Problem 14

  In preparation for this problem review Conceptual Example 5. A mixture of red light ($\lambda_{vacuum}$ = 661 nm) and blue light ($\lambda_{vacuum}$ = 473 nm) shines perpendicularly on a thin layer of gasoline (n_gas = 1.40) lying on water (n_water = 1.33). The gasoline layer has a uniform thickness of $1.69 \times 10^{-7} m$. Destructive interference removes one of the colors from the reflected light. By means of suitable calculations decide whether the gasoline looks red or blue in reflected light.

SOLUTION:
One of the colors vanishes due to destructive interference. For destructive interference, the wave which reflects from the air-gas interface (wave 1, say) must be completely out of phase with the wave which reflects from the gas-water interface (wave 2). When wave 1 reflects from the air-gas boundary, its phase is inverted because air has a lower index of refraction than gasoline. So, in order for wave 2 to be completely out of phase with wave 1, wave 2 must come out of the gasoline with the same phase it had going in. In order to have the same phase, the light must have travelled one wavelength (or some multiple of wavelengths); i.e.,

$$2t = \frac{\lambda_{vacuum}}{n_{gas}} ,$$

where t is the thickness of the gas layer. So, the wavelength in vacuum is

$$\lambda_{vacuum} = 2n_{gas}t = 473 nm .$$

This wavelength just happens to be the wavelength of blue light. So, since the blue light is destroyed, only the red light remains. The reflected light looks red.