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Chapter 35: Problem 54

35.54. White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

Short Answer

Expert verified
The thickness of the glass plate is approximately 733 nm.

Step by step solution

01

Understand the problem

We need to find the thickness of a glass plate where constructive interference occurs at two different wavelengths when light reflects off its surfaces. The problem gives us two wavelengths: 477.0 nm and 540.6 nm.
02

Define interference condition

For constructive interference with reflection, the condition can be described by the formula: \(2nt = m\lambda\), where \(n\) is the refractive index of glass, \(t\) is the thickness of the plate, \(m\) is an integer, and \(\lambda\) is the wavelength of light in air.
03

Establish equations for two wavelengths

For the wavelength \(\lambda_1 = 477.0\ \text{nm}\), we have the equation \(2nt = m\lambda_1\). For the wavelength \(\lambda_2 = 540.6\ \text{nm}\), the constructive interference could be with \(m-1\), so \(2nt = (m-1)\lambda_2\).
04

Insert refractive index and solve for thickness

Substitute the refractive index \(n = 1.52\) and the given wavelengths into the equations: \(2 \times 1.52 \times t = m \times 477.0\) and \(2 \times 1.52 \times t = (m-1)\times 540.6\). Now solve these simultaneous equations for \(t\) and eliminate \(m\) by setting the left sides equal and subtracting one equation from the other.
05

Solve the difference equation

By subtracting, we find \(m \times 477.0 - (m-1) \times 540.6 = 2 \times 1.52 \times t - 2 \times 1.52 \times t\), which simplifies to \(477.0 - 540.6 = -540.6 + m \times 477.0\). Solving for \(m\), we get \(m = \frac{540.6 - 477.0}{540.6}\). Find \(m\) and use it to find \(t\).
06

Calculate the glass plate thickness

Use \(m\) to find the actual thickness \(t\) with the formula \(t = \frac{m \times 477.0\ \text{nm}}{2 \times 1.52}\). Insert the value of \(m\) calculated.
07

Calculate final thickness

Insert the needed values, solve for \(t\), and confirm that the solution leads to an appropriate thickness based on the described phenomena.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more waves combine to produce a wave with greater amplitude. It's like two people clapping in sync to create a louder sound. For light waves, constructive interference is visible when the peaks of two light waves align perfectly. In our problem with the glass plate, this means that light reflecting from the top surface aligns perfectly with light reflecting from the bottom, amplifying the strength of that particular wavelength.
This phenomenon can be perfectly observed when the difference in path lengths traveled by the two waves is exactly a multiple of the wavelength. The condition for constructive interference is given by the equation: \(2nt = m\lambda \), where \( n \) is the refractive index, \( t \) the thickness of the plate, \( m \) an integer, and \( \lambda \) the wavelength of light in air. This ensures that the path difference aligns the peak of one wave with the peak of the next.
Refractive Index
The refractive index, \( n \), is a measure of how much light bends when it enters a material. It's like the blinkers on a car guiding the light through a new route, slowing them slightly. When light passes from one medium to another, like from air into glass in our problem, its speed changes due to the difference in refractive indices of the two media.
For glass with a refractive index of 1.52, this means light travels 1.52 times slower than in air. This change in speed is crucial for calculating the path length of light waves, which directly affects the condition for constructive or destructive interference. In our glass plate scenario, knowing the refractive index helps us determine the optical path length within the glass, key to solving the interference puzzle.
Wavelength Calculation
To solve the problem, we need to relate the physical path of light within the glass to its wavelength in air. This involves calculating how the wavelength changes due to the refractive index. When light enters a new medium, its speed and wavelength change proportionally according to the refractive index of the new medium.
For a wavelength \( \lambda \) in air to constructively interfere within the glass, we use the equation \( \lambda_{glass} = \frac{\lambda}{n} \). So, the wavelength in glass is shorter than in air. In our exercise, we were given two wavelengths in air (477.0 nm and 540.6 nm), and we used the refractive index to determine how these wavelengths behave within the glass, which is essential for knowing how to set up the interference conditions.
Optics Problem Solving
Solving optics problems like these require a blend of conceptual understanding and mathematical application. Start by thoroughly understanding the given problem and visualize the physical situation - light bouncing inside a thin film. With the right mental picture, apply the equations suitable for the phenomena involved鈥攈ere, the interference condition \(2nt = m\lambda \).
Use any additional given information, values like the refractive index or known wavelengths, to set up equations representing the physical scenario. In problems with two wavelengths, like ours, set up separate interference conditions for each wavelength. Subtract these conditions to eliminate unknowns like \( m \), then solve for your parameter of interest, such as the film's thickness. This systematic approach helps break down complex problems into manageable parts and leads to solutions grounded in physics laws.

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Most popular questions from this chapter

35.52. Red light with wavelength 700 \(\mathrm{nm}\) is passed through a two- slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

35.18. An FM radio station has a frequency of 107.9 \(\mathrm{MHz}\) and uses two identical antennas mounted at the same elevation, 12.0 \(\mathrm{m}\) apart. The antennas radiate in phase. The resulting radiation pattern has a maximum intensity along a horizontal line perpendicular to the line joining the antennas and midway between them. Assume that the intensity is observed at distances from the antennas that are much greater than 12.0 \(\mathrm{m}\) (a) At which other angles (measured from the line of maximum intensity) is the intensity maximum? (b) At which angles is it zero?

35\. 41. Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) 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 \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

35.28. Nonglare Glass. 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. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , 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 \(\mathrm{nm}\) ? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

35.25. Points \(A\) and \(B\) are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a \(12.5-\mathrm{MHz}\) signal horizontally. These transmitters are in phase with other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C\) . directly opposite the midpoint of \(A B\) . The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?
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