light of wavelength 700nm is incident on a pair of slits, forming fringes 3.0mm apart on the screen. What is the fringe spacing when light of wavelength 350nm is used and the slit separation is doubled?

Question

anonymous · Answer

Answers are 

a) 0.75mm
b) 1.5mm
c) 3.0mm
d) 6.0mm

anonymous · Answer

I have a couple of equations in the book but I don't even know which one calculates fringe spacing... Eish! Appreciate any help :)

sidsiddhartha · Answer

ok 
are familiar with bragg's law which is ----
$$dsin \theta=n \lambda$$

anonymous · Answer

Yep

anonymous · Answer

Theta is the angle between the fringes, d the distance  between the "lines" of the diffraction grating, n a whole number representing the order of the bright fringe, and lambda being the wavelength. Correct?

sidsiddhartha · Answer

yeah

sidsiddhartha · Answer

so 
$$\sin \theta=n \lambda/d$$
so considering n=1
look lambda is halfed and slit separation d is doubled so 
$$\delta [\sin(\theta)]=\frac{ 1 }{ 4 }$$
and it is generally spacing upon distence
so  fringe spacing when light of wavelength 350nm is used and the slit separation is doubled is=3/4=0.75 mm

anonymous · Answer

Okay two question more please: where does the three come from and so is sin theta equal to fringe spacing then?