Physics · Ch 6 — Optics

Interference

6.10

Interference

Interference is the phenomenon of superposition (addition) of two light waves, producing an increase in intensity at some points and a decrease at others. For two waves of the same frequency, amplitudes $a_1,a_2$ , and a constant phase difference $\phi$ between them ( $y_1=a_1\sin\omega t$ , $y_2=a_2\sin(\omega t+\phi)$ ), the resultant is $y=A\sin(\omega t+\theta)$ with resultant amplitude $A^2=a_1^2+a_2^2+2a_1a_2\cos\phi$ ; since intensity $I\propto A^2$ , the resultant intensity is $I\propto I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ . Constructive interference (maximum intensity, $I_{\max}\propto(a_1+a_2)^2$ ) occurs at $\phi=0,\pm2\pi,\pm4\pi,\ldots$ ; destructive interference (minimum intensity, $I_{\min}\propto(a_1-a_2)^2$ ) occurs at $\phi=\pm\pi,\pm3\pi,\ldots$ . For equal amplitudes ( $a_1=a_2=a$ , $I_1=I_2=I_0$ ), this simplifies to $I=4I_0\cos^2(\phi/2)$ , giving $I_{\max}=4I_0$ at $\phi=0,2\pi, …

Figure 6.52Superposition principle

What this figure shows. Two coherent light sources S1 and S2 send waves that overlap and meet at a common point P, having travelled the two different path lengths S1P and S2P to get there. The resultant disturbance at P is found by superposing (vector-adding) the two individual wave contributions, whose relative phase at P is set entirely by the path difference (S2P minus S1P) -- this simple two-source geometric picture is the starting point the general resultant-amplitude and resultant …

Misc Example 6.24Maximum-to-minimum intensity ratio for two sources of unequal amplitude

Worked out. Two coherent light sources have amplitudes a1 = 5 units and a2 = 3 units. The resultant amplitude is maximum, A(max) = a1 + a2 = 8 units, when the two waves arrive exactly in phase; it is minimum, A(min) = a1 - a2 = 2 units, when they arrive exactly out of phase. Since intensity is proportional to the square of amplitude, the ratio of maximum to minimum intensity is I(max)/I(min) = (A(max)/A(min))^2 = (8/2)^2 = 16, i.e. the brightest fringes are sixteen times more intense than the darkest ones -- a direct illustration of why unequal-amplitude interfering beams never give perfectly dark minima …

Misc Example 6.25Maximum-to-minimum intensity ratio for two sources of equal amplitude

Worked out. Two coherent light sources of equal amplitude a interfere; using I proportional to 4a^2 cos^2(phi/2), the resultant intensity is maximum, I(max) proportional to 4a^2, at phi = 0 (cos^2 = 1), and minimum, I(min) = 0, at phi = 180 degrees (cos^2 = 0). Because the minimum intensity comes out to exactly zero, the ratio I(max) : I(min) is undefined (or, informally, infinite) -- the crucial general lesson being that only when the two interfering amplitudes are exactly equal can destructive interference produce a perfectly dark fringe; any imbalance in amplitude (as in the previous example) leaves some residual light ev …

Misc Example 6.26Resultant intensity at a fixed phase difference of pi/3

Worked out. Two coherent light sources, each of intensity I0, interfere at a point where their phase difference is phi = pi/3 (60 degrees). Substituting into I = 4 I0 cos^2(phi/2) gives I = 4 I0 cos^2(30 degrees) = 4 I0 times (root3/2)^2 = 4 I0 times 3/4 = 3 I0. So at this particular intermediate phase difference -- neither fully in phase nor fully out of phase -- the resultant intensity of 3 I0 sits between the maximum possible value of 4 I0 (at phi = 0) and the minimum possible value of 0 (at phi = 180 degrees), exactly as the general cos-s …

Start your 14-day free trial to unlock the full solution →

← Previous · 6.9.4Proof for Laws of Refraction Using Huygens' Principle Next · 6.10.1 →Phase Difference and Path Difference