00:01
So this is a young dog that experiment.
00:05
So we need to get the conditions for destructive and constructive interference.
00:12
So for destructive interference, which is also the minimum, okay? so this is theta minimum, that's just m, oops, just m plus half, lambda.
00:40
Okay, therefore constructive interference which is also the maximum d -sum theta max that's just m -landa so once we have this we can also solve our equation and find n and what is how we find in m and d so this going to be like a simultaneous equation so let's look at the first equation so we're going to have so the angle for minimum is 4 .78.
01:34
So that's sign 4 .78 equals m plus half.
01:43
The wavelength is 650 nanometers.
01:51
This is equation 1.
01:54
Equation 2, which is this one, means that's d.
02:02
And the angle for maximum is 4 .1.
02:08
All right equals m times 650 nanometers.
02:20
So we got two equations.
02:23
We need to solve simultaneously to find m &d.
02:27
So what i'm going to do is to divide these two equations.
02:33
I'm going to do, for example, let's just do equation 1 divided by equation 2.
02:46
Now when we do that, these are going to cancel out.
02:50
So we're going to have sine 4 .78 divided by sine 4 .1.
03:01
Of course, the 650s are going to cancel out.
03:04
We're going to have m plus half divided by m.
03:10
So this is how we find m...