Derivation of Young's Slits Equation

 

For the central bright fringe, the path difference is 0. 

 

S2O - S1O = 0.

 

For the first bright fringe

 S2Q - S1Q = l

 

Notice that in the triangle S1S2Z, S2Z is l. 

 

We know that S1S2 is s. 

 

Therefore:

sin q = l/s.

 

For the triangle OPQ, tan f = w/D.  Although in this diagram, it is clear that q f, in the real thing, we can assume that q = f, as the real set up is very much longer.

 

We know that for small angles sin q = tan q.

 

Therefore:

   

Therefore:

 

 

To produce easily measurable fringes, D must be large i.e. in metres while s is small (<1 mm). The larger D is the less bright the fringes.  We can increase the width of the fringes by increasing the wavelength, or by decreasing the slit width.

 

BACK to Interference