Правка разделов «Statement», «Solution», «Answer»
en/12.2.13.md
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| @@ -1,32 +1,39 @@ | |||
| ### Statement | |||
| − | $12.2.13.$ | ||
| + | $12.2.13.$ | ||
| + | Estimate the size of the light spot on the moon from the laser beam. The | ||
| + | laser is located on the Ground, its beam radius is 10 cm, the wavelength is | ||
| + | 10−5cm. (The spot boundary is estimated from the condition that in the spot | ||
| + | area the rays coming from separate sections of the wave do not cancel each | ||
| + | other out.) | ||
| + | b. Estimate the size of the radar antenna emitting three-centimeter electro- | ||
| + | magnetic waves inside the angle of 0.01 rad. | ||
| + | |||
| ### Solution | |||
| First the Data | |||
| Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$. | |||
| − | Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$ | ||
| + | Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$ | ||
| use visible light $\lambda \approx 5\times10^{-5}\ \text{cm} = 5\times10^{-7}\ \text{m}$ | |||
| We will adopt $ \lambda = 5\times10^{-7}\ \text{m}$. | |||
| − | |||
| + | · Earth–Moon distance: | ||
| + | $L \approx 3.84\times10^{8}\ \text{m}$. | ||
| The laser beam emerges from a circular aperture of radius a. Diffraction produces an Airy pattern on the Moon. The first intensity minimum (where the waves begin to cancel) defines the limit of the central spot. For a circular aperture, the angle of the first minimum is: | |||
| $\theta \approx \frac{0.61\lambda}{a}$ | |||
| − | 3. Calculation of the spot radius | ||
| − | |||
| $R = L\,\theta = \frac{0.61\,\lambda L}{a}$. | |||
| − | Multiply | ||
| + | Multiply $\lambda L$: | ||
| − | $\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$ | ||
| + | $\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$ | ||
| Divide by a: | |||
| $\frac{\lambda L}{a} = \frac{192}{0.1} = 1920\ \text{m}$. | |||
| Apply the factor 0.61: | |||
| $R = 0.61 \times 1920 = 1171.2\ \text{m}$. | |||
| Rounding to order of magnitude: | |||
| $\boxed{R \approx 1\ \text{km}}$. | |||
| --- | |||
| @@ -44,21 +51,15 @@Solution | |||
| b) Radar antenna of 3 cm | |||
| − | |||
| − | Estimate the size of a radar antenna that emits electromagnetic waves of $\lambda = 3\ \text{cm} within an angle of 0.01\ \text{rad}$. | ||
| + | Data | ||
| − | Solution | ||
| − | |||
| − | 1. Data | ||
| − | |||
| − | |||
| + | Wavelength: $\lambda = 3\ \text{cm} = 0.03\ \text{m}$. | ||
| · Total angular divergence of the beam: $\theta \approx 0.01\ \text{rad}$. | |||
| A circular parabolic antenna of diameter D produces a beam whose angular width is limited by diffraction. The angular radius of the first minimum is: | |||
| $\theta \approx \frac{\lambda}{D}$. | |||
| − | 3. Calculation of the diameter and radius | ||
| Solve for D: | |||
| $D \approx \frac{\lambda}{\theta} = \frac{0.03}{0.01} = 3\ \text{m}$. | |||
| The radius of the antenna is half the diameter: | |||
| $R = \frac{D}{2} = 1.5\ \text{m}$. | |||
| @@ -69,7 +70,8 @@Solution | |||
| $\boxed{R \approx 1.5\ \text{m}}$. | |||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{R \approx 1\ \text{km}}$ | ||
| + | |||
| + | $\boxed{R \approx 1.5\ \text{m}}$ | ||
| @@ -1,32 +1,39 @@ | |||
| ### Statement | ### Statement | ||
| $12.2.13.$ |
$12.2.13.$ | ||
| Estimate the size of the light spot on the moon from the laser beam. The | |||
| laser is located on the Ground, its beam radius is 10 cm, the wavelength is | |||
| 10−5cm. (The spot boundary is estimated from the condition that in the spot | |||
| area the rays coming from separate sections of the wave do not cancel each | |||
| other out.) | |||
| b. Estimate the size of the radar antenna emitting three-centimeter electro- | |||
