0:00
Hi there.
00:01
So for this problem, we are told that fringes are absurd due to monogrammatic light in a mikkelson interferometer.
00:10
Now, we are given that the movable mirror is translated at distance of 0 .07 millimeters.
00:24
So we will have that.
00:26
That is a distance d, 0 .07 millimeters, which we can write as 0 .07 times 10 to minus 3 meters.
00:42
And a shift of 300 fringes is observed.
00:48
So we have that delta m of the fringes is 300.
00:55
Now for this, we need to calculate what is the wavelength of the light.
01:02
Now, what we can do in here is to use the following equation.
01:06
That is that the wavelength is equal to the product between two, the change in the distance, d, and the change in the fringes over the frame on the fringes.
01:22
So we just need to simply substitute those values in here.
01:27
So from this we obtain a wavelength of 4 .67 times 10 to the minus 7 meters that we can also write as 467 nanometers.
02:04
So that's the wavelength of the light...