Question

What is the meaning of wavefront? Explain the laws of reflection on the basis of Huygens' principle of wavefront.
OR
Determine the following for the given A.C. circuit: (i) impedance, (ii) power factor, and (iii) phase difference between voltage and current. (Series combination of resistance \( R = 500\ \Omega \), inductance \( L = 10\ \text{H} \) and capacitance \( C = 20\ \mu\text{F} \) connected to an A.C. source \( V = 200\sin 100t \).)

Hint:

Wavefront = surface of constant phase; use Huygens' secondary wavelets and congruent triangles to prove \( i = r \). For the circuit, find \( X_L = \omega L \) and \( X_C = 1/\omega C \), then \( Z = \sqrt{R^2+(X_L-X_C)^2} \), \( \cos\phi = R/Z \), \( \tan\phi = (X_L-X_C)/R \).

Answer 1

Option 1 (Wavefront and laws of reflection):
What a wavefront is: Picture a disturbance spreading through a medium. At any chosen instant, join every particle that is at the same stage of vibration (same phase); the surface so formed is the wavefront. It may be spherical (near a point source), cylindrical (near a line source) or plane (far from the source). The normal to the wavefront gives the direction of travel.
Huygens' construction: Treat each point of a given wavefront as a tiny secondary source sending out spherical wavelets that move forward at speed \(v\); after a time \(t\) their forward common tangent is the next wavefront.
Getting \(i = r\):
Step 1: A plane wavefront \(AB\) strikes a mirror \(XY\). Corner \(A\) arrives first, while the far corner \(B\) must still cover the distance \(BC\) to reach the surface.
Step 2: Let this take time \(t\), so \(BC = vt\). Meanwhile the secondary wavelet started from \(A\) has grown to radius \(AD = vt\) on the reflected side.
Step 3: The tangent \(CD\) drawn from \(C\) to this wavelet is the reflected wavefront, with \(CD = vt\).
Step 4: Right triangles \(ABC\) and \(CDA\) share the hypotenuse \(AC\) and have equal sides \(BC = AD = vt\), so they are congruent. This makes the angle of incidence equal to the angle of reflection, \(i = r\).
Step 5: The same geometry keeps the incident ray, the normal and the reflected ray in one plane. Both laws of reflection follow.
\[\boxed{\text{angle of incidence } i = \text{angle of reflection } r}\]

Option 2 (A.C. series circuit):
Step 1: From \(V = 200\sin 100t\), the peak voltage is \(V_0 = 200\) V and the angular frequency is \(\omega = 100\) rad/s. Given \(R = 500\ \Omega\), \(L = 10\) H, \(C = 20\times10^{-6}\) F.
Step 2: Express the reactances as phasors. Inductor: \(X_L = \omega L = 1000\ \Omega\) (voltage \(90^\circ\) ahead of current). Capacitor: \(X_C = 1/(\omega C) = 1/(100\times20\times10^{-6}) = 500\ \Omega\) (voltage \(90^\circ\) behind current).
Step 3: The net reactance is \(X = X_L - X_C = 1000 - 500 = 500\ \Omega\), which is positive, so the circuit is net inductive.
Step 4: The impedance is the phasor sum \(Z = \sqrt{R^2 + X^2} = \sqrt{500^2 + 500^2} = 500\sqrt2 \approx 707\ \Omega\).
Step 5: From the impedance triangle the power factor is \(\cos\phi = R/Z = 500/(500\sqrt2) = 0.707\) and the phase angle is \(\phi = \tan^{-1}(X/R) = \tan^{-1}(1) = 45^\circ\), with the applied voltage leading the current.
\[\boxed{Z \approx 707\ \Omega,\ \cos\phi \approx 0.707,\ \phi = 45^\circ}\]