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.

To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielec

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\text{MgF}_2

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Concept understanding — Index Matching

Index Matching

In ray optics, index matching means bringing two materials into contact so their refractive indices are equal (or very nearly equal). When $n_1 = n_2$ across a boundary, the boundary becomes optically invisible — light passes through as if the interface were not there at all.

Why a mismatched interface bends and reflects light

Whenever light crosses a boundary between media of index $n_1$ and $n_2$ , two things happen:

Refraction, governed by Snell's law:

n_1 \sin\theta_1 = n_2 \sin\theta_2

Partial reflection. For light at normal incidence, the fraction reflected is

R = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2

Both effects are driven by the difference $n_2 - n_1$ : a bigger mismatch means more bending and more reflected light.

What happens when the indices match

As $n_2 \to n_1$ :

Snell's law gives $\sin\theta_1 = \sin\theta_2$ , so $\theta_1 = \theta_2$ — the ray does not bend.
The reflection formula gives $R = 0$ — no light is reflected.

n_1 = n_2

The classic demonstration — the "disappearing" glass rod

Drop a clear glass rod into water: you still see it, because glass ( $n \approx 1.5$ ) and water ( $n \approx 1.33$ ) differ, so light reflects and refracts at the glass surface. But immerse the same rod in a liquid tuned to exactly $n = 1.5$ (a glycerine mixture, for example), and the submerged part vanishes from view — with no index difference, its surfaces no longer signal their presence to your eye.

Important

Index matching does not make the glass transparent — it already was. It removes the surface effects (reflection and refraction) by erasing the index difference at the boundary.

Where it is used

Oil-immersion microscopes: immersion oil ( $n \approx 1.5$ ) fills the gap between slide and objective that air would otherwise leave, removing reflection/refraction losses and letting steeply angled rays enter — sharpening the image.
Optical-fibre splices: an index-matching gel between two fibre ends removes the air gap so almost no signal reflects back at the joint. …

Why this formula?

Index Matching: Why the Formula Holds

Index matching is a powerful technique in combinatorics and probability — it's used to simplify sums over complicated index sets by cleverly re-indexing or pairing terms. The core idea is to match indices so that a double sum (or product) collapses into a simpler expression.

Let's build the reasoning step by step.

The Core Formula

The most common index matching identity is:

\sum_{i=1}^{n} \sum_{j=1}^{n} a_i b_j = \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{j=1}^{n} b_j \right)

This looks trivial — it's just the distributive law. But the why matters for deeper applications.

Why It Holds: The Distributive Law in Action

Step 1: Expand the outer sum

The left side means: for each fixed $i$ , sum over all $j$ , then sum over $i$ .

\sum_{i=1}^{n} \left( \sum_{j=1}^{n} a_i b_j \right)

Step 2: Factor out $a_i$ from the inner sum

Since $a_i$ does not depend on $j$ , it can be pulled out:

\sum_{i=1}^{n} \left( a_i \cdot \sum_{j=1}^{n} b_j \right)

Step 3: The inner sum is constant with respect to $i$

Let $S_b = \sum_{j=1}^{n} b_j$ . Then:

\sum_{i=1}^{n} a_i \cdot S_b = S_b \cdot \sum_{i=1}^{n} a_i

Step 4: Recognize the product

This is exactly:

\left( \sum_{i=1}^{n} a_i \right) \left( \sum_{j=1}^{n} b_j \right)

Key insight: The double sum over all $n^2$ pairs $(i,j)$ is just the product of the two separate sums. This works because the terms factor as $a_i b_j$ — no cross-dependence between $i$ and $j$ .

Why This Matters for Exam Problems

Index matching is used when you have double sums with constraints (like $i < j$ or $i \neq j$ ). The trick:

Start with the unconstrained double sum (all $i,j$ )
Subtract the diagonal terms ( $i=j$ ) or the off-diagonal terms
Use index matching to simplify

Example: Sum over $i < j$

We want $\sum_{1 \le i < j \le n} a_i b_j$ .

Derivation:

\sum_{i=1}^{n} \sum_{j=1}^{n} a_i b_j = \sum_{i=1}^{n} \sum_{j=1}^{i-1} a_i b_j + \sum_{i=1}^{n} a_i b_i + \sum_{i=1}^{n} \sum_{j=i+1}^{n} a_i b_j

The first and third terms are symmetric (just swap $i$ and $j$ ). So:

\left( \sum a_i \right) \left( \sum b_j \right) = \sum_{i=1}^{n} a_i b_i + 2 \sum_{i < j} a_i b_j

Thus:

\boxed{\sum_{i < j} a_i b_j = \frac{1}{2} \left[ \left( \sum a_i \right) \left( \sum b_j \right) - \sum a_i b_i \right]}

Why this works: The unconstrained double sum counts each unordered pair $(i,j)$ twice (once as $(i,j)$ and once as $(j,i)$ ), except the diagonal which appears once. Index matching lets us express the constrained sum in terms of the product.

The Deeper "Why": Symmetry and Factorization

The real power of index matching comes from symmetry: …

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For maximum transmission of light through a coated lens, we need destructive interference in the reflected light. This happens when the optical path difference in the film equals half a wavelength, leading to a minimum film thickness of $t = \frac{\lambda}{4n}$ . For $\lambda = 5500\ \text{\AA}$ and $n = 1.38$ , the required thickness is $t \approx 996\ \text{\AA}$ .

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.

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.
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:

2nt = \left(m + \frac12\right) \lambda, \quad m = 0, 1, 2, \dots

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

Solve for the thickness. …

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Concept understanding — Index Matching

Index Matching

Why a mismatched interface bends and reflects light

What happens when the indices match

The classic demonstration — the "disappearing" glass rod

Where it is used

Why this formula?

Index Matching: Why the Formula Holds

The Core Formula

Why It Holds: The Distributive Law in Action

Step 1: Expand the outer sum

Step 2: Factor out $a_i$ from the inner sum

Step 3: The inner sum is constant with respect to $i$

Step 4: Recognize the product

Why This Matters for Exam Problems

Example: Sum over $i < j$

The Deeper "Why": Symmetry and Factorization

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

Steps

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