Physics · Wave Optics · NEET
The two slits send coherent waves that overlap everywhere on the screen, not only at fringe centres. At any point the two waves have a phase difference phi, and the amplitude adds up to 2a cos(phi/2). Since intensity is proportional to amplitude squared, I = 4I0 cos^2(phi/2). As you move across the screen phi changes smoothly, so intensity rises and falls smoothly following a cos-squared curve. It is never zero-width lines; it is a continuous wave.
In ideal YDSE both slits are equally wide, so each gives the same intensity I0 and the peaks all reach 4I0. The graph shows equal humps because the cos^2 function repeats identically. This is the KEY difference from single-slit diffraction, where the central maximum is much taller than the side maxima. If the question shows equal humps, it is interference (YDSE); if the central hump is tallest, it is diffraction.
Let I0 be the intensity from ONE slit alone. When two equal coherent waves interfere constructively, amplitudes add: 2a, so intensity = (2a)^2 = 4 times a^2 = 4I0. So maximum intensity is 4I0, minimum is 0, and the AVERAGE is 2I0 (which equals I0 + I0, energy is conserved, just redistributed). Warning: many books write I = I_max cos^2(phi/2) where I_max already means 4I0. Read the symbol carefully.
Use phi = (2pi/lambda) x (path difference). So I = 4I0 cos^2( pi x (path difference) / lambda ). At path difference = 0, lambda, 2lambda you get bright (I = 4I0). At path difference = lambda/2, 3lambda/2 you get dark (I = 0). Example: at path difference = lambda/4, phi = pi/2, cos^2(pi/4) = 1/2, so I = 2I0 (half of maximum).
They are the same information shown two ways. The fringe pattern is what your eye sees: alternate bright and dark bands. The intensity graph plots I on the y-axis versus position (or phase difference) on the x-axis. Bright bands = peaks of the graph, dark bands = zeros of the graph. The graph makes it clear that brightness changes gradually, not suddenly.
The intensity at the maximum in a Young's double slit experiment is I0. Distance between two slits is d = 5 lambda, where lambda is the wavelength of light used. What is the intensity in front of one of the slits on a screen placed at distance D = 10 d?
In Young's double slit experiment, using monochromatic light of wavelength lambda, the intensity at a point where the path difference is lambda is K units. The intensity at a point where the path difference is lambda/3 will be
Try the real previous-year questions from this chapter — each with the answer and a full solution.
I = 4I0 cos^2(phi/2), where I0 is the intensity due to a single slit and phi is the phase difference between the two waves. Equivalently I = I_max cos^2(phi/2) with I_max = 4I0.
It is a series of identical cos-squared humps. Peaks reach 4I0 at phi = 0, 2pi, 4pi (bright fringes) and drop to zero at phi = pi, 3pi, 5pi (dark fringes). All peaks have equal height and equal spacing.
The average value of cos^2 over a full cycle is 1/2, so average I = 4I0 x 1/2 = 2I0. This equals I0 + I0, showing energy is only redistributed, not created or lost - conservation of energy holds.
In YDSE (interference) all bright humps are the same height. In single-slit diffraction the central maximum is much taller and the side maxima fall off quickly (about 4.5% of the central peak). Equal humps mean interference; one tall central hump means diffraction.
When cos^2(phi/2) = 1/2, phi = pi/2, which means path difference = lambda/4. So at a quarter-wavelength path difference the intensity is half the maximum (2I0).
