# Homework Help: Double Slit Interferance Intensity

1. Apr 11, 2015

### JJK1503

1. The problem statement, all variables and given/known data

A Helium laser, λ = 588 nm, shines on double-slits separated by 1.80 mm. An interference pattern is observed on a screen at a distance R from the slits. The point C on the screen is at the center of the principal maximum of the interference pattern. The point P is the point on the principal maximum at which the intensity of light is half that of the intensity at C. What is the value of the angle θ?

2. Relevant equations

I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )

small angle approximation sin(θ) = tan(θ) = θ

3. The attempt at a solution

On questions like these I typically find it easiest to pick some dummy value and solve. In this case i set I (initial) to 10.

The center of the center maxima is where intensity is the greatest. At this point θ = 0.
so,
I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )
= 10 * cos^2( (pi * d * sin(0)) / λ )
= 10 * cos^2 (0)
= 10

If I want the θ where I (final) = I(initial0 / 2. I need to find θ where I (final) = 10 / 2 = 5
so (with small angle approximation),
I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )
5 = 10 * cos^2( (pi * (1.8 * 10^-3) * θ) / (588 * 10^-9 )
0.5 = cos^2( (pi * (1.8 * 10^-3) * θ) / (588 * 10^-9 )
0.5 = cos^2 (9617.1204 * θ)
At this point I plug the above equation in to my calculator and have it solve for θ.

According to my calculator θ = 0.03743 deg

This seems like a reasonable result; However, the computer kicks it out as incorrect.

I am NOT looking for someone to give me this answer. However, ANY help with my method is GREATLY appreciated.

Last edited: Apr 11, 2015
2. Apr 12, 2015

### Seigi Yokota

Everything you are doing is correct. You have but one problem at the end though. You are probably typing it in your calculator wrong.

0.5 = cos^2 (9617.1204 * θ)
According to my calculator θ = 0.03743 deg

0.5 = cos^2 (9617.1204 * θ)
sqrt(0.5) = cos (9617.1204 * θ)
cos^-1(sqrt(.5))=9617.1204θ
θ=.004679 deg

You asked me to not supply you the answer but All im doing is being your calculator as you did everything else correctly ;)

3. Apr 12, 2015

### JJK1503

Thank you. I appreciate you taking the time to respond. I am unsure where I made my error. My initial thought is that I might have been in radian mode, but I suspect my result would have been much smaller had that been the case. Whatever the reason; it is reassuring to know that I do indeed understand the concept.

Thanks again.

4. Apr 12, 2015

### BvU

Why don't you just do away with the angle on the calculator in the first place: you know that $\cos{\pi\over 4}=\sqrt 2$ so you get $${\pi d \sin\theta\over \lambda} = {\pi\over 4} \quad \Rightarrow \quad \theta\approx\sin\theta = {\lambda\over 4 d}$$