Physics · Wave Optics · NEET
Intensity is proportional to the square of the amplitude, not the amplitude itself. When two waves overlap, their amplitudes add as vectors (using phase). Squaring the resultant amplitude gives R^2 = a1^2 + a2^2 + 2 a1 a2 cos(phi). Since I is proportional to amplitude squared, this becomes I = I1 + I2 + 2 root(I1 I2) cos(phi). The extra cross term 2 root(I1 I2) cos(phi) is called the interference term. It is present only for coherent sources. For incoherent light cos(phi) averages to zero over time, so it vanishes and you get plain I1 + I2.
Put I1 = I2 = I0 into the formula: I = I0 + I0 + 2 root(I0 I0) cos(phi) = 2 I0 (1 + cos phi) = 4 I0 cos^2(phi/2). At constructive points (phi = 0), I = 4 I0. No energy is created. At destructive points (phi = 180 degrees), I = 0. The energy from the dark fringes is simply redistributed to the bright fringes. The average intensity over the whole screen stays 2 I0, so total energy is conserved.
Amplitudes add (as vectors with phase); intensities do not add directly. The correct chain is: resultant amplitude R = root(a1^2 + a2^2 + 2 a1 a2 cos phi), then intensity I is proportional to R^2. A common mistake is adding intensities like I = I1 + I2, which is only true when you average out the interference (incoherent light). For coherent interference always work with the full formula.
phi is the phase difference (in radians). If you are given the path difference (delta x), convert first using phi = (2 pi / lambda) times delta x. For example, path difference lambda/2 gives phi = pi (180 degrees), a dark fringe. Path difference lambda gives phi = 2 pi, a bright fringe. Never plug a path difference directly into the cosine - always turn it into a phase angle first.
They are the same when both slits have equal intensity I0, because the maximum intensity I max = 4 I0 in that case. So I = I max cos^2(phi/2) is the general form: at phi = 0 you get I max, and that maximum equals 4 I0 for equal sources. Using I max is safer in problems where you are told the maximum value (like 'intensity at maximum is K') and asked for another point - just multiply K by cos^2(phi/2).
In Young's double slit experiment, using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where the path difference is lambda is K units. The intensity of light at a point where the path difference is lambda/3 will be
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (I_max - I_min)/(I_max + I_min) will be
Try the real previous-year questions from this chapter — each with the answer and a full solution.
I = I1 + I2 + 2 root(I1 I2) cos(phi), where I1 and I2 are the two intensities and phi is the phase difference. For equal intensities it becomes I = 4 I0 cos^2(phi/2).
Maximum intensity I_max = (root I1 + root I2)^2, occurring when cos(phi) = +1 (phi = 0). Minimum intensity I_min = (root I1 - root I2)^2, occurring when cos(phi) = -1 (phi = 180 degrees). For equal sources, I_max = 4 I0 and I_min = 0.
Intensity is the energy carried by the wave per unit area per unit time, and the energy of a wave is proportional to the square of its amplitude. So I is proportional to a^2. That is why amplitudes add first and then get squared, producing the 4 I0 result.
It disappears for incoherent sources, where the phase difference phi changes randomly with time. The time-average of cos(phi) is zero, so the term vanishes and the total intensity is just I1 + I2. This is why two separate bulbs do not produce interference fringes.
Use phi = (2 pi / lambda) times (path difference). A path difference of lambda gives phi = 2 pi (bright), and lambda/2 gives phi = pi (dark). Always convert path difference to phase before putting it in the cosine.
