New solution

Alexphysics edited Jun 16, 2026, 2:38 PM

View solution History

revision #18923 newer →

en/12.2.13.md +75 −0

		@@ -0,0 +1,75 @@
		+	### Statement
		+
		+	$12.2.13.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	First the Data
		+
		+	Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$.
		+
		+	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 f \lambda L$:
		+
		+	$\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$.
		+
		+	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:
		+
		+	$\boxed{R \approx 1\ \text{km}}$.
		+
		+	---
		+
		+	b) Radar antenna of 3 cm
		+
		+	Statement
		+	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
		+
		+	1. Data
		+
		+	· Wavelength: f \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:
		+
		+	$R = \frac{D}{2} = 1.5\ \text{m}$.
		+
		+	$\boxed{R \approx 1.5\ \text{m}}$.
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,75 @@
			### Statement

			$12.2.13.$ [Insert the problem statement]

			### Solution

			First the Data

			Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$.

			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 f \lambda L$:

			$\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$.

			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:

			$\boxed{R \approx 1\ \text{km}}$.

			---

			b) Radar antenna of 3 cm

			Statement
			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

			1. Data

			· Wavelength: f \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:

			$R = \frac{D}{2} = 1.5\ \text{m}$.

			$\boxed{R \approx 1.5\ \text{m}}$.


			#### Answer

			[Insert a concise answer or boxed result]