Physics · Wave Optics · NEET
No. Superposition is a general wave property. You first met it in Class 11 for waves on a string and sound waves, and it also works for water ripples, electric fields, and light. It says the medium point moves by the algebraic sum of what each wave alone would produce. In Wave Optics we simply apply this same rule to two light waves to explain interference.
For two waves of the same amplitude a and same frequency, meeting with a phase difference phi, the resultant amplitude is R = 2a cos(phi/2). If the two waves have different amplitudes a1 and a2, use the vector-addition form: R = sqrt(a1^2 + a2^2 + 2 a1 a2 cos phi). Both come straight from adding the two wave equations.
The plain sum 2a only happens when the two waves are perfectly in step (phi = 0), because cos(0) = 1. For any other phase difference the waves are shifted, so at a given instant one is not at its peak when the other is. The factor cos(phi/2) accounts for this shift. At phi = pi (fully opposite), cos(pi/2) = 0, so R = 0 and the waves cancel.
For coherent waves (fixed phase relation), you add the DISPLACEMENTS (amplitudes as vectors) first, then square to get intensity: I = 4I0 cos2(phi/2). For incoherent sources (rapidly changing phase), the cross term averages to zero and the INTENSITIES simply add: I = I1 + I2. This is why two separate bulbs never make a visible interference pattern.
Energy is not destroyed. Superposition only redistributes energy across the screen. Where the waves cancel (dark point) the energy is missing, but at points of constructive interference the intensity rises to 4I0, which is double the 2I0 you would expect from just adding two sources. Averaged over the whole pattern, the total energy is conserved.
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.
When two or more waves overlap at a point in a medium, the resultant displacement at that point is the algebraic (vector) sum of the displacements produced by each wave separately. Each wave travels as if the others are not present.
For equal amplitudes a with phase difference phi: R = 2a cos(phi/2). For unequal amplitudes a1 and a2: R = sqrt(a1^2 + a2^2 + 2 a1 a2 cos phi). The resultant intensity is proportional to R squared.
Maximum amplitude is R = a1 + a2 when phi = 0 (waves in phase, constructive). Minimum amplitude is R = |a1 - a2| when phi = pi (waves out of phase, destructive). For equal amplitudes the minimum is zero.
Coherent waves have a fixed phase, so amplitudes add: R = 2a and I_max is proportional to 4I0. Incoherent waves have a rapidly changing phase, so cos(phi) averages to zero and only intensities add, giving 2I0 everywhere with no pattern.
Resultant displacement is the instantaneous sum y = y1 + y2 and it changes with time. Resultant amplitude R = 2a cos(phi/2) is the fixed peak value of that combined wave at a given point, set only by the phase difference phi.
