Single slit diffraction: brightness ratio and 6th dark fringe distance

[hraghav]

Homework Statement

Light of wavelength λ = 517.4× 10^−9 m is incident on a metal plate that has two thin, parallel slits etched into it a distance of d = 5.44 × 10^−6 m apart. A distance of L=1.29 m away lies a viewing panel onto which the diffraction pattern can be displayed.
The metal plate with the two slits is replaced with a metal plate with a single slit that has a width of a= 3.96 × 10^−6 m. a) What is the ratio of the brightness at y = 1.26 m (from the central fringe) to the brightness of the central fringe?

b) The metal plate with the two slits is replaced with a metal plate with a single slit that has a width of a = 7.96 × 10^-6 m. How far from the central fringe is the 6th dark fringe (destructive interference)?

Relevant Equations

a) I = Io(sinβ / β)^2
β = (pi * a*sinθ) / λ
sinθ = y / L

b) asinθ=mλ

a) sinθ = y / L where y = 1.26 m and L = 1.29m
sinθ = 0.9767

β = (pi * a*sinθ) / λ where λ = 517.4 * 10^-9 m and a= 3.96 × 10^−6 m
β = 0.000012151 / 517.4 * 10^-9 = 23.485

I / Io = (sin(23.485) / 23.485)^2
I / Io = 0.0002879
which is 0.2879 * 10^-3

b) asinθ=mλ where m = 6 and λ = 517.4 * 10^-9 m
sinθ = 0.000003104 / a where a = 7.96 × 10^-6 m
sinθ = 0.39

y = Lsinθ. where L = 1.29 m
y = 0.5031 m
which is 50.31 * 10^-2 m

Both these answers are incorrect. Could someone please let me know where am I making an error?

 

[TSny]

Relevant Equations:
sinθ = y / L

Do you have the correct trig function here?

 

[TSny]

Homework Statement:
Relevant Equations: a) I = Io(sinβ / β)^2
β = (pi * a*sinθ) / λ

When you use this formula, is $\beta$ in degrees or in radians?

 

[hraghav]

When you use this formula, is $\beta$ in degrees or in radians?

I got my error in part a but still can't figure out part b

 

[TSny]

When you use the formula for obtaining the angle $\beta$, the angle will be in units of radians. So, be careful with the units.

Did you catch your mistake in writing $\sin \theta = y/L$?

EDIT: Never mind, you said you got part (a) correct. So, you must have made these corrections. In part (b) did you make the correction in $y = L \sin \theta$? Also, be careful if you've been switching your calculator back and forth between radian and degree mode.

 

[hraghav]

When you use the formula for obtaining the angle $\beta$, the angle will be in units of radians. So, be careful with the units.

Did you catch your mistake in writing $\sin \theta = y/L$?

EDIT: Never mind, you said you got part (a) correct. So, you must have made these corrections. In part (b) did you make the correction in $y = L \sin \theta$? Also, be careful if you've been switching your calculator back and forth between radian and degree mode.

Yes I understood what you mean, Thanks I got the correct answer