🚨 Hurry, space in our FREE summer bootcamps is running out. 🚨Claim your spot here.

Like

RC
Numerade Educator

Like

Report

Single Aperture - Example 2

In physics and optics, diffraction is the phenomenon in which waves, such as light or sound waves, spread out as they pass through a narrow opening, or aperture. The spreading occurs only for waves whose wavelength is comparable to or larger than the dimensions of the aperture.

Topics

No Related Subtopics

Discussion

You must be signed in to discuss.
Top Educators
LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Recommended Videos

Recommended Practical Videos

Recommended Quiz

Physics 103

Create your own quiz or take a quiz that has been automatically generated based on what you have been learning. Expose yourself to new questions and test your abilities with different levels of difficulty.

Recommended Books

Video Transcript

welcome to our second example video Looking at single aperture diffraction in this video, we'll continue to look at the single slit experiment. In this case, we're going to say that we have found at an angle of 15 degrees our first dark fringe. So angle is 15 degrees here. Given that we have a distance behind the filter of 90 centimeters, we have a with a equal to one are sorry 10 microns. So we have with the 10 microns of our aperture we have our length behind. We have 15 degrees here and we would like to find out what is going to be on the width of our central bright friends. So from here, Thio here will be our with. Now we've got a couple equations that we can consider here first we have data M which is equal to M times Lambda over a. You can see here knowing data and a we could solve for Lambda. We also know we have y m is equal to m times Lambda L over A. In this case, we have two unknowns. We don't know why and we don't know Lambda though we could find what why is because what we have is a triangle with one side n one angle, which means that up here we will have this as 90 centimeters divided by the cosine of 15 degrees. And this will give us our our high pot news here. Or we could do the opposite side and say that we have 90 centimeters. Remember this tangent of data which is equal to opposite over adjacent, So it will be 90 centimeters multiplied by tangent of 15 degrees will give us our Y side over here. So either calculation we could find the high partners or we could find this side over here. This side's a little more helpful partially because we know it is going to be two times this is going to give us our with so with will be equal to two times uh, 90 centimeters well applied by tangent of 15 degrees. Now, we could also go back to what was given in previously, which is that we had two times lambda l over a notice in this case that the only missing number that we have is l or Lambda. Sorry, so we could go back and solve for Lambda. But when you do this, make sure that you put data in terms of radiance. Okay, so theta must be and radiance. You can't just leave it as 15 degrees when you do this calculation. But when you do this, you should come up with the exact same answer as what we just found here.