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):
Meaning of wavefront: A wavefront is the locus of all points of a medium that are vibrating in the same phase at a given instant. The perpendicular drawn to the wavefront in the direction of travel is a ray. A point source gives a spherical wavefront and a far-off source gives a plane wavefront.
Huygens' principle: Every point on a wavefront acts as a fresh source of secondary wavelets which spread forward with the speed of the wave. The common tangent (envelope) of all these wavelets after a time \( t \) gives the new wavefront.
Laws of reflection:
Step 1: A plane wavefront AB falls on a plane reflecting surface MN, making the angle of incidence \( i \). Corner A touches the mirror first; the far corner B still has to reach it at C.
Step 2: If the wave takes time \( t \) to travel from B to C, then \( BC = vt \). In the same time a secondary wavelet from A grows to radius \( AD = vt \) on the reflected side (v = speed of the wave).
Step 3: Draw the tangent CD from C to this wavelet; CD is the reflected wavefront. In right triangles ABC and CDA, \( BC = AD = vt \), AC is common, and the angles at B and D are each \( 90^\circ \), so the triangles are congruent.
Step 4: Hence \( \angle BAC = \angle DCA \), i.e. the angle of incidence equals the angle of reflection: \( i = r \).
Step 5: Also the incident ray, the reflected ray and the normal all lie in the same plane. These two results are the laws of reflection.
\[\boxed{i = r}\]

Option 2 (A.C. series circuit):
Step 1 (data): Comparing \( V = 200\sin 100t \) with \( V = V_0\sin\omega t \) gives \( V_0 = 200 \) V and \( \omega = 100 \) rad/s. Also \( R = 500\ \Omega \), \( L = 10 \) H, \( C = 20\ \mu\text{F} = 20\times10^{-6} \) F.
Step 2 (reactances): Inductive reactance \( X_L = \omega L = 100 \times 10 = 1000\ \Omega \). Capacitive reactance \( X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 20\times10^{-6}} = \dfrac{1}{2\times10^{-3}} = 500\ \Omega \).
Step 3 (impedance): \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{500^2 + (1000-500)^2} = \sqrt{500^2 + 500^2} = 500\sqrt{2} \approx 707\ \Omega \).
Step 4 (power factor): \( \cos\phi = \dfrac{R}{Z} = \dfrac{500}{500\sqrt2} = \dfrac{1}{\sqrt2} \approx 0.707 \).
Step 5 (phase difference): \( \tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{500}{500} = 1 \Rightarrow \phi = 45^\circ \). Since \( X_L > X_C \), the circuit is inductive and the voltage leads the current by \( 45^\circ \).
\[\boxed{Z = 500\sqrt2 \approx 707\ \Omega,\quad \cos\phi = \tfrac{1}{\sqrt2} \approx 0.707,\quad \phi = 45^\circ}\]