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| ### Statement |
| ### Statement |
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| $12.2.12.$ [Insert the problem statement] |
| $12.2.12.$ |
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| Calculate the amplitude a of an elementary secondary Huygens-Fresnel wave. |
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| (The amplitude a is proportional to the amplitude A of the primary wave that came from the element ∆S, the area of this element, and is inversely propor- |
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| tional to r, i.e. $a =cA∆S$ |
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| . To determine c, compare the plane wave amplitude |
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| r |
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| at some point and the Fresnel amplitude at the same point, when the plane |
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| wave front is taken as an auxiliary surface.) |
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| ### Solution |
| ### Solution |
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| Huygens–Fresnel Principle |
| Huygens–Fresnel Principle |
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| Each point of a wavefront becomes an emitter of monochromatic secondary spherical waves. The total disturbance at a point P is obtained by integrating the contributions from all elements dS of the wavefront. For an element dS, the contribution is: |
| Each point of a wavefront becomes an emitter of monochromatic secondary spherical waves. The total disturbance at a point P is obtained by integrating the contributions from all elements dS of the wavefront. For an element dS, the contribution is: |
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| $dU_P = \frac{c A \, e^{ikr}}{r} \, dS$ |
| $dU_P = \frac{c A \, e^{ikr}}{r} \, dS$ |
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| where A is the amplitude of the primary wave at the wavefront, r is the distance from the element to P, $k = 2\pi/\lambda$, and c is the constant to be determined. |
| where A is the amplitude of the primary wave at the wavefront, r is the distance from the element to P, |
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| $k = 2\pi/\lambda$ |
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| and c is the constant to be determined. |
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| Consider a plane wave propagating in the z-direction. At t = 0, the wavefront coincides with the plane z = 0. We want to calculate the amplitude at a point P located on the z-axis at a distance R from the plane. The expected amplitude of the plane wave at P is: |
| Consider a plane wave propagating in the z-direction. At t = 0, the wavefront coincides with the plane z = 0. We want to calculate the amplitude at a point P located on the z-axis at a distance R from the plane. The expected amplitude of the plane wave at P is: |
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| $U_{\text{plane}}(P) = A e^{ikR}$. |
| $U_{\text{plane}}(P) = A e^{ikR}$. |
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| We use the plane z = 0 as an auxiliary surface. Each element dS in polar coordinates $(\rho,\phi) has dS = \rho d\rho d\phi$. |
| We use the plane z = 0 as an auxiliary surface. Each element dS in polar coordinates |
| The distance to P is |
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| $(\rho,\phi) has dS = \rho d\rho d\phi$. |
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| The distance to P is |
| $ r = \sqrt{\rho^2 + R^2}$. |
| $ r = \sqrt{\rho^2 + R^2}$. |
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| $U(P) = \iint_{\text{plane}} \frac{c A e^{ikr}}{r} dS |
| $U(P) = \iint_{\text{plane}} \frac{c A e^{ikr}}{r} dS= cA \int_0^{2\pi} d\phi \int_0^\infty \frac{e^{ik\sqrt{\rho^2+R^2}}}{\sqrt{\rho^2+R^2}} \rho d\rho$. |
| = cA \int_0^{2\pi} d\phi \int_0^\infty \frac{e^{ik\sqrt{\rho^2+R^2}}}{\sqrt{\rho^2+R^2}} \rho d\rho$. |
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| The angular integral gives $2\pi$. We make the substitution |
| The angular integral gives $2\pi$. |
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| We make the substitution |
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| $u = \sqrt{\rho^2+R^2} ⇒ \rho d\rho = u du$. The limits: $ \rho = 0 \rightarrow u = R, \rho \to \infty \rightarrow u \to \infty$. |
| $u = \sqrt{\rho^2+R^2} ⇒ \rho d\rho = u du$. The limits: $ \rho = 0 \rightarrow u = R, \rho \to \infty \rightarrow u \to \infty$. |
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| $U(P) = 2\pi c A \int_R^\infty \frac{e^{iku}}{u} u du = 2\pi c A \int_R^\infty e^{iku} du$. |
| $U(P) = 2\pi c A \int_R^\infty \frac{e^{iku}}{u} u du = 2\pi c A \int_R^\infty e^{iku} du$. |
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| The improper integral is evaluated by introducing a convergence factor |
| The improper integral is evaluated by introducing a convergence factor |
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| $ e^{-\epsilon u} and taking \epsilon \to 0^+$: |
| $ e^{-\epsilon u} and taking \epsilon \to 0^+$: |
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| $\int_R^\infty e^{iku} du = \lim_{\epsilon\to 0^+} \int_R^\infty e^{i(k+i\epsilon)u} du |
| $\int_R^\infty e^{iku} du = \lim_{\epsilon\to 0^+} \int_R^\infty e^{i(k+i\epsilon)u} du |
| = \lim_{\epsilon\to 0^+} \frac{1}{i(k+i\epsilon)} \left[ e^{i(k+i\epsilon)u} \right]_R^\infty |
| = \lim_{\epsilon\to 0^+} \frac{1}{i(k+i\epsilon)} \left[ e^{i(k+i\epsilon)u} \right]_R^\infty |
| = \frac{i}{k} e^{ikR}$. |
| = \frac{i}{k} e^{ikR}$ |
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| Therefore: |
| Therefore: |
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| $U(P) = 2\pi c A \cdot \frac{i}{k} e^{ikR}$. |
| $U(P) = 2\pi c A \cdot \frac{i}{k} e^{ikR}$ |
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| We equate this amplitude to that of the plane wave $A e^{ikR}$: |
| We equate this amplitude to that of the plane wave |
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| 2\pi c A \frac{i}{k} e^{ikR} = A e^{ikR}. |
| $A e^{ikR}$: |
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| $2\pi c A \frac{i}{k} e^{ikR} = A e^{ikR}$ |
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| Canceling A e^{ikR} (assumed non-zero), we obtain: |
| Canceling A e^{ikR} (assumed non-zero), we obtain: |
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| $2\pi c \frac{i}{k} = 1 \quad\Rightarrow\quad c = \frac{k}{2\pi i} = \frac{1}{i\lambda}$ |
| $2\pi c \frac{i}{k} = 1 \quad\Rightarrow\quad c = \frac{k}{2\pi i} = \frac{1}{i\lambda}$ |
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| since $k = 2\pi/\lambda$. It can also be written as: |
| since $k = 2\pi/\lambda$. It can also be written as: |
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| $c = -\frac{i}{\lambda}$. |
| $c = -\frac{i}{\lambda}$. |
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| --- |
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| 5. Final result |
| #### Answer |
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| The elementary amplitude of a Huygens–Fresnel secondary wave is: |
| The elementary amplitude of a Huygens–Fresnel secondary wave is: |
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| \boxed{a = \frac{1}{i\lambda} \frac{A \Delta S}{r} e^{ikr} = -\frac{i}{\lambda} \frac{A \Delta S}{r} e^{ikr}}. |
| $\boxed{a = \frac{1}{i\lambda} \frac{A \Delta S}{r} e^{ikr} = -\frac{i}{\lambda} \frac{A \Delta S}{r} e^{ikr}}$ |
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| The proportionality constant c is 1/(i\lambda) (or equivalently -i/\lambda). The factor 1/i = -i implies a phase shift of \pi/2 between the secondary wave and the primary one |
| The proportionality constant c is $1/(i\lambda) (or equivalently -i/\lambda)$ |
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| The factor $1/i = -i$ implies a phase shift of $\pi/2$ |
| #### Answer |
| between the secondary wave and the primary one |
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| [Insert a concise answer or boxed result] |
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