Double slit experiment problem help

[MarcL]

Homework Statement

Two narrow slits are .6 mm apart. The dark fringe of order 1 is 1.7 degrees from the central bright fringe. What is the wavelength of the light?

Homework Equations

D sinθ = (m + 1/2)λ --> chosen because it was destructive
m= order
d= distance between slit

The Attempt at a Solution

.6 (sin 1.7) = (1.5)λ
(6*10^-4)(sin1.7) /(1.5)=λ

I get , in nm ( after multiplying the answer I get 1*10^9) and I get an answer of 396666 nm. I plugged it in a few times in my calculator and I keep getting that answer. The website keeps saying it is wrong too, so I'm kinda lost

 

[PeterO]

Homework Statement

Two narrow slits are .6 mm apart. The dark fringe of order 1 is 1.7 degrees from the central bright fringe. What is the wavelength of the light?

Homework Equations

D sinθ = (m + 1/2)λ --> chosen because it was destructive
m= order
d= distance between slit

The Attempt at a Solution

.6 (sin 1.17) = (1.5)λ
(6*10^-4)(sin1.17) /(1.5)=λ

I get , in nm ( after multiplying the answer I get 1*10^9) and I get an answer of 396666 nm. I plugged it in a few times in my calculator and I keep getting that answer. The website keeps saying it is wrong too, so I'm kinda lost

I think the formula is D sinθ = (m - 1/2)λ

EDIT: Plus is the angle 1.7 or 1.17 degrees?

 

[rude man]

Well, what can m range over? The first dark fringe is the 1st-order fringe.

Plus, I ran your numbers, wrong though they are, and got a very different number for lambda.

 

[rude man]

I think the formula is D sinθ = (m - 1/2)λ

Depends on the range of m ...

 

[MarcL]

Well I don't think it is ( I was re-checking my notes and book ) but even if it was the answer is still wrong so I'm messing up somewhere :/

 

[PeterO]

Well I don't think it is ( I was re-checking my notes and book ) but even if it was the answer is still wrong so I'm messing up somewhere :/

Did you notice my edit? is the angle 1.7 or 1.17 degrees?

 

[MarcL]

Well I just ran it again and, again, I got 3.97 * 10^5 nm again ( or 3.97 *^10^-4 m)

 

[MarcL]

Oh sorry it is 1.7 oops, I plugged in 1.7 in my calculator.I must've done a typo I'm sorry!

 

[PeterO]

Well I just ran it again and, again, I got 3.97 * 10^5 nm again ( or 3.97 *^10^-4 m)

Is that correct or incorrect ?

What happens if you you use your formula and m = 0 ?

What happens if you use my formula with m = 1 ?

 

[PeterO]

Well I just ran it again and, again, I got 3.97 * 10^5 nm again ( or 3.97 *^10^-4 m)

Is your calculator set to radians or degrees ? [just grasping at straws]

 

[MarcL]

Radians :).

Plus if I use my formula with 0, I would get (6*10^-4)(sin 1.7)/ .5 = 1.19 *10^-3 m --> 1.19*10^-6 nm ( still wrong) and using your formula I get... the same thing because 1- 1/2 gives me .5 still... no?

 

[rude man]

Radians :).

Plus if I use my formula with 0, I would get (6*10^-4)(sin 1.7)/ .5 = 1.19 *10^-3 m --> 1.19*10^-6 nm ( still wrong) and using your formula I get... the same thing because 1- 1/2 gives me .5 still... no?

Time to replace the battery in your calculator? The expression above is correct - both of them!

 

[PeterO]

Radians :).

Plus if I use my formula with 0, I would get (6*10^-4)(sin 1.7)/ .5 = 1.19 *10^-3 m --> 1.19*10^-6 nm ( still wrong) and using your formula I get... the same thing because 1- 1/2 gives me .5 still... no?

Well if your calculator is set to radians, how are you hoping to get the sin of 1.7 degrees ?