How Many Constructive Interference Fringes Are Formed on the Screen?

AI Thread Summary
The discussion focuses on calculating the number of constructive interference fringes formed by 605-nm light passing through slits separated by 0.120 mm. The initial calculation suggests there are 397 fringes, including the central maximum. However, this is incorrect as it improperly assumes sin(theta) equals one. The correct approach requires a reevaluation of theta to accurately determine the maximum number of fringes, which is stated to be 265 according to the textbook. Understanding the geometry of the interference pattern is crucial for accurate calculations.
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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.
 
Physics news on Phys.org
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
 

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