Your physics study partner tells you that the width of the...

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength $\lambda$ onto a very narrow slit of width $a$ and measure the width $w$ of the central maximum in the diffraction pattern that is produced on a screen 1.50 m from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. (a) If $w$ is inversely proportional to $a$, then the product $aw$ is constant, independent of $a$. For the data in the table, graph $aw$ versus $a$. Explain why $aw$ is not constant for smaller values of $a$. (b) Use your graph in part (a) to calculate the wavelength $\lambda$ of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) $a$ = 0.78 $\mu$m and (ii) $a$ = 15.60 $\mu$m?

Transcript

00:01 Okay, so for this question, we can use the maximum intensity equation, which is a sine theta is equal to m landa.

00:09 Landa is the wavelength.

00:10 M is just the number.

00:12 Sign theta is the sine function, and a is the width of the slit.

00:17 Therefore, we have tension theta is equal to omega divided 2 over x, which is equal to omega divided by 2x.

00:23 Okay? it's showing the graph here.

00:25 So the x is the distance between the screen and the slit, and omega is the distance on the screen between the two minima on either side of the central maximum.

00:35 Okay? so since the angle theta in this question is very, very small, that means sine theta is approximately equal to theta, which is approximately equal to tangent theta.

00:44 So therefore we have sine theta is equal to omega over 2x.

00:48 Then we have a times omega 2x is equal m blanda.

00:51 So a omega over 2 s is equal to m lunda.

00:56 Then we have a omega is equal to 2x times m londa.

01:01 That said, m is equal to 1, so we have a omega is equal to 2x londar.

01:05 And as you can tell, a omega is equal to some kind of constant, which is 2x londer.

01:14 And this is the graph i drew for this question, which is the graph, a omega versus a.

01:21 And as you can tell, aw is not constant when the value of a is, small.

01:30 But as the value of an increase, you can tell that the value of aw is approximately constant.

01:39 Okay? so therefore, let's jump to question b.

01:47 Well, we know adw is equal to 2x landa.

01:49 So therefore landa is equal to aw over 2x...