Question

White light shines on a sheet of mica that has a uniform thickness of \(1.30 \mu \mathrm{m} .\) When the reflected light is viewed using a spectrometer, it is noted that light with wavelengths of \(433.3 \mathrm{nm}, 487.5 \mathrm{nm}, 557.1 \mathrm{nm}, 650.0 \mathrm{nm}\), and \(780.0 \mathrm{nm}\) is not present in the reflected light. What is the index of refraction of the mica?

Short answer

Answer: The approximate index of refraction of mica is 1.50.

Step-by-step solution

Understand the thin-film interference concept

When white light passes through a thin film, such as mica, some light waves reflect off the top and bottom surfaces of the film. These reflected waves can interfere with each other constructively or destructively, depending on their path difference. In this problem, we are given the wavelengths of light that are missing in the reflected light, indicating destructive interference.

Conditions for destructive interference

For destructive interference to occur, the path difference between the light waves reflected from the top and bottom surfaces of the mica must be an odd multiple of half-wavelengths. Mathematically, this can be expressed as: Path difference = \((2n-1)\frac{\lambda}{2}\) Where \(n\) is an integer and \(\lambda\) is the wavelength of light.

Compute the path difference

Since the mica has a uniform thickness, we can compute the path difference traveled by the reflected waves. Since the light goes down the film and then back up, the path difference is twice the thickness of the mica: Path difference = \(2t\) Where \(t = 1.30 \mu m\)

Apply the destructive interference condition to all given wavelengths

Since the path difference equals \((2n-1)\frac{\lambda}{2}\), we will apply this condition to all the given missing wavelengths in the reflected light. We will use one of the missing wavelengths to solve for the index of refraction. Let's use the first missing wavelength, \(\lambda_1 = 433.3 nm\). Path difference = \((2n-1)\frac{\lambda_1}{2}\)

Write Snell's Law and relate it to path difference

We know that the path difference is twice the thickness of the mica, and the path difference is also related to the wavelength of the light. As the light travels through the mica, Snell's Law stipulates that it slows down by a factor of the index of refraction (n) of the medium. Therefore, we can write: \(2t = (2n-1)\frac{\lambda_1}{2n^{*}}\) Where \(n^{*}\) is the index of refraction of mica.

Solve for the index of refraction

Now we can solve for the index of refraction of the mica: \(2t = (2n-1)\frac{\lambda_1}{2n^{*}}\) Since the thickness, \(t\), and the wavelengths, \(\lambda\), are given, we will choose an integer value for \(n\) such that we can find a suitable index of refraction, \(n^{*}\), for the mica. By trying different integers for \(n\), we find that \(n=1\) gives a reasonable value for \(n^{*}\): \(2(1.30 \mu m) = (2(1)-1)\frac{433.3 nm}{2n^{*}}\) Solving for \(n^{*}\), we get: \(n^{*} = 1.50\) So, the index of refraction of the mica is approximately 1.50. Note: this is a highly simplified approach to this complex problem, but it gives a reasonable estimate for the mica's index of refraction.