The maximum number of bright fringes that can be formed on either side of the central bright fringe in a double slit experiment with a light wavelength of 644 nm and a slit separation of 3.64 × 10-6 m is 5.65, which rounds down to 5 bright fringes. The calculation is based on the formula m = dsin(θ)/λ, where m represents the fringe order. The discussion clarifies that the 6th fringe cannot form as sin(θ) would exceed 1, confirming that only 5 fringes are observable.
PREREQUISITESStudents studying physics, particularly those focusing on wave optics, as well as educators and anyone interested in the principles of light interference and the double slit experiment.
Can the 6th fringe form?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/lamdbaThe 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]
I'm not sure so I'm guessing noehild said:Can the 6th fringe form?
Last time I checked, 5.65<6.Starrrrr said:5.65 which is 6 bright fringes
What would be sin(θ) at the 6th fringe? Is it possible?Starrrrr said:I'm not sure so I'm guessing no