Q.To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is $\text{MgF}_2$ ($n = 1.38$). What should the thickness of the film be so that at the center of the visible speetrum ($5500\ \text{\AA}$) there is maximum transmission.

The key idea is thin-film interference for anti-reflection coatings. Maximum transmission occurs when reflected light from the air-film and film-glass interfaces interferes destructively, cancelling the reflection.

1. For normal incidence, a phase change of $\pi$ (half-wavelength) occurs at both reflections (air to film, film to glass) because the film index $n=1.38$ lies between air ($n=1$) and glass ($n\approx1.5$). The two reflected waves are thus in phase at the point of reflection.
2. To get destructive interference, the extra path travelled inside the film must be an odd multiple of half-wavelengths in the film: $2t = (m+\frac12)\lambda_{\text{film}}$, where $\lambda_{\text{film}} = \lambda_0 / n$ and $\lambda_0 = 5500\ \text{\AA}$. …

The problem is about anti-reflection coatings — a beautiful application of wave optics. When light hits a lens surface, about 4% of it reflects off each air-glass interface. For a multi-element lens, this adds up to significant light loss and glare. The trick is to deposit a thin transparent film whose refractive index lies between that of air ($n=1$) and glass ($n \approx 1.5$). Here, $\text{MgF}_2$ with $n = 1.38$ is used.

Why does this work? Light reflects from two interfaces: air-to-film and film-to-glass. If these two reflected waves are exactly out of phase (by half a wavelength), they cancel each other — destructive interference. That reflected energy is not lost; it is redirected into the transmitted beam, boosting transmission. The condition for cancellation depends on the film thickness and the wavelength.

Let’s work through the calculation step by step.

1. Identify the phase changes on reflection.

   When light reflects off a boundary from a lower to a higher refractive index, it undergoes a phase shift of $\pi$ (equivalent to an extra path of $\lambda/2$). From air ($n=1$) to $\text{MgF}_2$ ($n=1.38$), the index increases, so the first reflected wave gets a $\pi$ shift. From $\text{MgF}_2$ to glass ($n \approx 1.5$), the index again increases, so the second reflected wave also gets a $\pi$ shift. Both reflections suffer the same phase change — so the net phase difference between them comes only from the extra distance travelled by the second wave inside the film.

2. Set up the condition for destructive interference.

   The second reflected wave travels an extra distance of $2t$ (down and back through the film). Inside the film, the wavelength is $\lambda/n$, where $\lambda$ is the vacuum wavelength. So the optical path difference (OPD) is $2nt$. For destructive interference, this OPD must equal an odd multiple of half-wavelengths in vacuum:

The smallest thickness (for $m=0$) gives the thinnest effective coating.

3. Solve for the thickness. …

Method: Finding Anti-Reflection Coating Thickness (Quarter-Wave Condition)

Use this method for any thin-film problem asking for the coating thickness that

gives maximum transmission (equivalently, minimum reflection) at a stated

wavelength.

Steps

Step 1: Work out the phase shift at each reflecting interface

Light reflects at two interfaces: air→film and film→substrate (e.g. glass). A

reflection off a boundary where the index INCREASES (going from lower to

higher $n$) carries an extra phase shift of $\pi$ (equivalent to $\lambda/2$ of

path); a reflection off a boundary where the index DECREASES carries no such

shift. Check both interfaces:

• If the film's index lies between the two neighbouring media (as it does for a coating chosen to sit between air and glass), BOTH reflections see an index increase, so BOTH pick up the same $\pi$ shift — these cancel out of the net phase comparison, leaving only the path travelled inside the film to matter.

Step 2: Write the optical path difference between the two reflected waves

The second reflected wave travels an extra distance $2t$ inside the film

(down to the second interface and back), so its optical path length is $2nt$

(using the film's own refractive index $n$, since light travels at $c/n$

inside it).

Step 3: Set the destructive-interference condition

Because both reflections carried the same $\pi$ phase shift (Step 1), those

shifts cancel between the two reflected waves, and the interference condition

depends only on the path travelled in the film. For the reflected light to

cancel (destructive interference in reflection = maximum transmission), the …