Lens Maker's Formula: Derivation and Application

Hello! Welcome to our next lesson on thin lenses.

In our last session, we mastered the thin lens formula and magnification, which allowed you to calculate image properties once the focal length ( $f$ ) was known. You practiced combining the math with the visual intuition from ray diagrams, a key skill for your NEET preparation.

Today, we will answer a fundamental question: What determines the focal length of a lens in the first place? We will delve into the physics of lens construction to derive and apply the Lens Maker's Formula. This formula is the bridge between the physical properties of a lens—its shape and material—and its optical behaviour.

1. The Lens Maker's Formula: Anatomy of a Lens

The focal length ( $f$ ) of a lens is not an arbitrary number; it's a direct result of two key factors:

The refractive index ( $n$ ) of the lens material relative to its surroundings.
The radii of curvature ( $R_{1}$ and $R_{2}$ ) of its two surfaces.

The Lens Maker's Formula mathematically connects these properties.

Lens Maker's Formula with Ray Diagram — This diagram shows the key components involved in the Lens Maker's Formula: the focal length (f), the refractive indices of the lens (n) and the surrounding medium, and the radii of curvature of the two surfaces (R1 and R2).

The formula is generally written as:
$\frac{1}{f} = (\frac{n _{2}}{n _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Where:

$f$ is the focal length of the lens.
$n_{1}$ is the refractive index of the surrounding medium (e.g., air, water).
$n_{2}$ is the refractive index of the lens material (e.g., glass).
$R_{1}$ is the radius of curvature of the first surface (where light enters).
$R_{2}$ is the radius of curvature of the second surface (where light exits).

If the lens is in air ( $n_{1} \approx 1$ ) and its material has a refractive index $n$ (i.e., $n_{2} = n$ ), the formula simplifies to the most common form you'll encounter:
$\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

2. The Derivation: From Two Surfaces to One Lens

To derive this formula, we model the lens as two separate spherical refracting surfaces. We then apply the formula for refraction at a single spherical surface, which we've seen before, to each surface in turn.

Key Assumptions

Before we begin, it's important to state the assumptions under which this formula holds true. These are standard for thin lens optics.

Lens Makers Formula & important Numerical I Chapter 8 Ray optics Class 12th Physics Boards NEET JEE

Let's watch a brief segment that clearly outlines the assumptions made when deriving the Lens Maker's formula. Understanding these limitations is important.

Watch from 03:18 to 05:35. Pay attention to the four key assumptions: the object is a point object, the lens is thin, the rays are paraxial, and the aperture is small.

In summary, we assume:

The lens is thin, so distances can be measured from a single optical center.
The aperture of the lens is small.
The object is a point object lying on the principal axis.
All rays are paraxial, meaning they make small angles with the principal axis.

The Derivation Steps

The core idea is to see the final image as the result of two consecutive refractions. The image formed by the first surface acts as the object for the second surface.

The following video provides a detailed, step-by-step derivation for a convex lens. This is a very important derivation for your board exams and provides the conceptual backbone for applying the formula.

Lens Makers Formula & important Numerical I Chapter 8 Ray optics Class 12th Physics Boards NEET JEE

This video from RADHIKA CLASSES walks through the entire derivation, from constructing the ray diagram to combining the equations. This will be the most detailed part of our lesson.

Watch from the beginning to 02:59 to recap the single-surface formula, and then from 05:35 to 33:34 for the main derivation. Focus on understanding the logic: Refraction at Surface 1: An object O forms an image I₁. Refraction at Surface 2: The image I₁ acts as a virtual object to form the final image I. Combining the equations to eliminate the intermediate image I₁. Note: The instructor correctly concludes that it's best to derive the general formula without substituting signs for u, v, R₁, R₂, and only apply sign conventions when solving numerical problems. This is the standard and least confusing method.