| magnetic waves inside the angle of 0.01 rad. | |||
| ### Solution | ### Solution | ||
| First the Data | First the Data | ||
| Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$. | Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$. | ||
| Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$ |
Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$ | ||
| use visible light $\lambda \approx 5\times10^{-5}\ \text{cm} = 5\times10^{-7}\ \text{m}$ | use visible light $\lambda \approx 5\times10^{-5}\ \text{cm} = 5\times10^{-7}\ \text{m}$ | ||
| We will adopt $ \lambda = 5\times10^{-7}\ \text{m}$. | We will adopt $ \lambda = 5\times10^{-7}\ \text{m}$. | ||
| · Earth–Moon distance: | |||
| $L \approx 3.84\times10^{8}\ \text{m}$. | |||
| The laser beam emerges from a circular aperture of radius a. Diffraction produces an Airy pattern on the Moon. The first intensity minimum (where the waves begin to cancel) defines the limit of the central spot. For a circular aperture, the angle of the first minimum is: | The laser beam emerges from a circular aperture of radius a. Diffraction produces an Airy pattern on the Moon. The first intensity minimum (where the waves begin to cancel) defines the limit of the central spot. For a circular aperture, the angle of the first minimum is: | ||
| $\theta \approx \frac{0.61\lambda}{a}$ | $\theta \approx \frac{0.61\lambda}{a}$ | ||
| 3. Calculation of the spot radius | |||
| $R = L\,\theta = \frac{0.61\,\lambda L}{a}$. | $R = L\,\theta = \frac{0.61\,\lambda L}{a}$. | ||
| Multiply |
Multiply $\lambda L$: | ||
| $\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$ |
$\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$ | ||
| Divide by a: | Divide by a: | ||
| $\frac{\lambda L}{a} = \frac{192}{0.1} = 1920\ \text{m}$. | $\frac{\lambda L}{a} = \frac{192}{0.1} = 1920\ \text{m}$. | ||
| Apply the factor 0.61: | Apply the factor 0.61: | ||
| $R = 0.61 \times 1920 = 1171.2\ \text{m}$. | $R = 0.61 \times 1920 = 1171.2\ \text{m}$. | ||
| Rounding to order of magnitude: | Rounding to order of magnitude: | ||
| $\boxed{R \approx 1\ \text{km}}$. | $\boxed{R \approx 1\ \text{km}}$. | ||
| --- | --- | ||
| @@ -44,21 +51,15 @@Solution | |||
| b) Radar antenna of 3 cm | b) Radar antenna of 3 cm | ||
| Data | |||
| Estimate the size of a radar antenna that emits electromagnetic waves of $\lambda = 3\ \text{cm} within an angle of 0.01\ \text{rad}$. | |||
| Solution | Wavelength: $\lambda = 3\ \text{cm} = 0.03\ \text{m}$. | ||
| 1. Data | |||
| · Total angular divergence of the beam: $\theta \approx 0.01\ \text{rad}$. | · Total angular divergence of the beam: $\theta \approx 0.01\ \text{rad}$. | ||
| A circular parabolic antenna of diameter D produces a beam whose angular width is limited by diffraction. The angular radius of the first minimum is: | A circular parabolic antenna of diameter D produces a beam whose angular width is limited by diffraction. The angular radius of the first minimum is: | ||
| $\theta \approx \frac{\lambda}{D}$. | $\theta \approx \frac{\lambda}{D}$. | ||
| 3. Calculation of the diameter and radius | |||
| Solve for D: | Solve for D: | ||
| $D \approx \frac{\lambda}{\theta} = \frac{0.03}{0.01} = 3\ \text{m}$. | $D \approx \frac{\lambda}{\theta} = \frac{0.03}{0.01} = 3\ \text{m}$. | ||
| The radius of the antenna is half the diameter: | The radius of the antenna is half the diameter: | ||
| $R = \frac{D}{2} = 1.5\ \text{m}$. | $R = \frac{D}{2} = 1.5\ \text{m}$. | ||
| @@ -69,7 +70,8 @@Solution | |||
| $\boxed{R \approx 1.5\ \text{m}}$. | $\boxed{R \approx 1.5\ \text{m}}$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{R \approx 1\ \text{km}}$ | ||
| $\boxed{R \approx 1.5\ \text{m}}$ | |||