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Chapter 35: Problem 40

Sensitive Eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of \(1.38,\) while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Short Answer

Expert verified
The minimum thickness of the film is approximately 103.45 nm. Other wavelengths can be reinforced or cancelled based on modified interference conditions for their specific wavelengths.

Step by step solution

01

Understanding Reinforcement Condition

To have a constructive interference (reinforcement of light) in the reflected light from a thin film, the path difference should be equal to an integer multiple of the wavelength: \[ m\lambda = 2nt + \frac{\lambda}{2} \] where \(m\) is an integer, \(\lambda\) is the wavelength of light in vacuum (600 nm here), \(n\) is the refractive index of the film (1.45 for the eyedrop), and \(t\) is the thickness of the film. The \(\frac{\lambda}{2}\) term represents the phase change of \(\pi\) on reflection from higher to lower index surface.
02

Calculating Wavelength in Medium

The wavelength of light in the medium of the eyedrop is shorter than in vacuum: \[ \lambda_{n} = \frac{\lambda_{0}}{n} \] Plugging in the values, we have: \[ \lambda_{n} = \frac{600 \text{ nm}}{1.45} = 413.79 \text{ nm} \]
03

Calculating Minimum Thickness of Film

Since we are looking for the minimum thickness and \(m=0\) gives the smallest non-trivial thickness for constructive interference:\[ 0 + \frac{1}{2} = 2nt \]Therefore, solving for \( t \):\[ t = \frac{\lambda_n}{4} = \frac{413.79}{4} \approx 103.45 \text{ nm} \] Thus, the minimum thickness of the film is approximately 103.45 nm.
04

Explanation of Additional Reinforced or Cancelled Wavelengths

Given the thickness and refractive index, wavelengths are reinforced when:\[ 2nt = m\lambda \] Calculate for other possible visible wavelengths (380 nm to 750 nm) using appropriate m values for each wavelength. Calculate cancellation using:\[ 2nt = \left(m + \frac{1}{2}\right)\lambda \]
05

Applying Thickness to Contact Lenses

With contact lenses having a refractive index of 1.50, the equation: \[ 2nt = m\lambda \] holds with the thickness found in step 3. Solving for other wavelengths using:\[ \lambda_{lenses} = \frac{600}{1.50} \approx 400 \text{ nm} \] The analysis is similar but adjusted for the index of 1.50, finding appropriate \(m\) values.

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Key Concepts

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

Refractive Index
The refractive index is a measure that describes how light travels through a medium. Represented by the symbol \( n \), this value compares the speed of light in a vacuum to its speed in a given substance. When light enters a medium, such as eyedrops on your eye, it slows down. The refractive index quantifies this reduction.

In our scenario, the eyedrops have a refractive index of 1.45, which is higher than the cornea鈥檚 1.38. This means light travels more slowly through the eyedrops than through the cornea.

When light encounters a transition between two transparent surfaces with different refractive indices, certain optical effects, such as reflection and refraction, occur. These effects are crucial for understanding phenomena like thin film interference evident in the colorful patterns sometimes seen on soap bubbles or, in this case, reflected from your eye.
Constructive Interference
Constructive interference occurs when waves combine to make a wave of greater amplitude. For light waves, this means that certain wavelengths will become more intense when they overlap correctly.

In the context of thin film interference, constructive interference results in bright reflections of specific colors. The path difference of light waves reflecting off the top and bottom surfaces of the thin film determines if interference will be constructive.

Constructive interference follows the condition \( m\lambda = 2nt + \frac{\lambda}{2} \), where \( m \) is an integer denoting the order of interference. The phrase \( \frac{\lambda}{2} \) accounts for the phase change during reflection at the higher refractive index interface. In this particular exercise, we focused on the 600 nm wavelength since it was notably enhanced.
Wavelength Calculation
Calculating the wavelength of light within a medium is important because it helps explain why certain colors are visible when light reflects off thin films.

The wavelength in the medium (e.g., eyedrops) is given by \( \lambda_{n} = \frac{\lambda_{0}}{n} \), where \( \lambda_{0} \) is the original wavelength in vacuum and \( n \) is the refractive index of the medium.

For the red light of 600 nm, the calculation goes:
  • First, find the wavelength within the eyedrops: \( \lambda_{n} = \frac{600 \text{ nm}}{1.45} \approx 413.79 \text{ nm} \)
  • This shorter wavelength in the medium is used to determine film thickness or other wavelengths that might also show constructive interference.

By understanding the wavelength in the medium, we can perform further calculations to find the film thickness that causes interference and predict which other wavelengths might be reinforced or canceled.

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Most popular questions from this chapter

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04\(\mu \mathrm{m}\) apart, and in line with an observer, so that one source is 2.04\(\mu \mathrm{m}\) farther from the observer than the other. (a) For what visible wavelengths \((380\) to 750 \(\mathrm{nm})\) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04\(\mu \mathrm{m}\) farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm 19.0^{\circ}\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wave-length of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(\frac{1}{10}\) the maximum intensity on the screen?

Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q ?\)
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