- #1
Amy Marie
- 13
- 0
Homework Statement
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 45.0 cm away from the slit, when light of wavelength 550 nm is used. (a) Find the slit width. (b) Calculate the angle θ of the first diffraction minimum
Homework Equations
a(sin θ) = m(λ)
a = slit width
λ = wavelength
θ = angle between ray and central axis
m = which minimum
The Attempt at a Solution
(a)
sin of angle θ to fifth minimum:
(5)(550 * 10[-9] m)/(a) = 0.00000275/a
By the small angle approximation, 0.00000275/a also gives the angle to the fifth minimum and the tangent of that angle.
Distance from center of diffraction pattern to fifth minimum:
[(0.00000275)(0.45 m)]/a = 0.0000012375/a
Distance from center of diffraction pattern to first minimum:
[(550*10[-9])(0.45 m)] = 0.0000002475/a
Solving for a:
0.000350 m = 0.0000012375/a - 0.0000002475/a
a = 0.002828 m
(b)
(0.002828 m)(sin θ) = 550 * 10[-9] m
sin θ = 0.000194 rad
θ = 0.000194 rad
When I checked, my answer for the slit width was correct, but the angle wasn't.
Can anyone please find what went wrong with the angle in part b?
Thank you!