Physics · Wave Optics · NEET
Both work, but you must be careful. Amplitudes add and subtract directly: A(max) = a1 + a2 and A(min) = a1 - a2. Intensity is proportional to amplitude squared, so I(max) is proportional to (a1 + a2)^2 and I(min) is proportional to (a1 - a2)^2. You cannot just add intensities. If you are given intensity ratio I1/I2, first take the square root to get the amplitude ratio a1/a2 = sqrt(I1/I2), then add or subtract.
Because intensity is not the same as amplitude. Intensity measures energy carried by the wave, and energy is proportional to the square of the amplitude (I is proportional to A^2). At a bright fringe the two waves arrive in step, so their amplitudes add to give A = a1 + a2. Squaring this gives Imax proportional to (a1 + a2)^2. Forgetting the square is the most common mistake.
Take the square root. If I1/I2 = n, then a1/a2 = sqrt(n) because I is proportional to A^2. Example: if I1/I2 = 9, then a1/a2 = 3. Now Imax/Imin = (a1 + a2)^2/(a1 - a2)^2 = (3 + 1)^2/(3 - 1)^2 = 16/4 = 4. Always convert intensity ratio to amplitude ratio first.
This is the fringe visibility or contrast. It tells you how clear the bright and dark fringes look. Using Imax = (a1 + a2)^2 and Imin = (a1 - a2)^2, it simplifies to 2(a1 a2)/(a1^2 + a2^2), which equals 2 sqrt(n)/(n + 1) when the intensity ratio is n. When the two sources are equal (n = 1), visibility = 1, meaning perfect contrast. When one source is much weaker, visibility drops toward 0 and fringes fade.
When a1 = a2, the minimum amplitude is a1 - a2 = 0, so Imin proportional to 0^2 = 0. This means perfectly dark fringes. Also Imax = (a1 + a2)^2 = (2a1)^2 = 4 a1^2 = 4I0, four times a single-slit intensity. This is the standard equal-slit YDSE case from NCERT: I = 4 I0 cos^2(phi/2).
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (Imax - Imin)/(Imax + Imin) will be
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Imax/Imin = (a1 + a2)^2/(a1 - a2)^2, where a1 and a2 are the amplitudes of the two waves. If you are given intensities, use a1/a2 = sqrt(I1/I2) first.
Imax is proportional to (a1 + a2)^2 and Imin is proportional to (a1 - a2)^2. For equal sources (a1 = a2 = a), Imax = 4I0 and Imin = 0, where I0 is single-source intensity.
Intensity is proportional to amplitude squared, so I1/I2 = (a1/a2)^2. To get amplitude ratio, take the square root of the intensity ratio.
Fringe visibility = (Imax - Imin)/(Imax + Imin) = 2 sqrt(n)/(n + 1) where n is the intensity ratio. It is 1 (best contrast) for equal sources and drops as the sources become unequal.
For two equal slits each of intensity I0, amplitude adds to 2a at a bright fringe. Intensity is proportional to amplitude squared, so (2a)^2 = 4a^2, giving Imax = 4I0. This matches I = 4 I0 cos^2(phi/2).
