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Hi there.
00:01
So for this problem, we are told that after a laser beam passes through two thin parallel sleeves, the first completed dark fringes occurs at an angle theta that is equal to 19 degrees.
00:16
And with the original direction of the bin, as view on the screen far from the slips.
00:21
So for part a of this question, we are asked about what is the ratio of the distance between the sleeps to the wavelength of the.
00:30
The light illuminating these leads.
00:33
Now, at the first star frames, we have that we can use the following equation, that is the distance d, sine of theta, is equal to the wavelength divided by two.
00:53
Now, what we are going to do in here is to solve for the ratio d over the wavelength.
01:02
So this is going to, to be equal to 1 over 2 times the sign of theta.
01:08
And in this case, we substitute the angle that is 19 degrees.
01:12
So from here, we obtain a value of 1 .54.
01:18
So that's a solution for part a of this problem.
01:24
Now, for part b, we are asked about, what is the smallest angle relative to the original direction of the laser beam, at which the intensity of the light is 1 over 10 the maximum intensity on the screen.
01:43
So we are given the condition that the intensity is equal to the initial intensity divided by 10.
01:52
So the intensity at any angle is given by the following equation, that is i .0, the maximum intensity, cosine to the square of pi times the distance d, sine of theta, divided by the wavelength.
02:13
So what we are going to do in here is to simplify this further.
02:19
As you can see, we can simplify this, because are in both sides of this equation, and we can take the square root of this.
02:26
So we're going to obtain that this is the cosine of pi times this, sine of theta, divided by the angle, by the wavelength...