White light illuminates an oil film on water. Viewing it directly from above, it looks red. Assume that the reflected red light has a wavelength of 615 nm in air, and that the oil has a thickness of 2.430e-7 m. What is the refractive index of the oil? Assume that the refractive index of water is greater than that of the oil.

To find the refractive index of the oil, we can use the concept of interference in thin films. When white light is incident on an oil film on water, some of the light is reflected from the top surface of the oil film, and some is reflected from the bottom surface where oil and water meet. The reflected light from the two surfaces can interfere constructively or destructively, resulting in different colors being observed.

In this case, we are observing the film from directly above, so we are seeing the reflected light from the top surface only. We are told that the observed color appears red, which corresponds to a wavelength of 615 nm in air. We also know the thickness of the oil film is 2.430e-7 m.

Now, to find the refractive index of the oil, we can use the equation for the wavelength of light in a medium:

λ = λ0 / n

where λ is the wavelength in the medium, λ0 is the wavelength in vacuum (or air), and n is the refractive index of the medium.

We can rewrite this equation as:

n = λ0 / λ

Substituting the given values, we have:

n = 615 nm / λ

Since we know the thickness of the oil film (2.430e-7 m), we can use it to convert the wavelength of light in air to the wavelength in the oil:

λ = λ0 / n = 615 nm / n

By rearranging the equation and substituting the values, we get:

n = 615 nm / (2.430e-7 m)

Calculating this expression gives us the refractive index of the oil.