# Wave interference-double slit

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1. Jul 23, 2015

### Hannahj1

1. The problem statement, all variables and given/known data
A student performing Young's experiment with a single-colour source finds the
distance between the first and the seventh nodal lines to be 6.0 cm. The screen
is located 3.0 m from the two slits. The slit separation is 2.2 x 10^2 nanometres. Calculate
the wavelength of the light.

2. Relevant equations
x/L = wavelength/ d

3. The attempt at a solution
I found a solution online and they made the distance 2.2 x 10^-4 metres instead of 2.2 x 10^-7 if you do the conversion. I think they multiplied by something but i don't know why they would need to.

2. Jul 23, 2015

### Dr. Courtney

The slit separation of 2.2 x 10^2 nanometres does not make any sense.

3. Jul 24, 2015

### blue_leaf77

Why not? If you are referring to the practical limitation, we already have technology to fabricate sub-micro structure with a very good precision.
What number did you get using the original value of slit separation? I believe the number you got won't qualify to being called "light". Changing the slit separation by three orders of magnitude will make the wavelength lies in the visible region, which is in conform with the calling of the illuminating EM wave as light.

Last edited: Jul 24, 2015