Problem 40 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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: \[ \lambda_{n} = \frac{600 \text{ nm}}{1.45} = 413.79 \text{ nm} \]

Step 3: 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.

Step 4: 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:\[ 2nt = \left(m + \frac{1}{2}\right)\lambda \]

Step 5: 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.

Key Concepts

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’s 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.