Question

A glass lens is coated on one side with a thin film of magnesium fluoride (MgF₂) to reduce reflection from the lens surface. The index of refraction of MgF₂ is 1.38; that of the glass is 1.50. What is the least coating thickness that eliminates (via interference) the reflections at the middle of the visible spectrum ($\lambda$ = 550 nm)? Assume that the light is approximately perpendicular to the lens surface.

Answer 1

99.6 nm

Answer 2

Reflections occur at the air-MgF₂ and MgF₂-glass interfaces. Both reflections involve a phase shift of $\pi$ because the light is reflecting from a medium with a lower refractive index to one with a higher refractive index ($n_{\text{air}} < n_{\text{MgF}_2}$ and $n_{\text{MgF}_2} < n_{\text{glass}}$).

For destructive interference (elimination of reflection), the optical path difference must be an odd multiple of half the wavelength: $2n_{\text{MgF}_2} L = (m + 1/2)\lambda$ where $L$ is the film thickness, $n_{\text{MgF}_2}$ is the refractive index of the film, $\lambda$ is the wavelength of light, and $m$ is an integer ($0, 1, 2, \dots$).

For the least thickness, we take $m=0$: $2n_{\text{MgF}_2} L = \frac{\lambda}{2}$ $L = \frac{\lambda}{4n_{\text{MgF}_2}}$ Substituting the given values: $\lambda = 550$ nm and $n_{\text{MgF}_2} = 1.38$: $L = \frac{550 \text{ nm}}{4 \times 1.38} \approx 99.6 \text{ nm}$