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

step 1 Okay, so this problem lays out a lot of our things. The distance between the slits is going to b

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 $50.0 .(\mathrm{b})$ 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 $27.2,$ calculate the wavelength of the light. (d) Compute the angle for the 50 th-order bright fringe from Equation 27.2 (e) Find the position of the 50 th-order bright fringe on the screen from Equation $27.5 .$ (f) Comment on the agreement between the answers to parts (a) and (e).

Key Concepts

Young's Double-Slit Experiment

This concept explains the experimental arrangement where monochromatic, coherent light passes through two narrow, closely spaced slits and produces an interference pattern on a screen. The interference pattern, consisting of bright and dark fringes, demonstrates the wave nature of light and forms the foundation for studying wave interference phenomena.

Constructive Interference Condition

This principle defines the condition for bright fringes in an interference pattern. It states that constructive interference occurs when the path difference between light from the two slits is an integer multiple of the wavelength, commonly expressed as d sin(?) = n?. This equation is crucial for determining the angular positions and therefore the spatial locations of the bright fringes.

Small Angle Approximation

When the angles involved in the interference pattern are very small, as is common in many practical setups, sin(?) and tan(?) can be approximated by ? (in radians). This simplifies the calculations relating angular displacement to fringe position on the screen, making it easier to estimate fringe locations using linear relationships.

Fringe Position Calculation

This concept describes the process of determining the position of interference fringes on a screen based on the geometry of the experimental setup. It involves using the relationship between the angle given by d sin(?) = n? and the distance from the slits to the screen to convert angular positions into linear positions, facilitating the prediction of fringe locations for different orders.

Transcript

00:01 Okay, so this problem lays out a lot of our things.

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

00:15 The distance between the slits and the screen is 1 .8 meters.

00:21 We're also given that the first bright mark, where m is equal to 1, the first bright interference pattern, is going to be at 4 .52 millimeters.

00:42 So for part a, it says, assuming that they're linear, i just multiply that y bright by 50 instead of by 1 and see what you get.

00:53 And so if i multiply 4 .52 millimeters by 50, i'm just going to get 226 millimeters, which is 0 .226 meters.

01:10 And that's for m is equal to 50.

01:14 Okay, so yay, we're done.

01:19 But then we get to the answer choice b where it's like we're going to use all the formulas to figure out what it actually would be.

01:26 So for part b, it wants us to set up to solve for the tangent.

01:36 So we have tangent theta of our first fringe is going to be 4 .52 millimeters divided by 1 .8 meters.

01:53 So that tangent would equal 2 .51 times 10 to the negative third.

02:08 Part c, we're going to use that information to find the wavelength of the light that we're working with.

02:16 And so we need to do d times sine of that theta.

02:20 So i'm going to do, let's do 2 .4.

02:25 Times 10 to the negative 4 times sign of the inverse tangent of 2 .51 times 10 to the negative 3rd.

02:44 Okay, that'll give us the angle, then we're taking the sign of the angle just to kind of keep everything together.

02:50 And that's going to be equal to m, which is 1 times our wavelength.

02:55 So we can get that our wavelength is going to be around 603 nanometers.

03:07 Okay, part d asks us to compute the angle for the 50th order fringe.

03:19 So now that we have the wavelength, we're just going to multiply that wavelength by 50 and divide by our d.

03:26 So we have sine theta is equal to 50 times 603 nanometers, which is times 10 to the negative 9th, over our distance d, which is 2 .4 times 10 to the negative 4...

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