Edits to “Solution”, “Answer”
en/13.4.25.md
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| − | ### Statement | ||
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| − | $13.4.25.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| ### Problem | |||
| $13.4.25^*.$ In an optical communication system, a laser beam transmitting information has the shape of a cone with a vertex angle of $10^{-4}$ rad (divergence angle). At the receiving device, the light energy is focused on a photocell using a $1$ m lens. It turned out that when the distance between the transmitter and receiver changed from $5$ to $10$ km, the signal received from the photocell decreased by a factor of two (due to light absorption in the atmosphere). By what factor will the signal change when the distance increases from $10$ to $20$ km? | |||
| ### Solution | |||
| Let's determine the diameter of the laser beam at different distances. | |||
| At a distance $r_1 = 5$ km, the diameter is $d_1 = \alpha r_1 = 10^{-4} \cdot 5000 = 0.5$ m. | |||
| At a distance $r_2 = 10$ km, the diameter is $d_2 = \alpha r_2 = 10^{-4} \cdot 10000 = 1.0$ m. | |||
| At a distance $r_3 = 20$ km, the diameter is $d_3 = \alpha r_3 = 10^{-4} \cdot 20000 = 2.0$ m. | |||
| Since the lens diameter is $D = 1$ m, at distances of $5$ km and $10$ km, the entire laser beam enters the lens ($d_1 < D$ and $d_2 = D$). The signal reduction by a factor of $2$ between $5$ km and $10$ km is solely due to absorption in the $5$ km thick atmospheric layer. | |||
| According to the Beer-Lambert-Bouguer law, absorption is exponential. Therefore, passing through every $5$ km of the atmosphere attenuates the signal by half. | |||
| When the distance increases from $10$ to $20$ km, the light travels an additional $10$ km (two $5$ km layers). Thus, due to absorption in the atmosphere, the signal will decrease by a factor of $2^2 = 4$. | |||
| Furthermore, at $20$ km, the beam diameter $d_3 = 2$ m exceeds the lens diameter $D = 1$ m. The fraction of energy captured by the lens is proportional to the ratio of their areas: | |||
| $$\frac{S_{\text{lens}}}{S_{\text{beam}}} = \frac{D^2}{d_3^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ | |||
| This means the signal decreases by another factor of $4$ due to the geometric expansion of the beam. | |||
| @@ -32,11 +26,7 @@Solution | |||
| The total signal reduction factor is the product of both effects: | |||
| $$N = 4 \cdot 4 = 16$$ | |||
| − | Note: there is a typo in the official answer in the problem book. | ||
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| ### Answer | |||
| $N = 16$ | |||
| − | #### Answer | ||
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| + | <i>Note: there is a typo in the official answer in the problem book.</i> | ||
| @@ -1,9 +1,3 @@ | |||
| ### Statement | |||
| $13.4.25.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | ### Problem | ||
| $13.4.25^*.$ In an optical communication system, a laser beam transmitting information has the shape of a cone with a vertex angle of $10^{-4}$ rad (divergence angle). At the receiving device, the light energy is focused on a photocell using a $1$ m lens. It turned out that when the distance between the transmitter and receiver changed from $5$ to $10$ km, the signal received from the photocell decreased by a factor of two (due to light absorption in the atmosphere). By what factor will the signal change when the distance increases from $10$ to $20$ km? | $13.4.25^*.$ In an optical communication system, a laser beam transmitting information has the shape of a cone with a vertex angle of $10^{-4}$ rad (divergence angle). At the receiving device, the light energy is focused on a photocell using a $1$ m lens. It turned out that when the distance between the transmitter and receiver changed from $5$ to $10$ km, the signal received from the photocell decreased by a factor of two (due to light absorption in the atmosphere). By what factor will the signal change when the distance increases from $10$ to $20$ km? | ||
| ### Solution | ### Solution | ||
| Let's determine the diameter of the laser beam at different distances. | Let's determine the diameter of the laser beam at different distances. | ||
| At a distance $r_1 = 5$ km, the diameter is $d_1 = \alpha r_1 = 10^{-4} \cdot 5000 = 0.5$ m. | At a distance $r_1 = 5$ km, the diameter is $d_1 = \alpha r_1 = 10^{-4} \cdot 5000 = 0.5$ m. | ||
| At a distance $r_2 = 10$ km, the diameter is $d_2 = \alpha r_2 = 10^{-4} \cdot 10000 = 1.0$ m. | At a distance $r_2 = 10$ km, the diameter is $d_2 = \alpha r_2 = 10^{-4} \cdot 10000 = 1.0$ m. | ||
| At a distance $r_3 = 20$ km, the diameter is $d_3 = \alpha r_3 = 10^{-4} \cdot 20000 = 2.0$ m. | At a distance $r_3 = 20$ km, the diameter is $d_3 = \alpha r_3 = 10^{-4} \cdot 20000 = 2.0$ m. | ||
| Since the lens diameter is $D = 1$ m, at distances of $5$ km and $10$ km, the entire laser beam enters the lens ($d_1 < D$ and $d_2 = D$). The signal reduction by a factor of $2$ between $5$ km and $10$ km is solely due to absorption in the $5$ km thick atmospheric layer. | Since the lens diameter is $D = 1$ m, at distances of $5$ km and $10$ km, the entire laser beam enters the lens ($d_1 < D$ and $d_2 = D$). The signal reduction by a factor of $2$ between $5$ km and $10$ km is solely due to absorption in the $5$ km thick atmospheric layer. | ||
| According to the Beer-Lambert-Bouguer law, absorption is exponential. Therefore, passing through every $5$ km of the atmosphere attenuates the signal by half. | According to the Beer-Lambert-Bouguer law, absorption is exponential. Therefore, passing through every $5$ km of the atmosphere attenuates the signal by half. | ||
| When the distance increases from $10$ to $20$ km, the light travels an additional $10$ km (two $5$ km layers). Thus, due to absorption in the atmosphere, the signal will decrease by a factor of $2^2 = 4$. | When the distance increases from $10$ to $20$ km, the light travels an additional $10$ km (two $5$ km layers). Thus, due to absorption in the atmosphere, the signal will decrease by a factor of $2^2 = 4$. | ||
| Furthermore, at $20$ km, the beam diameter $d_3 = 2$ m exceeds the lens diameter $D = 1$ m. The fraction of energy captured by the lens is proportional to the ratio of their areas: | Furthermore, at $20$ km, the beam diameter $d_3 = 2$ m exceeds the lens diameter $D = 1$ m. The fraction of energy captured by the lens is proportional to the ratio of their areas: | ||
| $$\frac{S_{\text{lens}}}{S_{\text{beam}}} = \frac{D^2}{d_3^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ | $$\frac{S_{\text{lens}}}{S_{\text{beam}}} = \frac{D^2}{d_3^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ | ||
| This means the signal decreases by another factor of $4$ due to the geometric expansion of the beam. | This means the signal decreases by another factor of $4$ due to the geometric expansion of the beam. | ||
| @@ -32,11 +26,7 @@Solution | |||
| The total signal reduction factor is the product of both effects: | The total signal reduction factor is the product of both effects: | ||
| $$N = 4 \cdot 4 = 16$$ | $$N = 4 \cdot 4 = 16$$ | ||
| Note: there is a typo in the official answer in the problem book. | |||
| ### Answer | ### Answer | ||
| $N = 16$ | $N = 16$ | ||
| #### Answer | <i>Note: there is a typo in the official answer in the problem book.</i> | ||