In a double slit interference arrangement the fringe spacing is W when the wavelength of the radiation is lambda, the distance between the double slit is S and the distance between the slits and the plane of the observed fringes is D, in which of the following cases would the fringe spacing be the same as W.

Wavelength dist between slits dist between slits and fringes
A 2lambda 2S 2D
B 2, 4S 2D
C 2, 2S 4D
D 4, 2S 2D

I thought C because it would have to travel double the distance, but the mark scheme says B can anyone explane why this is

qn 3d

r=(I+7j-5k)+(lambda)(3i-j-2k)

find the ratio AC:CB

co-ords of A -5,9,-9 B 25,-1,11 C 4,6,-3

I thought I could do (-5+9-9)*(4+6-3) : (4+6-3)*(25-1+11) and workout the ratio that way I get 1:7
but mark scheme gets 3:7 how and why


5a, for the last part of the table when x=2 I always get y=0.766 y=4(x^0.5)e^-x

mark scheme gets 1.083







qn 6)find the integral of 2sin3x sin2x dx I tried parts but went wrong some where








Andrew Cropley

[(3/2)ln(2x-3)+ln(x+2)] (with 6 as the top and 2 as the bottom)

Any help would be greatly appreiated