00:01
For this problem, we have a really thin triangle.
00:05
And at some distance, three centimeters, we're seeing the bandwidth for the 420 nanometers.
00:17
So at some further point down, at 680 nanometers, it wants to know how far down we are from the top.
00:26
So we can set up a ratio.
00:28
We can say 3 centimeters over 420 nanometers.
00:34
Is going to be equal to x over 680 nanometers since their thicknesses are related to their wavelengths in a linear fashion.
00:53
Solving that just gives us an x, an x is the distance from the top to the line for red.
01:05
I guess these are the wrong colors, but the 680 is for red.
01:09
And so x is going to be around 4 .86 centimeters.
01:20
Okay.
01:20
And so this angle is going to be a really tight angle.
01:23
So we can figure out theta.
01:26
Let's figure out our thicknesses first.
01:28
So we can use 2nt equals m lambda, or m plus 1ā2 times lambda.
01:41
And since this is the first time they're appearing, our m is going to be zero.
01:45
So our thickness is going to equal 1 .5 times our lambda over 2 times n, which is equal to lambda divided by four times 1 .33, which is 5 .32...