Let's break down the logic from the video:

Recall the Spherical Surface Formula: For an object at distance $u$ and image at distance $v$ for a surface with radius $R$ separating media $n_{1}$ and $n_{2}$ :
$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$
First Surface (ABC): Light travels from medium $n_{1}$ to $n_{2}$ . The object is at $u$ . If the second surface weren't there, an image would form at a distance $v_{1}$ .
$\frac{n _{2}}{v _{1}} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R _{1}} \dots (1)$
Second Surface (ADC): The image $I_{1}$ from the first surface now acts as the object for this surface. The object distance is $v_{1}$ . Light travels from $n_{2}$ back to $n_{1}$ . The final image is formed at $v$ .
$\frac{n _{1}}{v} - \frac{n _{2}}{v _{1}} = \frac{n _{1} - n _{2}}{R _{2}} \dots (2)$
Combine the Equations: Adding equations (1) and (2), the $\frac{n _{2}}{v _{1}}$ terms cancel out:
$\frac{n _{1}}{v} - \frac{n _{1}}{u} = (n_{2} - n_{1}) (\frac{1}{R _{1}}) + (n_{1} - n_{2}) (\frac{1}{R _{2}})$
$n_{1} (\frac{1}{v} - \frac{1}{u}) = (n_{2} - n_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{v} - \frac{1}{u} = (\frac{n _{2}}{n _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
Final Step: We know from the thin lens formula that $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . Equating the two gives us the Lens Maker's Formula:
$\frac{1}{f} = (\frac{n _{2}}{n _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Derivation of Lens Maker Formula and Lens Formula — This image provides a static summary of the derivation process we just covered, from the initial ray diagram to the final formulas.

3. Applying the Formula: Sign Convention for Radii

For NEET problem-solving, correctly applying the sign convention is everything. The convention for $u$ and $v$ is the same as before. The new challenge is assigning signs to $R_{1}$ and $R_{2}$ .

Rule: Light travels from left to right.

Measure the radius of curvature from the optical center.
If the center of curvature ( $C$ ) is to the right of the surface, the radius ( $R$ ) is positive.
If the center of curvature ( $C$ ) is to the left of the surface, the radius ( $R$ ) is negative.
For a plane surface, the radius of curvature is infinity ( $R = \infty$ ).

Let's apply this to standard lenses:

Lens Type	First Surface (R₁)	Second Surface (R₂)
Biconvex (Double Convex)	Center is on the right $⟹ R_{1} > 0$	Center is on the left $⟹ R_{2} < 0$
Biconcave (Double Concave)	Center is on the left $⟹ R_{1} < 0$	Center is on the right $⟹ R_{2} > 0$
Plano-convex	$R_{1} = \infty$ (plane side first)	$R_{2} < 0$
Plano-concave	$R_{1} = \infty$ (plane side first)	$R_{2} > 0$

Test your understanding!

A biconcave lens has radii of curvature of 20 cm and 30 cm. What are the values of $R_{1}$ and $R_{2}$ with the correct signs?

Show answer

For a biconcave lens, the first surface is concave to the incident light, so its center of curvature is on the left. The second surface is convex to the incident light, so its center of curvature is on the right.
Therefore, $R_{1} = - 20$ cm and $R_{2} = + 30$ cm.

4. Problem-Solving for NEET

Let's work through some typical problems. This is where your focus on mathematical application will pay off.

Example 1: Finding Refractive Index

Problem: A double convex lens has radii of curvature of 10 cm and 15 cm. Its focal length in air is 12 cm. What is the refractive index of the glass?

This problem is solved in the video. Let's solve it here, focusing on the steps.

Lens Makers Formula & important Numerical I Chapter 8 Ray optics Class 12th Physics Boards NEET JEE

Let's watch a numerical example where we use the lens maker's formula to find the refractive index.

Watch the problem solved from 48:34 to 52:32. Notice how the sign convention for R₁ and R₂ is applied before substituting into the formula.

Solution:

List knowns with signs:
- Focal length (convex lens): $f = + 12$ cm.
- Medium is air: $n_{1} = 1$ . We need to find $n_{2} = n$ .
- First surface (convex): $R_{1} = + 10$ cm.
- Second surface (concave shape relative to path): $R_{2} = - 15$ cm.
Apply the formula:
$\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{12} = (n - 1) (\frac{1}{10} - \frac{1}{- 15})$
$\frac{1}{12} = (n - 1) (\frac{1}{10} + \frac{1}{15})$
$\frac{1}{12} = (n - 1) (\frac{3 + 2}{30}) = (n - 1) (\frac{5}{30}) = (n - 1) (\frac{1}{6})$
Solve for n:
$n - 1 = \frac{6}{12} = 0.5$
$n = 1.5$
The refractive index of the glass is 1.5.

Example 2: Changing the Medium

This is a classic and very important question type for competitive exams.

