A thin film of oil \((n=1.50)\) of thickness \(0.40 \mu \mathrm{m}\) is spread over a puddle of water \((n=1.33) .\) For which wavelength in the visible spectrum do you expect constructive interference for reflection at normal incidence?

: When light is reflected from a medium of higher to a lower refractive index, it undergoes a phase shift of 180 degrees. Light reflects off both the oil-air and oil-water interfaces, but since both interfaces involve a transition from a higher to lower refractive index, both reflections result in a 180-degree phase shift. These two phase shifts cancel each other, resulting in a net phase shift of 0.

: To find the phase shift due to the path difference between rays reflecting from the top and the bottom surfaces of the film, we need to find the extra distance traveled by the ray reflecting from the bottom surface. Since the rays are incident at the normal, this is simply twice the thickness of the film. Extra distance = 2 x thickness = 2 x \(0.40 \mu \mathrm{m}\) = \(0.80 \mu\mathrm{m}\)

: As we know that constructive interference occurs when the phase difference between interfering waves is a multiple of 2π or when the path difference is a multiple of the wavelength. Since the extra distance traveled by the ray reflecting from the bottom surface of the film is \(0.80 \mu\mathrm{m}\), we can use this value to find the wavelength in the visible spectrum (which ranges from 400 nm to 700 nm) that exhibits constructive interference: Extra distance = m * wavelength_inside_film We also need to account for the fact that the wavelength inside the film is different from the wavelength in the air, due to the refractive index of the oil. To convert the wavelength in the air (which we are trying to find) to the wavelength inside the film, we use the following equation: wavelength_inside_film = \(\frac{wavelength_{air}}{n}\) Combining the above two equations, we get: \(0.80 \mu\mathrm{m} = m * \frac{wavelength_{air}}{1.50}\) Now we need to find an integer value of m that results in a wavelength in the visible range. For m=1: \(wavelength_{air} = 1.50 * 0.80 \mu\mathrm{m} = 1.2 \mu\mathrm{m}\) But this value is outside the visible range. For m=2: \(wavelength_{air} = 0.75 * 0.80 \mu\mathrm{m} = 0.60 \mu\mathrm{m}\) This value lies within the visible range (600 nm). Therefore, the expected wavelength for constructive interference in the visible spectrum when the light is incident at normal incidence is 600 nm.