00:01
Let's calculate the wavelength for question one.
00:04
For question one, for destructive interference, because we have a dark fringe, we can use our formula, d -sign theta, is equal to m -plus -1 -half.
00:15
Solving for lambda, lambda is d -sign theta over m -plus -1 -half.
00:21
Now we substitute our values.
00:24
We have 0 .042 times 10 to the negative 3 times sine of 3 .8 degrees.
00:31
Divided by 5 plus 1 1 half.
00:36
And now we can solve for our wavelength.
00:38
We find our wavelength to be 506 nanometers.
00:44
For question two, what's the fringe separation? our fringe separation is given by lambda l over d...