Maximum Bright Fringes for Double Slit Experiment

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The discussion centers on calculating the maximum number of bright fringes in a double slit experiment with a wavelength of 644 nm and a slit separation of 3.64 × 10^-6 m. The calculation yields a value of 5.65, suggesting that a maximum of 5 bright fringes can be observed on either side of the central fringe. Participants debate whether the 6th fringe can form, concluding that it cannot, as the sine of the angle for the 6th fringe exceeds 1, making it impossible. The consensus is that the correct answer is 5 bright fringes, not 6. Understanding the limitations of the sine function is crucial in this context.
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Homework Statement


At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength 644 nm falls on a double slit whose slit separation is 3.64 × 10-6 m?

Homework Equations


m=dsintheta/lamdba

The Attempt at a Solution


m=(3.64x10e-6)(1)/644x10e-9) = 5.65 which is 6 bright fringes right but the system telling me it's wrong?[/B]
 
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What is the width of each slit? Without this information your question cannot be answered.
 
That's the full question.
 
Starrrrr said:

Homework Statement


At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength 644 nm falls on a double slit whose slit separation is 3.64 × 10-6 m?

Homework Equations


m=dsintheta/lamdba

The Attempt at a Solution


m=(3.64x10e-6)(1)/644x10e-9) = 5.65 which is 6 bright fringes right but the system telling me it's wrong?[/B]
Can the 6th fringe form?
 
ehild said:
Can the 6th fringe form?
I'm not sure so I'm guessing no
 
Starrrrr said:
5.65 which is 6 bright fringes
Last time I checked, 5.65<6.
 
Starrrrr said:
I'm not sure so I'm guessing no
What would be sin(θ) at the 6th fringe? Is it possible?
 
The brightness fringe should not be rounded from the equation. so yes the answer from the equation is 5.65. Which makes the greatest fringe 5 not 6.
 

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