Calculating the distance between fringes in an alternate universe

[Vitani1]

Homework Statement

In an alternative universe, the Planck’s constant h= 6.625x103 J •s . A resident of the universe, stands in front of a window made of two narrow parallel slits 0.6 m apart and throws tiny golf balls (m = 66.25 g), one at a time, at the slits with a speed of 5 m/s. A wall is 12 m behind the window.
(a)Describe the pattern you expect to see on the wall as the number of golf balls hitting the wall increases.
(b)Calculate the distance between neighboring golf ball fringes on the wall.

Relevant Equations

lambda = h/p

It's straightforward to calculate the wavelength of the balls which is 20,000m. I said that because this is the case and then the pattern must not be a an interference pattern as with electrons.

The second question relies on the formula d = n(lambda)/2. Setting n = 1 for two golf balls will give me 10,000m.

I am asking for help because these are large numbers. Obviously I don't want the answer but just some feedback.

Thanks,

John

 

[TSny]

It's straightforward to calculate the wavelength of the balls which is 20,000m. I said that because this is the case and then the pattern must not be a an interference pattern as with electrons.

Yes, you get a very large wavelength. I'm wondering if the question meant to take $h = 6.625 \times 10^{-3} \rm J\cdot s$ rather than $h = 6.625 \times 10^3 \rm J\cdot s$ .

The second question relies on the formula d = n(lambda)/2. Setting n = 1 for two golf balls will give me 10,000m.

Does $d$ represent the distance between fringes on the screen? If so, I don't understand this equation. Shouldn't the distance between fringes depend on the distance from the slits to the screen?

 

[Vitani1]

Actually I just met with my professor and he made a typo. It is in fact to the power of -3. Also, yes, I agree with you. There is a sin term in the original formula which I set equal to 1 because I assumed the balls were being shot at the screen on a trajectory perpendicular to its length.

 

[Vitani1]

I ended up calculating this angle and using some geometry to find this distance between fringes. Thanks for the help.