Problem: A convex lens made of glass ( $n_{g} = 1.5$ ) has a focal length of 20 cm in air ( $n_{a} = 1$ ). What will its new focal length be if it is completely immersed in water ( $n_{w} = 1.33$ )?

The key here is to use the formula twice: once for air and once for water.

Lens Makers Formula & important Numerical I Chapter 8 Ray optics Class 12th Physics Boards NEET JEE

Now for a critical NEET-style problem: finding the change in focal length when a lens is moved to a different medium. The approach used here is very efficient.

Watch from 53:09 to 58:20. The strategy is to first write the equation for the lens in air to get a value for the term with the radii, and then use that in the equation for the lens in water.

Solution Strategy:

In Air: Write the Lens Maker's formula for the lens in air. This allows us to find a value for the geometrical part $(\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .
$\frac{1}{f _{ai r}} = (n_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{20} = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.5 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
So, $(\frac{1}{R _{1}} - \frac{1}{R _{2}}) = \frac{1}{20 \times 0.5} = \frac{1}{10}$ . Let's call this equation (A).
In Water: Now write the formula for the lens in water. The radii don't change, but the relative refractive index does. The term we use is $\frac{n _{2}}{n _{1}} = \frac{n _{g l a ss}}{n _{w a t er}}$ .
$\frac{1}{f _{w a t er}} = (\frac{n _{g}}{n _{w}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
Substitute and Solve: Substitute the value of the radii term from equation (A).
$\frac{1}{f _{w a t er}} = (\frac{1.5}{1.33} - 1) (\frac{1}{10})$
$\frac{1}{f _{w a t er}} = (\frac{1.5 - 1.33}{1.33}) (\frac{1}{10}) = (\frac{0.17}{1.33}) (\frac{1}{10}) \approx \frac{0.128}{10} = 0.0128$
$f_{w a t er} = \frac{1}{0.0128} \approx 78.1 cm$

Notice that the focal length increased significantly, meaning the lens became less powerful when placed in water. This is because the refractive index contrast between the lens and its surrounding medium decreased.

Test your understanding!

What would happen to the focal length of the glass lens ( $n_{g} = 1.5$ ) if it were submerged in a liquid with a refractive index of 1.5? What if the liquid's refractive index was 1.7? (This is a conceptual question).

Show answer

Liquid with n = 1.5: The relative refractive index would be $\frac{n _{g}}{n _{l i q u i d}} = \frac{1.5}{1.5} = 1$ . The term $(\frac{n _{g}}{n _{l i q u i d}} - 1)$ becomes zero. This makes $\frac{1}{f} = 0$ , so the focal length $f$ becomes infinite. The lens would act like a flat piece of glass and would not bend light at all; it would essentially become invisible.
Liquid with n = 1.7: The relative refractive index would be $\frac{n _{g}}{n _{l i q u i d}} = \frac{1.5}{1.7} < 1$ . The term $(\frac{n _{g}}{n _{l i q u i d}} - 1)$ becomes negative. This would reverse the sign of the focal length. A convex lens (converging in air) would behave like a concave lens (diverging), and vice versa!

Conclusion

Excellent work today! You have now connected the macroscopic optical property of a lens (focal length) to its underlying physical design (curvature and material). This is a crucial piece of the puzzle for a complete understanding of lens optics.

Key Takeaways:

The Lens Maker's Formula is $\frac{1}{f} = (\frac{n _{l e n s}}{n _{m e d i u m}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .
The derivation involves applying the refraction formula for a single spherical surface twice.
The image from the first surface acts as the object for the second surface.
The sign convention for radii of curvature ( $R_{1}, R_{2}$ ) is critical and depends on the location of their centers of curvature relative to the lens.
The focal length of a lens is dependent on the surrounding medium.

Preview of the Next Lesson:
We have thoroughly analyzed single lenses. The next logical step is to see what happens when we combine them. In our next lesson, we will learn how to calculate the equivalent power and focal length for lenses in contact. This will allow us to analyze more complex optical systems, moving us one step closer to understanding instruments like microscopes and telescopes.

Recommended resources

1. The Lens Maker's Formula: Anatomy of a Lens

2. The Derivation: From Two Surfaces to One Lens

Key Assumptions

The Derivation Steps

3. Applying the Formula: Sign Convention for Radii

4. Problem-Solving for NEET

Example 1: Finding Refractive Index

Example 2: Changing the Medium

Conclusion