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Two very narrow slits are spaced 1.80 μm apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with λ = 550 nm? (Hint: The angle θ in Eq. (35.5) is not small.)

Short Answer

Expert verified
The distance between the first and second dark lines is approximately 3.1 mm.

Step by step solution

01

Understand the Problem

We need to calculate the distance between the first and second dark lines in the interference pattern from two slits using the given wavelength λ=550 nm, slit separation d=1.80 μm, and distance to the screen L=35 cm. The dark lines appear where destructive interference occurs, given by the path difference dsinθ=(m+12)λ, where m is an integer for dark lines.
02

Use the Condition for Dark Lines

For the first dark line, m=0 so the condition becomes:dsinθ1=(0+12)λ=12λFor the second dark line, m=1 so:dsinθ2=(1+12)λ=32λ
03

Calculate the Angles θ1 and θ2

Solve for θ1 and θ2 using the equations:sinθ1=12λd=27518000.1528sinθ2=32λd=82518000.4583Find θ1=arcsin(0.1528) and θ2=arcsin(0.4583).
04

Calculate Position of Dark Lines on the Screen

Use y=Ltanθ to find the position y1 for the first dark line and y2 for the second:y1=35×tan(θ1)y2=35×tan(θ2)Compute y1 and y2.
05

Find the Distance Between the Dark Lines

The distance Δy between the first and second dark lines is:Δy=y2y1Calculate the difference between the two positions obtained in Step 4 to find Δy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive interference
When we talk about interference patterns, "destructive interference" is a crucial concept. This happens when two or more waves overlap in such a way that they cancel each other out. The high points of one wave meet the low points of another, resulting in reduced or zero amplitude. This effect is very easy to observe in light waves, with the result being areas that appear dark on a screen.

In the context of our double-slit experiment, destructive interference results in dark lines on the screen where the waves from the two slits perfectly cancel each other out. This occurs when the path difference between the two waves is a half-multiple of the wavelength, specifically dsinθ=(m+12)λwhere:
  • d is the distance between the slits,
  • λ is the wavelength of the light,
  • θ is the angle of the dark fringe from the horizontal,
  • m is an integer (0, 1, 2, ...).
This formula helps us find where the darkness happens, indicating complete wave cancellation on our pattern.
Double-slit experiment
The double-slit experiment, first conducted by Thomas Young, is a famous demonstration of the wave properties of light. In this experiment, coherent light — light of a single wavelength — shines on a pair of closely-spaced slits. As the light passes through these slits, it spreads out or diffracts.

The resulting light waves overlap and interfere with each other, producing distinct patterns on a screen behind the slits. We see a series of bright and dark lines. Bright lines occur from constructive interference, while dark lines, where no light appears, result from destructive interference.

This experiment is not only foundational in understanding wave properties of light but also delves into the principles of superposition, where the resulting wave pattern is a sum of individual wave patterns. As in our problem, by knowing slit spacing and wavelength, interference patterns teach us about wave interactions on a detailed level.
Wavelength calculation
Wavelength calculation is an integral part of solving interference problems. Wavelength, denoted as λ, measures the distance between consecutive peaks of a wave. It's a key parameter that determines where the interference pattern's bright and dark spots will appear.

In calculating where dark lines form in interference patterns (as in our example), we use the formula specific to destructive interference:dsinθ=(m+12)λHere, m is the order of the dark fringe, and knowing the position θ helps us locate these dark spots precisely on the screen.

To find where these lines appear on the screen, converting angles to linear position using y=Ltanθ, where L is the distance from the slits to the screen, is crucial. All these calculations revolve around understanding wavelength, showing its central role in predicting wave interference patterns effectively.

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Most popular questions from this chapter

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at ±35.20 on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at ±19.46 instead. What is the index of refraction of this liquid?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

The index of refraction of a glass rod is 1.48 at T =20.0C and varies linearly with temperature, with a coefficient of 2.50 × 105/C. The coefficient of linear expansion of the glass is 5.00 × 106/C. At 20.0C the length of the rod is 3.00 cm. A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 C/min. The light source has wavelength λ = 589 nm, and the rod initially is at T = 20.0C. How many fringes cross the field of view each minute?

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. When λ0 = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

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