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Chapter 24: Problem 19

A coating is applied to a lens to minimize reflections. The index of refraction of the coating is \(1.55\) and that of the lens is \(1.48\). If the coating is \(177.4 \mathrm{~nm}\) thick, what wavelength is minimally reflected for normal incidence in the lowest order?

Short Answer

Expert verified
Calculation gives the wavelength \(\lambda = 550.34 \times 10^{-9} \) m or \(\lambda = 550.34 \) nm. So, the minimal wavelength for normal incidence reflection in the lowest order is approximately 550.34 nanometers.

Step by step solution

01

Understand types of interference and conditions for minimum reflection

In this problem, we are dealing with destructive interference; that's when the crest of one wave coincides with trough of another leading to minimal reflection. For such an interference to happen in a thin film, the optical path difference between the two waves (one reflected from the top surface and the other from the bottom surface of the film) should be equal to half the wavelength of the light.
02

Refraction Index and Film Thickness

The refractive index (\(n\)) of the coating is given as \(n=1.55\) and the thickness (\(t\)) of the film is given as \(t=177.4\) nm or \(t=177.4 \times 10^{-9}\)m. We'll plug these values into the destructive interference formula \(2 t n = m \lambda\).
03

Calculation

Solve the equation \(2 t n = m \lambda \) for the wavelength. Given \( m=1 \) because it's the lowest order for reflection, \( n=1.55\) for the refractive index of the coating, and \( t=177.4 \times 10^{-9}\) m for the thickness of the film, you have: \[ \lambda = \frac{2 t n}{m} = \frac{2 \times 177.4 \times 10^{-9} \times 1.55}{1} \]

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Key Concepts

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

Destructive Interference
When light waves interact, they can either amplify or cancel each other out. This is known as interference. In the case of thin film interference, we're often interested in destructive interference. It's called destructive because the light waves reduce each other's intensity.
Here's how it works:
  • Two waves reflect from two different surfaces of a thin film.
  • For the waves to cancel out, their crests and troughs must align oppositely.
In destructive interference, the optical path difference between the waves should be equal to half the wavelength. This condition ensures that the waves are out of phase, leading to minimal reflection from the surface.
Refractive Index
The refractive index is vital in thin film interference. It measures how much light bends, or refracts, when entering a different medium. For a coating applied to a lens, there are two refractive indices at play:
  • One for the coating ( n = 1.55 ).
  • Another for the lens ( n = 1.48 ).
When light passes through these materials, the refractive index influences how the waves interfere. A higher refractive index indicates that light travels slower in the material, which is crucial for calculating how much the waves shift and interact as they pass through and reflect back.
Optical Path Difference
The optical path difference (OPD) determines how far light travels in a material compared to air. This difference is crucial for the interference patterns in thin films.
  • The OPD is given by the formula: \( 2tn \), where \( t \) is the thickness of the film, and \( n \) is the refractive index of the material.
  • In our problem, this means calculating based on a coating that's 177.4 nm thick with a \( n = 1.55 \).
The mathematical relationship used for destructive interference can be described as:\(2tn = m\lambda\) where \( m = 1 \) for minimal reflection, indicating the first order of interference. This formula helps us find the specific wavelength where reflection is minimized. It's important because it shows how physical dimensions of materials affect light behavior, making it a core concept in optical applications.

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Most popular questions from this chapter

Ecp Astronomers observe the chromosphere of the Sunt with a filter that passes the red hydrogen spectral line of wavelength \(656.3 \mathrm{~nm}\), called the \(\mathrm{H}_{\alpha}\) line. The filter consists of a transparent dielectric of thickness \(d\) held between two partially aluminized glass plates. The filter is kept at a constant temperature. (a) Find the minimum value of \(d\) that will produce maximum transmission of perpendicular \(\mathrm{H}_{a}\) light if the dielectric has an index of refraction of \(1.378 .\) (b) If the temperature of the filter increases above the normal value increasing its thickness, what happens Lo the transmitted wavelength? (c) The dielectric will also pass what near-visible wavelength? One of the glass plates is colored red to absorb this light.

A beam of monochromatic light is diffracted by a slit of width \(0.600 \mathrm{~mm}\). The diffraction pattern forms on a wall \(1.30 \mathrm{~m}\) beyond the slit. The width of the central maximum is \(2.00 \mathrm{~mm}\). Calculate the wavelength of the light.

Sunlight is incident on a diffraction grating that has 2750 lines \(/ \mathrm{cm}\). The second-order spectrum over the visible range \((400-700 \mathrm{~nm})\) is to be limited to \(1.75 \mathrm{~cm}\) along a screen that is a distance \(L\) from the grating. What is the required value of \(L\) ?

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 is placed \(1.80 \mathrm{~m}\) away from the slits. The first-order bright fringe is \(4.52 \mathrm{~mm}\) from the center of the central maximum. (a) Draw a picture, labeling the angle \(\theta\) and the legs of the right triangle associated with the first-order bright fringe. (b) Compute the tangent of the angle \(\theta\) associated with the first-order bright fringe. (c) Find the angle corresponding to the first-order bright fringe and compute the sine of that angle. Are the sine and tangent of the angle comparable in value? Does your answer always hold true? (d) Calculate the wavelength of the light. (e) Compute the angle of the fifth-order bright fringe. (f) Find its position on the screen.

In a Young's interference experiment, the two slits are separated by \(0.150 \mathrm{~mm}\) and the incident light includes two wavelengths: \(\lambda_{1}=540 \mathrm{~nm}\) (green) and \(\lambda_{2}=\) \(450 \mathrm{~nm}\) (blue). The overlapping interference patterns are observed on a screen \(1.40 \mathrm{~m}\) from the slits. (a) Find a relationship between the orders \(m_{1}\) and \(m_{2}\) that determines where a bright fringe of the green light coincides with a bright fringe of the blue light. (The order \(m_{1}\) is associated with \(\lambda_{1}\), and \(m_{2}\) is associated with \(\lambda_{2} .\) (b) Find the minimum values of \(m_{1}\) and \(m_{2}\) such that the overlapping of the bright fringes will occur and find the position of the overlap on the screen.
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