Double Slit Interferance Intensity

[JJK1503]

Homework Statement

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?

Homework Equations

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

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

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 into 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.

 

[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 degYou asked me to not supply you the answer but All I am doing is being your calculator as you did everything else correctly ;)

 

[JJK1503]

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

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.

 

[BvU]

Check your calculator settings. Does the computer want the angle in radians, degrees or grads ?
Can't be a coincidence that 3743/4679 = 0.800 (but how that ratio comes about is a mystery to me -- should be the inverse ).

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 $${\pi d \sin\theta\over \lambda} = {\pi\over 4} \quad \Rightarrow \quad \theta\approx\sin\theta = {\lambda\over 4 d}$$