Young's Double Slit Question

[Wheelwalker]

Homework Statement

605-nm light passes through a pair of slits and creates an interference pattern on a screen 2.0 m behind the slits. The slits are separated by 0.120 mm and each slit is 0.040 mm wide. How many constructive interference fringes are formed on the screen? (Many of these fringes will be of very low intensity.)

Homework Equations

d*sin(theta)=m*lambda

The Attempt at a Solution

By solving the above equation for m and setting sin(theta) equal to one, I can find the number of fringes on one side of the central maximum, then multiply that number by two and add one (for the central maximum).

So, m=(d*sin(theta))/lambda=((0.120*10^-3)*(1))/605*10^9=198.35

This means there are 198 fringes on each side of the central maximum, correct? So, 198*2=396. And then, accounting for the central maximum, 396+1=397.

This answer is wrong. The correct answer in the back of the book is 265 fringes.

 

[H.fulls]

Have a look at theta again. You can't just set sin theta to 1.. Have a look at a diagram of the experiment to see what theta actually corresponds to