Red light of wavelength 630 nm is incident on a pair of slits. The interference pattern is proje

step 1 This is problem number 42 of chapter 23. In this problem, we're asked to draw a diagram of the s

Red light of wavelength 630 nm is incident on a pair of...

Key Concepts

Double-Slit Interference

This concept involves the phenomenon where waves emanating from two coherent sources (the slits) superpose to create a pattern of alternating bright and dark bands on a screen due to constructive and destructive interference. The pattern depends on the wavelength of the light, the separation of the slits, and the distance from the slits to the screen.

Interference Conditions and Path Difference

Constructive interference occurs when the path difference between the two waves equals an integer multiple of the wavelength. This condition (m?, where m is an integer) is crucial for determining the positions of the bright fringes on the screen. The concept of path difference is key to understanding how varying distances traveled by the light from each slit leads to phase differences that result in interference.

Ray Diagram Construction

A ray diagram is a graphical representation used in optics to trace the paths of light from the source through the slits to the screen. It helps in visualizing the geometry of the experiment, including the paths that lead to the constructive and destructive interference patterns due to differences in the distance traveled by light from each slit.

Fringe Spacing

Fringe spacing refers to the distance between adjacent bright (or dark) fringes in the interference pattern. It is determined by the wavelength of the light, the distance to the screen, and the separation between the slits. Understanding fringe spacing is essential for predicting and analyzing the detailed structure of the interference pattern on the screen.

Determination of Slit Separation

By measuring the position of a given bright fringe relative to the central maximum and knowing the wavelength and distance to the screen, one can calculate the separation between the slits. This calculation employs the interference condition and underscores the relationship between the physical geometry of the slit pair and the resulting interference pattern.

Transcript

00:01 This is problem number 42 of chapter 23.

00:05 In this problem, we're asked to draw a diagram of the situation.

00:10 And this situation is a double slit experiment.

00:13 So let's go ahead and draw the double slits.

00:22 And the separation distance between these slits, we're going to denote it with the letter, norcase d.

00:30 And the distance between the slit and the screen, we're going to denote that distance l.

00:35 Let's go ahead and put the screen.

00:51 In this problem, the wavelength used is lambda equal to 630 nanometers.

00:58 So let's go ahead and convert that to meters by multiplying it by 10 to negative 9 meters for every 1 nanometer.

01:07 So it's become 630 times 10 to the negative 9 meters.

01:14 The separation distance between the splits here is d equal to...

01:19 Well, we don't know that one, so that's going to be an unknown.

01:24 We are given the length l here at 6 meters, and we are given the distance to the fourth right fringe, 2 .8 centimeters.

01:39 So let's go ahead and convert that to meters.

01:42 10 to the negative 2 meters for every 1 centimeter.

01:47 So this is 2 .8 times 10 to the negative 2 .2.

01:52 And our units here are meters.

01:56 Okay, so here this distance l must be much further than the separation distance.

02:02 And we have the position of this fourth bright fringe.

02:08 This is the position from the central maximum.

02:10 So the central maximum is going to be given if we look at the center of the separation here.

02:17 We draw a line from there to the screen.

02:20 That's the central maximum.

02:22 This y -sep 4 is a distance from this central maximum to this position so let's say that it's y4 somewhere up here and now we can see an angle formed between this starting location and this y4 so let's go ahead and label this this is going to be y4 and this distance is l the separation distance and then we see that the angle formed between the blue and green line is theta so from the diagram we see that tangent of theta is equal.

03:09 So this is actually theta 4 because it's the fourth order bright fringe.

03:13 This is going to be y4 over l.

03:20 So let's go ahead and take the arc tangent of both sides to see if small angle approximations are valid.

03:33 So in degrees, let's see what this comes out to.

03:35 This is going to be arc tangent of y4 over l.

03:39 So y4 is 2 .8 times 10 to the negative 3.

03:44 Meters and this is why we converted to meters so that when we put the 6 .0 meters here in the denominator the meters castes out with meters so let's plug this into the calculator let's see what we get we get 0 .267 approximately that and we're going to save that onto the calculator in case we need it later but that is a small angle so this is going to be in degrees let me see let me check the mode here yes that's in degrees so so small angle approximations are valid.

04:30 So this is approximately this much degrees.

04:37 Okay, since small angle approximations are valid, let's go ahead and recall the formulas for young's double -slit experiment.

04:44 We have d sine of theta.

04:49 M is going to be equal to m lambda, where m is equal to 0, 1, 2, 3, 4, etc.

05:01 Okay, we can solve this expression for sine of theta.

05:12 Sign of theta m is equal to m lambda over d.

05:20 Now, due to small angle approximations, we can say that sine of theta 4 is approximately equal to tangent of theta 4.

05:38 And this is valid for small angles.

05:41 So here we can say that this is going to be equal to y sub m over.

05:48 L.

05:49 This is for small angle approximation.

05:51 So in this problem that would translate to if we go ahead and substitute the part for sign this is the part for sign.

06:00 And here let me put an approximate here for small angles...

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