Double Slit Diffraction fringes

[foxx0775]

Homework Statement

A parallel beam of mono-chromatic light (500nm) is incident on 2 long narrow slits: slit width is 0.1mm and the distance between slit centres is 0.3mm.

a) How many bright fringes cover the central diffraction maximum?

Homework Equations

Condition for constructive interference:
$$\Delta = 2d.cos\theta = 2m(\lambda\div2)$$

Reduces to:

$$2d=m\lambda$$ as $$cos\theta=1$$ ?

The Attempt at a Solution

I get 1200 once I convert both$$d$$and $$\lambda$$ into metres. Is this the right thing to do? I'm not sure this is the amount that cover the central maximum, what does this bit mean?

- - - Part b)

Homework Statement

Part b) Calculate the angle corresponding to the first diffraction minima

Homework Equations

$$sin\theta= ((2m-1)\lambda)\div2)\div d$$

The Attempt at a Solution

Taking m as 1 because it is the first diffraction minima

Reduces to:

$$\theta = arcsin[ (5\times10^{-7}) / (6\times10^{-4}) ]$$

I get $$\theta$$ = 0.048 degrees

is this correct?

There is part c) but haven't attempted it yet, overall the whole question is allocated 14 marks.

 

[Mr confusion]

we need to find the range of the diffraction central bright band first by using 1st minima condition a sin theta= lambda(m=1 for first minima) and a=slit width
then we determine the condition for interferance maxima (in the same angular range)by using d sin theta= m lambda where d= distance between two slits,m= 1,2,3... for the right half.
dividing these 2 equations , we will get m which will indicate the number of interferance bright bands within the first diffraction band ON ONE SIDE ONLY.
remember that we still need to add the left half and the central bright interferance band.
in general , the number of bright bands will be 2m+1.