Question about Intensity in an Interference Pattern

[Haramura]

Homework Statement

Two slits spaced d 0.0720 mm apart are 0.800 m from a screen. Coherent light of wavelength λ passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is 3.00 mm. The intensity at the peak of the central maximum is 0.0500 W/m^2.

What is the intensity at point on the screen that is 1.50 mm from the center of the central maximum?

PS: wavelength calculated: 2.7 *10^-7

Homework Equations

Δ=Dλ/
=0 (cos (πd/λ) * sin θ )^2
sin θ ≈ y/D

The Attempt at a Solution

Δ=Dλ/
3 *10^-3 = 0.8 * λ / 0.072*10^-3
λ = 2.7 *10 ^-7m
this one should be correct coz my attempt in part a (which is a similar question but just change to 2mm from 1.5mm )is correct.
=0 (cos (πd/λ) * sin θ )^2
sin θ ≈ y/D = 1.875*10^-3 (sin θ=0.1074)
(cos (πd/λ) * sin θ )^2
= cos^2(π* 0.072*10^-3 * 1.875*10^-3/2.7 *10 ^-7)
= 0^2
I=0
in part (a) I used degree but it's wrong, then I used radian and got the correct ans. but now I'm wrong again.
even I use actual no. of sin θ=0.1074, I still don't get the correct ans.

besides using maths calc, as I remember destructive interference happens in the middle of two bright regions, but now I'm unsure about it.

 

[haruspex]

=0 (cos (πd/λ) * sin θ )^2

I don't understand your notation. What is that leading zero?

 

[Haramura]

I don't understand your notation. What is that leading zero?

that should be I = I₀ (cos (πd/λ) * sin θ )^2
and
Δy =Dλ/ d
sorry for not preview first

 

[haruspex]

that should be I = I₀ (cos (πd/λ) * sin θ )^2
and
Δy =Dλ/ d
sorry for not preview first

Ok.
I strongly recommend never plugging in the numbers until the final step. Keep everything symbolic until then. It has many benefits.
For such small angles, you can approximate sin(θ) as θ.
If the first minimum is at 3mm, and that corresponds to angle θ, what angle does 1.5mm correspond to?
For θ to be the first minimum, what must the value of d sin(θ)/λ be?

 

[Haramura]

Ok.
I strongly recommend never plugging in the numbers until the final step. Keep everything symbolic until then. It has many benefits.
For such small angles, you can approximate sin(θ) as θ.
If the first minimum is at 3mm, and that corresponds to angle θ, what angle does 1.5mm correspond to?
For θ to be the first minimum, what must the value of d sin(θ)/λ be?

I try to keep it, but every time I get cos (πd/λ) * sin θ = π/2 and the final result is still 0.
To me, sin θ means to be 1.5mm/ 0.8m.
I used 3mm as Δy in Δy=Dλ/ d, I no longer use it in I = I₀ (cos (πd/λ) * sin θ )^2
or should I use 3mm instead of 1.5mm?

 

[haruspex]

I try to keep it, but every time I get cos (πd/λ) * sin θ = π/2 and the final result is still 0.
To me, sin θ means to be 1.5mm/ 0.8m.
I used 3mm as Δy in Δy=Dλ/ d, I no longer use it in I = I₀ (cos (πd/λ) * sin θ )^2
or should I use 3mm instead of 1.5mm?

Some confusion re θ. I was defining θ as the angle to the first minimum, so that would be 3mm/0.8m.
For that, yes, (πd/λ) * sin θ = π/2.
If θ is the angle to 3mm, what is the angle to 1.5mm?

 

[Dominic Schuh]

I believe the wavelength may be wrong, I got 5.4*10^-7.