Ch 35: Interference

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 35: Interference

Problem 20a

Chapter 35, Problem 20a

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I₀. What is the distance on the screen from the center of the central maximum to the first minimum

Verified step by step guidance

Step 1: Recognize that this is a double-slit interference problem. The first minimum occurs where the path difference between the light from the two slits is equal to half a wavelength (λ/2). This condition corresponds to destructive interference.

Step 2: Use the formula for the position of minima in a double-slit interference pattern:

y = \frac{m}{d} \cdot \frac{λ}{L}

, where

y

is the distance from the central maximum to the minimum on the screen,

m

is the order of the minimum (for the first minimum,

m = 1

d

is the slit separation,

λ

is the wavelength of the light, and

L

is the distance from the slits to the screen.

Step 3: Substitute the given values into the formula. Here,

d = 0.260 mm = 0.260 x 10 ⁻ 3 m

λ = 660 nm = 660 x 10 ⁻ 9 m

, and

L = 0.900 m

. For the first minimum,

m = 1

Step 4: Rearrange the formula to solve for

y

y = \frac{m}{d} \cdot \frac{λ}{L}

. Substitute the values:

y = \frac{1}{0.260} x 10 ⁻ 3 x \frac{660}{} x 0.900

Step 5: Simplify the expression to find the value of

y

. This will give the distance from the central maximum to the first minimum on the screen. Ensure all units are consistent during the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Double-Slit Experiment

The double-slit experiment demonstrates the wave nature of light through interference patterns created when coherent light passes through two closely spaced slits. The resulting pattern consists of alternating bright and dark fringes on a screen, where bright fringes correspond to constructive interference and dark fringes to destructive interference.

Interference and Path Difference

Interference occurs when two or more waves overlap, leading to a new wave pattern. For the double-slit setup, the path difference between light waves from the two slits determines whether they interfere constructively or destructively. The first minimum occurs where the path difference equals half the wavelength, resulting in destructive interference.

Angular Position of Minima

The angular position of the minima in a double-slit interference pattern can be calculated using the formula sin(θ) = mλ/d, where m is the order of the minimum, λ is the wavelength, and d is the distance between the slits. For the first minimum (m=1), this relationship helps determine the distance from the central maximum to the first minimum on the screen.

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Related Practice

Textbook Question

Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

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Textbook Question

In a two-slit interference pattern, the intensity at the peak of the central maximum is I₀. At a point in the pattern where the phase difference between the waves from the two slits is 60.0°, what is the intensity?

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Textbook Question

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I₀. What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I₀/2?

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Textbook Question

Coherent light of frequency 6.32 × 10¹⁴ Hz passes through two thin slits and falls on a screen 85.0 cm away. You observe that the third bright fringe occurs at ±3.11 cm on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

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Textbook Question

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. If the glass has a refractive index of 1.62 and you use TiO₂, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm?

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Textbook Question

What is the thinnest film of a coating with n = 1.42 on glass (n = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

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Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I0. What is the distance on the screen from the center of the central maximum to the first minimum

Verified video answer for a similar problem:

Key Concepts

Double-Slit Experiment

Interference and Path Difference

Angular Position of Minima

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I₀. What is the distance on the screen from the center of the central maximum to the first minimum