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Chapter 36: Problem 37

Coat Glass We wish to coat flat glass \((n=1.50)\) with a transparent material \((n=1.25)\) so that reflection of light at wavelength \(600 \mathrm{~nm}\) is eliminated by interference. What minimum thickness can the coating have to do this?

Short Answer

Expert verified
The minimum thickness of the coating is 240 nm.

Step by step solution

01

Identify the Problem Requirements

Determine the required thickness of a transparent coating with a refractive index of 1.25 applied to flat glass with a refractive index of 1.50, in order to eliminate reflection of light at wavelength 600 nm.
02

Use the Thin Film Interference Condition

For destructive interference to eliminate reflection, the condition for the minimum thickness (constructive interference in the reflected light waves) is given by \[\frac{2nt}{m} = \frac{\frac{\text{wavelength}}{n}}{2}\]. Here, m=1 for the first minimum thickness, n is the refractive index of the coating, and t is the thickness.
03

Apply the Formula

Rearrange the formula \[\frac{2t}{\text{wavelength / n}} = m\] to find t. Substituting the values: \[\frac{2 \times 1.25 \times t}{600 \text{ nm}} = 1\].
04

Solve for t

Solve for the thickness t:\[t = \frac{600 \text{ nm}}{2 \times 1.25} = 240 \text{ nm}\].

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Key Concepts

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

destructive interference
Destructive interference occurs when two waves cancel each other out. This happens when the crest of one wave aligns with the trough of another, resulting in a lower overall amplitude. In thin film interference, destructive interference is used to eliminate reflection by ensuring that the reflected waves from different surfaces are out of phase. By carefully choosing the thickness of the coating, it is possible to cause these waves to interfere destructively for a specific wavelength of light.

To achieve this, we need to consider the path difference between the reflected waves. The path difference should be an odd multiple of half the wavelength in the medium. This causes the waves to cancel each other. The formula for this condition is:
\[2nt = (m + 0.5) \frac{\text{wavelength}}{n} \]
where:
    - n is the refractive index of the coating
    - t is the thickness of the coating
    - m is the order of the interference (m = 0, 1, 2,...)
    - wavelength is the wavelength of light in the coating

    For the first order of interference (m = 0), the condition simplifies further.
refractive index
The refractive index is a measure of how much light is bent, or refracted, when entering a material. It is denoted by the symbol 'n'.

The higher the refractive index, the more light slows down and bends when entering the material. In the context of thin film interference, understanding the refractive index of both the film and the substrate (the material the film is coating) is crucial.

Let's break down the refractive indices in our original exercise:
    - The refractive index of the glass is given as 1.50.
    - The refractive index of the transparent coating material is 1.25.

    When light passes from a material with a higher refractive index to a material with a lower refractive index, there is a phase change of half a wavelength. This phase change must be considered when calculating the necessary thickness for achieving destructive interference.
minimum thickness calculation
To eliminate reflection using destructive interference, we need to determine the minimum thickness of the coating. Given:
    - Refractive index of the coating, n = 1.25
    - Wavelength of light, \( \text{wavelength} = 600 \text{ nm} \)

    We use the condition for destructive interference which, for minimum thickness, simplifies to:
    \[ 2nt = (0.5)( \text{wavelength}) \]
    Rearranging the formula to solve for t (the thickness), we get:
    \[ t = \frac{ \text{wavelength} }{2 \times 1.25} \]
    Substituting the values:
    \[ t = \frac{600 \text{ nm}}{2 \times 1.25} = 240 \text{ nm} \]
    Therefore, the minimum thickness of the coating required to eliminate reflection at the wavelength of 600 nm is 240 nm. This precision ensures that we achieve the desired destructive interference to minimize reflected light effectively.

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Most popular questions from this chapter

Two Interference Patterns In a double-slit experiment the distance between slits is \(5.0 \mathrm{~nm}\) and the slits are \(1.0 \mathrm{~m}\) from the screen. Two interference patterns can be seen on the screen: one due to light with wavelength \(480 \mathrm{~nm}\), and the other due to light with wave length \(600 \mathrm{~nm}\). What is the separation on the screen between the third-order \((m=3)\) bright fringes of the two interference patterns?

Same Frequency Two waves of the same frequency have amplitudes \(1.00\) and \(2.00\). They interfere at a point where their phase difference is \(60.0^{\circ}\). What is the resultant amplitude?

Wedge-Shaped A broad beam of light of wavelength \(630 \mathrm{~nm}\) is incident at \(90^{\circ}\) on a thin, wedge-shaped film with index of refraction 1.50. An observer intercepting the light transmitted by the film sees 10 bright and 9 dark fringes along the length of the film. By how much does the film thickness change over this length?

Mica Flake A thin flake of mica \((n=1.58)\) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe \((m=7)\) before the mica was used. If \(\lambda=550 \mathrm{~nm}\), what is the thickness of the mica? (Hint: Consider the wavelength of the light within the mica.)

Sketch Intensity Light of wavelength \(600 \mathrm{~nm}\) is incident normally on two parallel narrow slits separated by \(0.60 \mathrm{~mm}\). Sketch the intensity pattern observed on a distant screen as a function of angle \(\theta\) from the pattern's center for the range of values \(0 \leq \theta \leq\) \(0.0040 \mathrm{rad}\).
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