00:01
Okay, so we want to find path differences here, our phase differences here.
00:05
And so phase difference, phi, is given by 2 pi over lambda times d sine theta.
00:15
And if we're going to use a small angle approximation, tan theta will be approximately equal to sine theta, will be approximately equal to theta.
00:24
And so thine theta, therefore, will be approximately equal to y over l.
00:29
That makes this 2 pi over lambda times d, y over l.
00:37
All right, so let's work out each of the parts, part a.
00:41
It's 2 pi over 500 nanometers times d is 1 .2 times 10 to minus 4 meters.
00:53
Why, in this case, we're given a degree, so sign 0 .5 degrees.
01:04
Over l or i guess l is y over l is sine theta so this gives us 13 .2 radiance because the phase difference is in radiance when calculated like this.
01:23
Part b again you have 2 pi over 500 nanometers.
01:31
Now you don't have a angle instead of y over l...