Question

PART 2: Multiple slits - Interference Remove the single slit disk, and install the multiple slit disks. You will now sements as in part 1 to explore the nature of the multiple slit interference patit this case, however, you will mark the maxima, not the minima. Siit disks. You will now do much the same Use the double slits of separation d = 0.25 mm (and a = 0.04mm). Increase distance L ben the screen and the slit as much as possible. Mark the positions of the 1 and 2 ordet and measure the separation between them. From this data, again calculate the wavelength of the laser light. Distance between the two marks Wavelength 2 Order of the maxima m = 1 m=2 Compare to your previous value obtained for 2. What will happen to the interference patter if you increase the slit separation d? After writing down your prediction, try it and see if you are correct. What happens when you send the laser beam through more than two slits? Describe your observations.

Answer 1

In double slit experiment, two coherent sources of light interfere with each other and produces alternative bright (maxima) and dark (minima) fringes on the screen.

If d is the separation between two slits and L is the distance of slit from screen,then the path difference under two approximations that is, d<<L and d/lambda >>1 is given by

Under these conditions θ is small, thus we can use the approximation sin θ approx tan θ = γ/L.

∴ path difference, Δz = γ/L

Now,position of fringes on screen. For maxima,it can given as,

Δz = nλ (n = 0, ±1, ±2, . . . .)

y (bright) = (nλ\d)L (n = 0, ±1, ±2, . . . .)

Now,find this and substitute there. You've to give us mark position in cm,to find wavelength.

If you increase slit separation d,then both bright fringes will be at a smaller distance from each other. Thus,interference pattern will be more clear and distant.