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Alexphysics edited Jun 15, 2026, 3:15 PM

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		+	### Statement
		+
		+	$12.2.1.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Huygens–Fresnel Principle (recall)
		+
		+	At time t, the position of the wavefront is known.
		+	Each point on that wavefront becomes a source of secondary spherical waves of radius c\tau (where \tau is the considered time interval).
		+	The new wavefront at time t+\tau is the envelope surface of all those secondary spheres.
		+
		+	Note: c is the speed of light in vacuum.
		+
		+	---
		+
		+	Part a)
		+
		+	Initial assumption:
		+
		+	At time t = 0, the wavefront is the plane x = 0 (perpendicular to the x-axis, propagating to the right).
		+
		+	We apply Huygens–Fresnel:
		+
		+	· Each point P(0, y, z) on the plane x = 0 emits a secondary spherical wave that reaches a radius c\tau in time \tau.
		+	· The spheres have centers on the plane x = 0 and all have the same radius R = c\tau.
		+
		+	Geometry of the envelopes:
		+	The envelope of a set of spheres of equal radius whose centers lie on a plane is another parallel plane, located at a distance equal to the radius.
		+
		+	The spheres centered at (0, y, z) just touch the plane x = c\tau.
		+	No sphere reaches beyond x = c\tau.
		+	The plane x = c\tau is tangent to all spheres (because the minimum distance from any center (0, y, z) to the plane x = c\tau is exactly c\tau).
		+
		+	Result:
		+
		+	The new wavefront at time \tau is the plane x = c\tau. Therefore, in time \tau the wavefront has advanced a distance c\tau in the direction perpendicular to itself.
		+
		+	Speed:
		+
		+	v = \frac{\text{displacement}}{\text{time}} = \frac{c\tau}{\tau} = c
		+
		+	\boxed{\text{The plane wavefront moves with speed } c \text{ perpendicular to the plane.}}
		+
		+	Part b)
		+
		+	We apply Huygens–Fresnel:
		+
		+	Each point P on the spherical surface of radius R emits a secondary spherical wave of radius c\tau after time \tau.
		+
		+	Geometry of the envelope:
		+	The centers of the secondary spheres lie on the original sphere of radius R.
		+	All secondary spheres have the same radius c\tau.
		+
		+	The envelope of all these spheres is another sphere concentric with the original one, whose radius is R + c\tau.
		+
		+	Justification:
		+	A point Q is on the outer envelope if there exists a center P on the original sphere such that Q lies on the corresponding secondary sphere and, moreover, the direction O \to P \to Q is radial. In that case:
		+
		+	OQ = OP + PQ = R + c\tau
		+
		+	Since this holds for all directions, the envelope is the sphere of radius R + c\tau.
		+
		+	Increase in radius in time \tau:
		+
		+	\Delta R = (R + c\tau) - R = c\tau
		+
		+	\boxed{\text{The radius of a spherical wavefront increases by } c\tau \text{ during a time } \tau.}
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$12.2.1.$ [Insert the problem statement]

			### Solution

			Huygens–Fresnel Principle (recall)

			At time t, the position of the wavefront is known.
			Each point on that wavefront becomes a source of secondary spherical waves of radius c\tau (where \tau is the considered time interval).
			The new wavefront at time t+\tau is the envelope surface of all those secondary spheres.

			Note: c is the speed of light in vacuum.

			---

			Part a)

			Initial assumption:

			At time t = 0, the wavefront is the plane x = 0 (perpendicular to the x-axis, propagating to the right).

			We apply Huygens–Fresnel:

			· Each point P(0, y, z) on the plane x = 0 emits a secondary spherical wave that reaches a radius c\tau in time \tau.
			· The spheres have centers on the plane x = 0 and all have the same radius R = c\tau.

			Geometry of the envelopes:
			The envelope of a set of spheres of equal radius whose centers lie on a plane is another parallel plane, located at a distance equal to the radius.

			The spheres centered at (0, y, z) just touch the plane x = c\tau.
			No sphere reaches beyond x = c\tau.
			The plane x = c\tau is tangent to all spheres (because the minimum distance from any center (0, y, z) to the plane x = c\tau is exactly c\tau).

			Result:

			The new wavefront at time \tau is the plane x = c\tau. Therefore, in time \tau the wavefront has advanced a distance c\tau in the direction perpendicular to itself.

			Speed:

			v = \frac{\text{displacement}}{\text{time}} = \frac{c\tau}{\tau} = c

			\boxed{\text{The plane wavefront moves with speed } c \text{ perpendicular to the plane.}}

			Part b)

			We apply Huygens–Fresnel:

			Each point P on the spherical surface of radius R emits a secondary spherical wave of radius c\tau after time \tau.

			Geometry of the envelope:
			The centers of the secondary spheres lie on the original sphere of radius R.
			All secondary spheres have the same radius c\tau.

			The envelope of all these spheres is another sphere concentric with the original one, whose radius is R + c\tau.

			Justification:
			A point Q is on the outer envelope if there exists a center P on the original sphere such that Q lies on the corresponding secondary sphere and, moreover, the direction O \to P \to Q is radial. In that case:

			OQ = OP + PQ = R + c\tau

			Since this holds for all directions, the envelope is the sphere of radius R + c\tau.

			Increase in radius in time \tau:

			\Delta R = (R + c\tau) - R = c\tau

			\boxed{\text{The radius of a spherical wavefront increases by } c\tau \text{ during a time } \tau.}


			#### Answer

			[Insert a concise answer or boxed result]