# Interference Intensity

1. Nov 2, 2015

### Isaac Pepper

1. The problem statement, all variables and given/known data
Monochromatic light of wavelength 463 nm from a distant source passes through a slit that is 0.0350 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ = 0∘) is 9.20*10-5W/m2 .
What is the intensity at a point on the screen that corresponds to θ = 1.20∘.
2. Relevant equations
I=I0*[sin(πa(sinθ)/λ) / (πasinθ/λ)]2

3. The attempt at a solution
I = 9.20*10-5 * [sin(π*0.035*10-3*(sin1.20)) / 463*10-9 / π*0.035*10-3(sin1.20)/463*10-9 ] 2
= 2.802*10-8 W.m-2

This is the answer that I have found but unfortunately it has come up as wrong - I have made sure that my calculator is in degrees and not radians, and have tried multiple times to see if I have made a mistake whilst inputting in the calculator.
Any idea where I have made a mistake ?

PS: As I am new could you also give me any insight as to how to write my calculations out more clearly on this website, because I am aware that it is very tedious to read the calculations, thanks !

2. Nov 3, 2015

### BvU

So what do you get for $\ {\displaystyle{ \pi \, a \sin\theta\over \lambda}}\$ ?

You can use LaTeX by putting latex source between $\#\#$ or between 

A good way to learn is to use the right mouse button while over a typeset formula and select "Show Math As | TeX Commands"​

 your "I have made sure that my calculator is in degrees" makes me suspicious about $\ \ \sin \left ( \displaystyle { \pi \, a \sin\theta\over \lambda} \ \right ) \ \$ !! 0.087 is dead wrong !

Oh and for such small angles $\sin\theta \approx \theta$ and you can leave your calculator on radians

Last edited: Nov 3, 2015
3. Nov 3, 2015

### Isaac Pepper

For ${ \pi \, a \sin\theta\over \lambda}$
I have got :
π*0.035*10-3*sin(1.2) / 463*10-9 ≈ 4.97

which would therefore give me$\ \ \sin \left ( \displaystyle { \pi \, a \sin\theta\over \lambda} \ \right ) ≈ -0.97 \ \$And if I apply the whole formula : I=I0*[sin(πa(sinθ)/λ) / (πasinθ/λ)]2

Is that correct?

Sorry, about the problems using LaTeX, I'm still getting used to it and am trying to edit them out!

Edit : The answer is indeed correct, thank you ! I was just confused as to why I needed to use radians instead of degrees, I think I understand but an explanation would still be welcome, thank you <3

Last edited: Nov 3, 2015
4. Nov 3, 2015

### BvU

If you are really a beginner, you are doing quite well with LaTeX.

The natural unit for angles is radians. (even in a miscreant like Excel !).

In the expression for $I/I_0$ the $\displaystyle { a \sin\theta\over \lambda} \$ is a phase difference in number of wavelengths - so in natural units, not in degrees. See also here (but this is a calculation for a minimum).