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

step 1 So monochromatic light is incident upon a pair of slits. The distance between the slits is 2 .4

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 $\mathrm{m}$ from the slits. The first-order bright fringe is at a position $y_{\text { bright }}=$ 4.52 $\mathrm{mm}$ measured from the center of the central maximum. From this information, we wish to predict where the fringe for $n=50$ would be located. (a) Assuming the fringes are laid out linearly along the screen, find the position of the $n=50$ fringe by multiplying the position of the $n=1$ fringe by 50.0 . (b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum. (c) Using the result of part (b) and Equation $37.2,$ calculate the wavelength of the light. (d) Compute the angle for the 50 th-order bright fringe from Equation $37.2 .$ (e) Find the position of the 50 th-order bright fringe on the screen from Equation $37.5 .$ (f) Comment on the agreement between the answers to parts (a) and (e).

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 $\mathrm{m}$ from the slits. The first-order bright fringe is at a position $y_{\text { bright }}=$ 4.52 $\mathrm{mm}$ measured from the center of the central maximum. From this information, we wish to predict where the fringe for $n=50$ would be located. (a) Assuming the fringes are laid out linearly along the screen, find the position of the $n=50$ fringe by multiplying the position of the $n=1$ fringe by 50.0 . (b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum. (c) Using the result of part (b) and Equation $37.2,$ calculate the wavelength of the light. (d) Compute the angle for the 50 th-order bright
fringe from Equation $37.2 .$ (e) Find the position of the 50 th-order bright fringe on the screen from Equation $37.5 .$ (f) Comment on the agreement between the answers to parts (a) and (e).

Key Concepts

Double-Slit Interference

This concept explains how coherent light passing through two closely spaced slits undergoes superposition, resulting in an interference pattern of alternating bright and dark fringes on a distant screen. It lays the foundation for understanding how waves combine constructively and destructively.

Constructive Interference Condition

This principle describes the condition under which two or more waves meet in phase to yield a bright fringe. For double-slit experiments, constructive interference occurs when the difference in path length between the two waves equals an integer multiple of the wavelength.

Fringe Spacing and Angular Position

Fringe spacing refers to the spatial separation between consecutive bright (or dark) fringes on the screen. The angular position of these fringes, given by trigonometric relationships such as d sin(theta) = n * lambda, is crucial for determining where each fringe will appear in the interference pattern.

Small-Angle Approximation

This approximation simplifies calculations by assuming that for small angles, sin(theta) and tan(theta) are nearly equal to theta in radians. It is widely used in interference and diffraction problems to relate fringe positions on the screen to the angles at which they occur.

Wavelength Determination

Using measured positions of specific fringes and the corresponding angular relationships, the wavelength of light can be computed. This involves applying the interference condition along with fringe positioning equations, highlighting the wave nature of light.

Transcript

00:01 So monochromatic light is incident upon a pair of slits.

00:05 The distance between the slits is 2 .4 times 10 to the negative 4 meters.

00:13 And we have an interference pattern on the screen.

00:16 The screen is 1 .80 meters away.

00:19 So we call that l.

00:20 The first order, so m is equal to 1, bright fringe is at a position of 4 .52 millimeters.

00:30 So from this information, we want to predict where the fringe for the 50th bright line would be.

00:40 So first they ask us to just take our answer for the first one, which is y1, and multiply it by 50.

00:50 So y of 50 would be 50 times 4 .52.

00:59 Oops, is that millimeters? yes.

01:03 So 4 .52 millimeters and that's going to be equal to 226 millimeters.

01:12 Okay.

01:13 And then in part b, that's the answer to part a, find the tangent of the angle of the first order bright fringe.

01:19 So we can do that by basically looking at our trig here.

01:24 If this is our angle here and this is what we call y and this what we call l, the tangent of theta is.

01:32 Just y over l.

01:34 So in this case the tangent of our angle is 4 .52 millimeters over 1 .8, but 1 .8 is in meters, so i need to have the units be consistent.

01:47 So it doesn't matter which one you change.

01:49 I'm just going to choose to change the denominator to 1800.

01:53 So the tangent of my angle, which is what they're looking for, is .00251 or 2 .51, or 2 .51, 10 ,000.

02:04 To 10 to the negative 3.

02:08 So then the next part we are used, we're asked to find the results of part b and equation 372, we need to find the wavelength of the light.

02:16 So equation 372 is just the d -sign theta is equal to m lambda for a right spot.

02:29 So we need to find what the angle is.

02:31 Well, this is the tangent of the angle.

02:33 So if we just do inverse hand on our calculators, the angle itself is 0 .144 degrees using three sig figs.

02:42 So now we want to find the, we want to find the wavelength of this light using this equation.

02:52 So we can plug in our d.

02:54 Our d is 2 .4 times 10 to the negative 4 meters times the sign of the angle, and that's equal to 1 times the wavelength, because this is for the first.

03:10 Bright spot.

03:12 So plugging that into your calculator, just 2 .4 times 10 to the negative 4 times the sign of 0 .144, we have a wavelength of 6 .03 times 10 to the negative 7 meters, which in nanometers is 603...

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