1.Speaker end must be antinode, try putting you hand on the speaker. The antinode will blow away the sand creating a valley , the node is still ,so the sand builds up higher. 48 cm contains 7.5 N-N. One N-N is 6.4 cm. One wavelength is 12.8 cm .
2. path difference = n x wavelength. Let the other path be L. Path difference = L - 6 . If L - 6 = n x wavelength. If L is at least 6m, then choose n=0. L = 6+0 =6.0 m (edited)
2b. At X, the amplitude arriving there is A +A =2A. Its resultant intensity is I. If the intensities arriving at X is now reduced to one quarter. The amplitude arriving there must be 0.5A + 0.5 A. The new resultant intensity is I'. Taking I=k(square of 2A) and I' = k (square of 1A). We have I' = one quarter I.
3. a) Fringe separation the same but higher intensity(brighter)
b) Fringe separation decrease because wavelength decrease.
c) The amplitude from the two slits are not the same , the dark fringe is not completely dark and the bright fringe will not be as bright. The contrast between bright and dark fringe is not good. The fringe separation remain the same as that of red light.
4. First order on each side of the central maxima will constitute twice the angle tetha. Therefore, need divide angle by two. Answer will be 450 nm and 599 nm.
b) Maximum order can be found if maximum angle of 90' is substituted. Round down n.
450nm (violet) , n=4
599nm(red) , n=3.
If both colours of the same order can be seen , answer should be n=3.
