaning the person is | ||
| + | If a person sees clearly in water, it means their eye somehow compensates for this lack of corneal refraction. This is possible if, in the air, their eye excessively refracts light (due to an overly strong cornea or an elongated eyeball), meaning the person is $nearsighted$ (myopic). In water, their "extra" refractive power compensates for the loss of corneal refraction, and the image falls exactly on the retina again. | ||
| #### Answer | |||
| Nearsighted | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,22 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.13.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $13.3.13.$ Is a person who sees normally in water nearsighted or farsighted? | $13.3.13.$ Is a person who sees normally in water nearsighted or farsighted? | ||
| ### Solution | ### Solution | ||
| In a person with normal vision, the eye focuses light exactly on the retina thanks to the cornea and the lens. The refractive index of the cornea ($n_{\text{cornea}} \approx 1.376$) and the lens ($n_{\text{lens}} \approx 1.386 - 1.406$) provide the necessary refractive power ($\approx 60 \text{ D}$), most of which ($\approx 40 \text{ D}$) is produced by the cornea due to the significant difference in the refractive indices of air ($n_{\text{air}} \approx 1$) and the cornea. | In a person with normal vision, the eye focuses light exactly on the retina thanks to the cornea and the lens. The refractive index of the cornea ($n_{\text{cornea}} \approx 1.376$) and the lens ($n_{\text{lens}} \approx 1.386 - 1.406$) provide the necessary refractive power ($\approx 60 \text{ D}$), most of which ($\approx 40 \text{ D}$) is produced by the cornea due to the significant difference in the refractive indices of air ($n_{\text{air}} \approx 1$) and the cornea. | ||
| Water has a refractive index of $n_{\text{water}} \approx 1.33$, which is very close to that of the cornea. Because of this, the cornea almost completely stops refracting light in water (since the difference $n_{\text{cornea}} - n_{\text{water}}$ is very small). The main focusing is now provided solely by the lens, but its optical power ($\approx 20 \text{ D}$) is insufficient to focus light on the retina. As a result, a normal eye in water becomes hypermetropic (farsighted) — the image is focused behind the retina. | Water has a refractive index of $n_{\text{water}} \approx 1.33$, which is very close to that of the cornea. Because of this, the cornea almost completely stops refracting light in water (since the difference $n_{\text{cornea}} - n_{\text{water}}$ is very small). The main focusing is now provided solely by the lens, but its optical power ($\approx 20 \text{ D}$) is insufficient to focus light on the retina. As a result, a normal eye in water becomes hypermetropic (farsighted) — the image is focused behind the retina. | ||
| If a person sees clearly in water, it means their eye somehow compensates for this lack of corneal refraction. This is possible if, in the air, their eye excessively refracts light (due to an overly strong cornea or an elongated eyeball), meaning the person is |
If a person sees clearly in water, it means their eye somehow compensates for this lack of corneal refraction. This is possible if, in the air, their eye excessively refracts light (due to an overly strong cornea or an elongated eyeball), meaning the person is $nearsighted$ (myopic). In water, their "extra" refractive power compensates for the loss of corneal refraction, and the image falls exactly on the retina again. | ||
| #### Answer | #### Answer | ||
| Nearsighted | Nearsighted | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||