Monochromatic light of wavelength λis incident on a pair of slits separated by 2.40 ×10^-4 m, an

Monochromatic light of wavelength λis incident on a pair of...

Monochromatic light of wavelength $\lambda$ is incident on a pair of slits separated by $2.40 \times 10^{-4} \mathrm{m},$ and forms an interference pattern on a screen placed 1.80 $\mathrm{m}$ away from the slits. The first-order bright fringe is 4.52 $\mathrm{mm}$ mm from the center of the central maximum. (a) Draw a picture, labeling the angle $\theta$ and the legs of the right triangle associated with the first-order bright fringe. (b) Compute the tangent of the angle $\theta$ associated with the first-order bright fringe. (c) Find the angle corresponding to the first-order bright fringe and compute the sine of that angle. Are the sine and tangent of the angle comparable in value? Does your answer always hold true? (d) Calculate the wavelength of the light. (e) Compute the angle of the fifth-order bright fringe. (f) Find its position on the screen.

Key Concepts

Double-Slit Interference

This concept describes the phenomenon where coherent light passes through two closely spaced slits and produces an interference pattern on a screen. The overlapping light waves from the two slits interfere constructively, creating bright fringes, or destructively, creating dark fringes, which can be predicted using the interference condition d sin? = m?.

Constructive Interference Condition

In a double-slit experiment, bright fringes occur when the path difference between the waves from the two slits is an integer multiple of the wavelength. This is expressed mathematically as d sin? = m?, where m is the order of the fringe. This relationship is fundamental for determining the positions and angles of the bright fringes on the screen.

Trigonometric Relations in Interference

Trigonometry is used to relate the physical geometry of the experiment to the interference pattern. For example, the tangent of the angle (tan?) can be determined from the ratio of the fringe displacement to the screen distance (tan? = y/L), and for small angles, sin? is nearly equal to tan?, simplifying calculations.

Small Angle Approximation

When the fringe displacement is much smaller than the distance to the screen, the angle ? is small. Under these conditions, the small angle approximation (sin? ? tan? ? ? in radians) can be used to simplify the mathematical analysis of the interference pattern without significant loss of accuracy.

Fringe Position and Geometry

This concept involves understanding the geometric relationships in the experimental setup, including drawing right triangles with the screen distance and fringe displacement to find angles. It also encompasses converting between angular measurements and linear positions on the screen, essential for calculating fringe positions and corresponding light wavelengths.

Transcript

00:01 Part a of our question says to draw a picture labeling the angle theta and the legs of the right triangle associated with the first order bright fringe.

00:08 So what i have drawn here is that right triangle in green.

00:13 The distances between like the screen as well as between the first fringe and the starting point, y1, as well as the distances between the two slits d are all labeled in red.

00:25 So that'll be our drawing or kind of the overhead view of what's going on.

00:31 Here in this picture.

00:33 For part b, it says to compute the tangent of the angle theta associated with the first order bright fringe.

00:39 So if we go back here, you can see that the tangent, which is the opposite over the adjacent of that angle theta 1, are all values that we have.

00:49 So the tangent of theta 1, which is what we're asked to compute by our trigonometric identities of a right angle is y1 over l.

01:02 We were given y1 and we were given l.

01:04 So we're just going to plug those values into this expression and we find that the tangent of that angle is equal to 2 .5 .1 times 10 of the minus 3.

01:22 We can box that in as our solution for part b.

01:27 Part c says that it wants us to find the angle corresponding to the first order bright fringe and compute the sign of that angle.

01:39 Are the sign and the cosine, or excuse me, are the sign and the tangent of the angle comparable in value? it says, does your answer always hold true? okay.

01:49 So for theta 1, all we have to do is take the inverse tangent of the value that we just found, right? so the inverse tangent of y1 over l, and we find that theta 1 is equal to 0 .144 degrees.

02:12 And we can box it in as our solution for c.

02:23 Or at least for the first part of c.

02:24 Next, it asks us to compute the sign of the angle theta 1 that we just found.

02:31 Plugging that value in, we find that this is equal to 2 .51.

02:36 Oh, i forgot the 1 up here.

02:38 So this is 2 .51 times 10 to the minus 3.

02:43 This is 2 .51 times 10 to the minus 3, which is approximately the same as the tangent of theta 1.

02:55 So the answer to that question is, yes, they are comparable in value, but only for small values of the angle theta 1.

03:17 And i'm going to have to draw out theta 1 here.

03:22 And that's the last part of our solution for part c...