00:01
So here comes another double -slid interference problems.
00:06
But the unique part of this problem is that we immerse the whole setup in water.
00:14
What does that mean? and if you've already gone, if you're already done with chapter 23, you know that all lights, we have a different wavelengths in water, because water have an index of refraction of 1 .3.
00:34
And because of that, the wavelength is 407 nanometer in air, but it will be different in the water.
00:45
So that is why we're immersing the whole set of water.
00:49
That is what this question wants us to do.
00:51
It wants us to consider this extra index.
00:54
But other than that, this really is our, standard double slit interference problem.
01:03
So let's first write down the equation.
01:10
I always like to draw this graph because i think it's helpful to visualize the setup.
01:19
We have d sine theta equals m lambda.
01:25
But this lambda is not for 70 actually i'm gonna call it lambda w for standing for the wavelength in water and this is the actual lambda of 47 nanometer divided by n so the wavelength will become a lot smaller when it's in water so let us now we are asked to find how far parts of fringes on the screen let us again i'm plugging l and x, sine theta equals, or approximately equals x over l.
02:09
So we can rearrange it...