∙A diffraction grating has 5580 lines/cm. When a beam of...

$\bullet$ A diffraction grating has 5580 lines/cm. When a beam of monochromatic light goes through it, the second pair of bright spots occurs at 26.3 $\mathrm{cm}$ from the central spot on a screen 42.5 $\mathrm{cm}$ past the grating. (a) What is the wavelength of this light? (b) How far from the central spot does the next pair of bright spots occur on the screen?

$\bullet$ A diffraction grating has 5580 lines/cm. When a beam of
monochromatic light goes through it, the second pair of bright
spots occurs at 26.3 $\mathrm{cm}$ from the central spot on a screen
42.5 $\mathrm{cm}$ past the grating. (a) What is the wavelength of this
light? (b) How far from the central spot does the next pair of
bright spots occur on the screen?

Key Concepts

Diffraction Grating Equation

The diffraction grating equation, d·sin? = m·?, is fundamental in analyzing how waves interfere when they pass through multiple slits. It relates the spacing between the grating lines (d), the diffraction angle (?), the order of the maximum (m), and the wavelength of light (?). Understanding this relationship is essential when determining one parameter given the others in diffraction experiments.

Grating Spacing and Lines per Unit Length

The grating spacing is inversely related to the number of lines per unit length, meaning that a higher line density results in a smaller spacing. This spacing is critical as it directly influences the angles at which the interference maxima occur. Converting lines per unit length to actual distance between the slits is a basic step in applying the grating equation.

Geometric Interpretation of Diffraction Patterns

The position of the diffraction maxima on a screen can be related to the diffraction angle using geometric relationships. By considering the distance between the central maximum and the higher-order spots along with the distance from the grating to the screen, trigonometric functions (like tangent) are used to determine the angle of deviation. This bridging of physical measurements and theoretical angles is crucial to solving problems in optics.

Diffraction Orders

Diffraction orders refer to the series of bright fringes observed in a diffraction pattern, labeled by an integer m. The order number signifies the number of wavelengths by which the path lengths differ. Higher orders typically appear at larger angles from the central maximum, and each order provides additional information about the light's wavelength and the grating structure.

Transcript

00:01 Okie -dokey.

00:03 And this problem, we have a diffraction grading that's 5580 lines per centimeter.

00:17 We know that the second bright spot, so i'm going to call that y2, occurs at 26 .3 centimeters away from the central bright spot.

00:30 And we know that the screen is a distance of 42 .5 centimeters away.

00:40 We want to know what is the wavelength of the light that's going through that grading.

00:46 And then what's the distance from the center bright spot to the third bright spots? so now in this case, unlike we've been dealing with for a lot of our diffraction problems, the y distances and r are very similar to each other.

01:03 So these are going to be big angles.

01:05 So we can't use the assumption that sine theta is approximately tan theta is approximately theta.

01:15 That's not valid here.

01:18 So we need to actually go in and look at what our angle is going to be.

01:24 So this is my diffraction grading and my screen is over here.

01:29 This is r.

01:31 We have our bright spot one.

01:33 Bright spot.

01:35 Bright spot.

01:36 Bright spot.

01:38 So my second bright spot, y2, is going to make an angle theta 2.

01:50 So the tangent of that angle is opposite over adjacent.

01:56 So that's y2 over r.

02:02 So theta is just the inverse tangent of y2 over r.

02:12 So now we can use that with our bright spot equation, d sine theta, is equal to n lambda.

02:23 So if i plug in for theta, we get d sign of inverse tangent y2 over r.

02:37 I like to leave things symbolizing.

02:40 For everything personally just because it helps mitigate rounding errors.

02:47 Okay, so we're looking for lambda first so that we just need to divide all this by m and that gives us our equation for lambda.

02:59 So let's hop over to another page.

03:04 All right, so where we left off? we said that lambda was equal to d sine of inverse tangent y over r all divided by.

03:27 So if i plug in my numbers, d is the reciprocal of the, like, lines per centimeter or whatever.

03:39 So for every one centimeter, there are 5580 lines.

03:48 So notice this has a unit of a distance now, so that's the width of our slits in our grading.

03:56 Sign of inverse tangent...

Please give Ace some feedback