Finding separation of bright spots

[omc1]

Homework Statement

A scientist decides to do the double-slit experiment in a vertical tube, with the slits at the top and a viewing screen at the bottom as shown. With red light from a HeNe laser (632 nm) a nice pattern forms on the viewing screen, with the bright spots 1.00 cm apart.
Now the scientist fills the tube up with water. What is the separation of the bright spots now?

Homework Equations

i used v=c/n v=λf y=m(λ)L/d

The Attempt at a Solution

λ=c/nf plugged that into y(initial)=m(λ)L/d, i used 1.33 as the index number. y(final)=m(λ)L/d to find the new separation, the m, L, d cancel out leaving y=λ or y(final)=c/nf c=3x10^8, n=1.33, f=632x10^-9 meters, i got 3.5x10^14 which seems to me to be really large...please help its due tonight, thanks!

 

[rude man]

Optical path difference changes from d sin(theta) to n*d sin(theta) = m(lambda). Careful about lambda in water ...

 

[omc1]

wait, i don't see how that equation applies here? i don't see how sin(theta) works here?

 

[haruspex]

y(initial)=m(λ)L/d, ... y(final)=m(λ)L/d

But the lambdas are different, right?

the m, L, d cancel out leaving y=λ

I don't understand how you arrived at that. y is the band separation, lambda is the wavelength of the light. Why should they be equal?

f=632x10^-9 meters

f is frequency, surely, not wavelength.

 

[rude man]

wait, i don't see how that equation applies here? i don't see how sin(theta) works here?

In air:
If θ_m is the angle subtended at maximum number m, then d sin θ_m = mλ.

In water:
if θ'_m is the angle subtended at maxium number m, then n*d sinθ' = mλ'.

For both air and water, for m = 0 we get the center maximum. For m = 1 we get the first separation.

The ratio of sinθ'_m/sinθ_m is the same as the ratio of screen spacings in air vs. in water.

You know n, λ and λ' so you know the sine ratio and therefore the spacing ratio. Use m = 1 for the spacings for both cases.