Consider Young's experiment set-up that is formed of two very
thin parallel slits Fi and F2, separated by a distance a = 1 mm ,
and a screen of observation (E) placed parallel to the plane of
the slits at a distance D = 2 m from the mid point I of F1F2 and a
thin slit F, equidistant from Fi and F2, situated on the straight
line (A) whose intersection with (E) is the point O.
The object of this exercise is to study the interference pattern
observed on the screen (E) in different situations.
(4)
A- First situation
The slit F is illuminated with a monochromatic light of wavelength 2 = 0.64 um in air.
1) Describe the interference pattern observed on (E).

You need to be more specific.  i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time,  then constant velocity is a straight line . . i.e. a line that has  constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant  velocity is a piece of line that lies on the x-axis . . it's value is zero. 

Ray 2 should cross through the focal point on the right side of the lens.

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.