Young's double slit experiment with two sets of slits

AI Thread Summary
The discussion focuses on the application of the interference minima formula in Young's double slit experiment with two sets of slits. A participant attempts to derive the distance 'd' for first-order minima but encounters a discrepancy in their solution compared to the expected answer. There is confusion regarding the use of the interference minima condition versus diffraction effects from the first slit. Participants emphasize the need to verify the formulas used, indicating that the initial formula may be incorrect. The conversation highlights the complexities of applying interference and diffraction principles in optical experiments.
Apashanka das
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1. The problem statement, all variables and given/known da
15187116671331187855915.jpg
15187116671331187855915.jpg

Homework Equations


Interference minima is asinΘ=nλ/2 where n=1,3,5,7...
[/B]

The Attempt at a Solution


Putting ab/(2sqrt(b^2/4+d^2))=λ/2 for first order minima
And solving for d I get (b/2)*sqrt((2a/λ)^2-1) in which only the 2 factor differs in front of a in the square root
from the correct ans which is option 2 in the question
I may be wrong but where I didn't get it?
 

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Why did you start with the condition for interference minima? The passage of light through the first slit is diffraction.
 
Apashanka das said:
1. The problem statement, all variables and given/known daView attachment 220386 View attachment 220386

Homework Equations


Interference minima is asinΘ=nλ/2 where n=1,3,5,7...
[/B]
Check the formula in red.
 
ehild said:
Check the formula in red.
yes I have used this formula
 
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