Inteference Fringes of Double Slit Experiment in Water

[mitchy16]

Homework Statement

Suppose a double-slit experiment is immersed in water (with an index of refraction of 1.33). When in the water, what happens to the interference fringes?

Homework Equations

λ = λ₀ / n
y = (λmL) / d
d = distance between slits
L = distance to viewing screen
n = index of refraction

The Attempt at a Solution

So the wavelength in the water would be:
λ_water = λ_air / 1.33
1.33λ_water = λ_air

And then:
y_water = ( (m) (1.33λ_water) (L) / d )

The answer is supposedly they will be more closely spaced, but I am not sure why that is correct because wouldn't the distance be 1.33 times that of the original distance?

 

[jtbell]

λ_water = λ_air / 1.33

According to this equation the wavelength in water is less than the wavelength in air, no?

 

[jtbell]

y_water = ( (m) (1.33λ_water) (L) / d )

What you're saying here is the same as y_water = ( (m) (λ_air) (L) / d ).

If you want the value of y in water, you need to use the value of λ in water, instead.

 

[mitchy16]

What you're saying here is the same as y_water = ( (m) (λ_air) (L) / d ).

If you want the value of y in water, you need to use the value of λ in water, instead.

Yes! Thank you, I realize my mistakes now! I retried and it worked out, I appreciate the help!