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When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is \(296 \mathrm{nm}\). What is the vacuum wavelength of the light in \(\mathrm{nm} ?\)

Short Answer

Expert verified
The vacuum wavelength of the light is approximately 525.44 nm.

Step by step solution

01

Understanding Interference Conditions

Destructive interference in thin films occurs when the path difference between rays leads to a half-wavelength phase difference, specifically when \(2nt = (m + 1/2)\lambda_0\), where \(n\) is the refractive index, \(t\) is the thickness, \(m\) is an integer (order of interference), and \(\lambda_0\) is the wavelength in vacuum.
02

Identifying Parameters

From the problem statement, you have the refractive index \(n = 1.33\) and film thickness \(t = 296 \ \mathrm{nm}\). The thickness \(t\) corresponds to the second smallest nonzero thickness resulting in destructive interference, implying \(m = 1\).
03

Applying the Interference Condition

Using the condition for destructive interference for a soap film in air (with air film-air medium interfaces), and considering the film is single n-type, we'll set up the equation \(2nt = (m + 1/2)\lambda_0\), substituting the known values: \(2(1.33)(296 \ \mathrm{nm}) = (1 + 1/2)\lambda_0\).
04

Solving for Vacuum Wavelength \(\lambda_0\)

Substitute \(t\) and \(m\) into the equation and solve for \(\lambda_0\): \[2(1.33)(296 \ \mathrm{nm}) = \left(1 + \frac{1}{2}\right)\lambda_0 \788.16 \ \mathrm{nm} = \frac{3}{2}\lambda_0 \\lambda_0 = \frac{2(788.16)}{3} \ \approx 525.44 \ \mathrm{nm}\]
05

Verifying Solution

Check if the steps logically follow and confirm the result by substitution if necessary. All physical steps and computations are coherent with interference rules and computations, thus validating the solution.

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Key Concepts

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

Destructive Interference
Destructive interference in thin films is a fascinating phenomenon that occurs due to the overlay of light waves. When light hits a thin film, like a soap film, some of it reflects off the top surface, while the rest travels through it, reflects off the bottom surface, and then exits. The key to destructive interference is the condition where the path difference between these two light waves leads them to cancel each other out. This happens when the path difference is a half-wavelength or an odd multiple of a half-wavelength. In simpler terms, this means the peaks of one wave align with the troughs of the other, resulting in complete cancellation of the reflected light at certain thicknesses.
For destructive interference in a thin film, the general condition is given by the formula:
  • \(2nt = (m + 1/2)\lambda_0\)
Here, \(n\) is the refractive index, \(t\) is the film thickness, \(\lambda_0\) is the wavelength of light in a vacuum, and \(m\) is an integer indicating the order of interference. This formula allows us to determine the conditions under which no light is reflected, meaning the film appears darker at these specific thicknesses.
Refractive Index
The refractive index \(n\) is a crucial property that describes how much light slows down as it travels through a medium compared to its speed in a vacuum. For a soap film surrounded by air, the refractive index is given as 1.33, which means that light travels 1.33 times slower in the soap film than in a vacuum. This decrease in speed leads to a reduction in wavelength when light enters the medium, a key factor in achieving the conditions necessary for destructive interference.
Understanding the refractive index helps us to accurately calculate the path difference of light waves reflected from the surfaces of the thin film. Since the speed of light is fundamental to determining how far light waves travel in a given time, the refractive index directly affects the interference patterns we observe.
In our example, the soap film causes the light to slow down, leading to a specific path length increase, which is essential for determining the film thickness that results in destructive interference.
Monochromatic Light
Monochromatic light refers to light consisting of a single wavelength or color. When solving problems involving thin film interference, it simplifies the analysis significantly because we only need to consider how this one wavelength interacts with the film.
The use of monochromatic light is essential in experiments involving interference because it ensures that brightness variations due to interference are clearly visible and not confused by the presence of multiple wavelengths. For instance, a soap film under white light appears colorful because different wavelengths interfere at different thicknesses, forming a spectrum of colors.
By using monochromatic light, such as red or green laser light, we can achieve precise control over the interference process, allowing us to perform accurate calculations involving film thickness and refractive indices like in the exercise at hand.
Vacuum Wavelength
Vacuum wavelength \(\lambda_0\) is the wavelength of light as it would appear in a vacuum, where there is no medium to alter its speed and wavelength. This concept simplifies many calculations because, in a vacuum, the speed of light is constant and at its maximum, i.e., approximately \(3 \, x \, 10^8 \, m/s\).
In our exercise, determining the vacuum wavelength helps to find the conditions under which the soap film causes destructive interference. The calculation is made by considering the second smallest nonzero film thickness for which this interference occurs. Given the refractive index and thickness, we rearranged the interference formula to solve for \(\lambda_0\), leading to a vacuum wavelength of approximately 525.44 nm.
Understanding the vacuum wavelength is essential because it provides a baseline measurement, unaffected by a medium. By comparing it to the wavelength within a medium, we can delve deeper into how materials affect light and how thin films can be utilized to create colorful patterns or reduce glare.

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Most popular questions from this chapter

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is \(\lambda_{\mathrm{A}}=645 \mathrm{nm} .\) The other wavelength is \(\lambda_{\mathrm{B}}\) and is unknown. On a viewing screen, the light with wavelength \(\lambda_{\mathrm{A}}=645 \mathrm{nm}\) produces its third-order bright fringe at the same place where the light with wavelength \(\lambda_{\mathrm{B}}\) produces its fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength?

A soap film \((n=1.33)\) is \(375 \mathrm{nm}\) thick and coats a flat piece of glass \((n=1.52) .\) Thus, air is on one side of the film and glass is on the other side, as the figure illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through air and strikes the film nearly perpendicularly. Concepts: (i) What, if any, phase change occurs when light, traveling in air, reflects from the air-film interface? (ii) What, if any, phase change occurs when light, traveling in the film, reflects from the film-glass interface? (iii) Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum? Calculations: For what wavelengths in the range of 380 to 750 nm does constructive interference cause the film to look bright in reflected light?

How many dark fringes will be produced on either side of the central maximum if light \((\lambda=651 \mathrm{nm})\) is incident on a single slit that is \(5.47 \times 10^{-6} \mathrm{m}\) wide?

A spotlight sends red light (wavelength \(=694.3 \mathrm{nm}\) ) to the moon. At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{m},\) and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.

The wavelength of the laser beam used in a compact disc player is 780 nm. Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

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