A problem using Young's equation from the double-slit experiment

[erik-the-red]

Question:

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm and the interference pattern is observed on a screen 4.00 m from the slits.

A

What is the width (in mm) of the central interference maximum?

y_m = R \cdot \frac{m \cdot \lambda}{d}

I thought I would get the width of the central interference maximum by doubling y_m = 4.00 \cdot \frac{4.00 \cdot 1 \cdot 4.00 \cdot 10^(-9)}{.00200} = .016.

That's not correct, though.

Why was my approach incorrect?

 

[quickslant]

for central maximum doesn't m=0?

 

[erik-the-red]

But, does y_0 = 0 make sense physically?

 

[rsk]

Your substitution doesn't look right to me

YOu appear to have one too many 4.0s in there.

400 nm is either 400 x 10^-9 or 4 x 10^-7 - you appear to have used 4.0 x 10^-9

2mm is 0.0002m, not 0.002

[Edited to correct should read 0.2 = 0.0002 not 0.002]

Incidentally, I don't know whether that's why you've gone wrong, because either using the extra 4 or not, I can see the answer being 0.016. 4x4 is 16 so divided by 2 should give you an 8 - using an extra 4 would give you a 32 in your answer.

 

[radou]

2mm is 0.0002m, not 0.002

2 [mm] = 0.002 [m].

 

[rsk]

2 [mm] = 0.002 [m].

My typing mistake - he's got 0.2 mm in the original, which gives 0.0002

 

[Doc Al]

Why was my approach incorrect?

It looks like you found the distance of the first maximum (m = 1) from the center and then doubled it. (There are typos in your expression.) That gives you the distance between the two first non-central maxima, which is not the same thing as the width of the central maximum. A good estimate of the width of the central maximum would be the distance between the first minima. (For double slit interference, the minima are centered between adjacent maxima.)

 

[erik-the-red]

Thanks for all the responses. Indeed, my tex equation had numerous typos, I apologize for those.

I was finally able to do the problem by using d \cdot \sin(\theta) = (m + \frac{1}{2}) \cdot \lambda and its corresponding constructive counterpart.

I used the equation y_m = R \cdot \sin(\theta_m), and I got my answers for part A and part B (not shown).