Question

A double-slit pattern is observed on a screen 3m from the slits. Given that the light is incident normally and has a wavelength of 390nm, what is the minimum slit seperation for a point 3.8mm from the centre of the middle bright fringe to be
A) A maximum
b) a minimum

Answer 1

What have you got so far?

Answer 2

I feel like I'm missing data. I can't use \[dsin \theta= x _{1}-x2\]

Answer 3

the question doesn't even specify if they are in phase or out of phase :/

Answer 4

Who are 'they' ? And no, you are not missing data, you just need to think a little - I'm sure, you can do it..

Answer 5

the two light rays... if they are out of phase the equation will change.

Answer 6

Its okay I figured it out :) thanks thoug

Answer 7

Well, I'd claim the problem implies, there is only one light source behind the double-slit plate.

Also, if different areas of the source are emitting with different phases - independent from one another - which is the case for most sources, the phases of the resulting waves in slit 1 and slit 2 will vary (statistically). However, this will _not_ mess with the intensity of the wave at a given point on the screen, given the variations are synchronous, cause the phase-difference of the outgoing waves (Huygens principle) from S1 and S2 is time-invariant. In this case the two slits are forming two coherent light-sources that will show a time-independent interference-pattern on the screen.

(Sry, english is a 2nd language - hope that makes sense...)

Answer 8

You did? Congratulations ;)