Diffraction Question: Find Slit Width & Angle θ

[Amy Marie]

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!

 

[Ivan Karamazov]

I think you may have used an incorrect equation. For destructive interference, the equation should be

dsinΘ = (m + 1/2)λ​

this will yield a different value for the angle.

If I am correct, then I think that the anomaly of you having the correct answer for one component and yet not the other arises because conceptually the difference in slit produces both maximum and minimum fringes and their angles for the same value of d; so by calculating the maximum constructive interference angle you have the same distance as the minimum destructive interference angle.

 

[Amy Marie]

I think you may have used an incorrect equation. For destructive interference, the equation should be

dsinΘ = (m + 1/2)λ​

this will yield a different value for the angle.

If I am correct, then I think that the anomaly of you having the correct answer for one component and yet not the other arises because conceptually the difference in slit produces both maximum and minimum fringes and their angles for the same value of d; so by calculating the maximum constructive interference angle you have the same distance as the minimum destructive interference angle.

I still couldn't get it to work, but thank you!

 

[Miles Whitmore]

I think you may have used an incorrect equation. For destructive interference, the equation should be

dsinΘ = (m + 1/2)λ​

This equation is actually for double-slit destructive interference. $a\sin\theta = m\lambda$ is indeed the correct equation for single-slit minima. Since your slit width is correct, I'm not sure where your error lies since it looks like you calculated the angle correctly...​

 

[Amy Marie]

This equation is actually for double-slit destructive interference. $a\sin\theta = m\lambda$ is indeed the correct equation for single-slit minima. Since your slit width is correct, I'm not sure where your error lies since it looks like you calculated the angle correctly...​

Thank you for taking the time to look at this!