Derive lens maker's formula: 1/f = (μ-1)(1/R₁ - 1/R₂)

Question

Derive the lens maker's formula:

$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

(NCERT Class 12, Chapter 9)

Solution — Step by Step

Consider refraction at the first surface

A ray from object at distance $u$ (in the rarer medium) refracts at the first surface of radius $R_1$ . Using the single surface refraction formula:

$\frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1} \quad \cdots (i)$

Here $v_1$ is the image distance from surface 1 (inside the lens).

Consider refraction at the second surface

The image from surface 1 acts as a virtual object for surface 2. The ray goes from the denser medium ( $\mu$ ) to the rarer medium (1), so:

$\frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2} \quad \cdots (ii)$

where $v$ is the final image distance.

Add the two equations

Adding (i) and (ii), the $\mu/v_1$ terms cancel:

$\frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Apply the definition of focal length

When $u \to \infty$ (parallel rays), $v = f$ :

$\frac{1}{f} - 0 = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\boxed{\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$

Why This Works

A thin lens is essentially two refracting surfaces back-to-back. We apply the single-surface refraction formula at each surface and combine the results. The thinness assumption lets us use the same distances from both surfaces (since the thickness is negligible).

The cancellation of $v_1$ is the elegant part — the intermediate image position drops out, leaving a clean relationship between $u$ , $v$ , and the lens parameters ( $\mu$ , $R_1$ , $R_2$ ).

The formula tells us that focal length depends on both the refractive index and the curvatures. A more curved lens ( $R$ values smaller) or higher $\mu$ gives a shorter focal length (stronger lens).

Alternative Method — Using the thin lens equation directly

If you already know $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (thin lens equation), then combining with the result from Step 3:

$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

follows immediately. This is circular for a derivation question but useful as a quick verification.

🎯 Exam Insider

This is a guaranteed 5-mark derivation in CBSE. Draw a clear ray diagram showing the two refracting surfaces, mark $R_1$ , $R_2$ , $u$ , $v_1$ , $v$ , and the refractive index $\mu$ . State the sign convention you're using (Cartesian is standard). The diagram and sign convention together carry 2 marks.

Common Mistake

⚠️ Common Mistake

The biggest source of errors: sign convention. When applying the formula at surface 2, students often use the wrong sign for $v_1$ or $R_2$ . For a biconvex lens with Cartesian convention: $R_1 > 0$ (centre of curvature on the transmission side) and $R_2 < 0$ (centre of curvature on the incidence side). Getting $R_2$ 's sign wrong flips the formula to $1/R_1 + 1/|R_2|$ instead of $1/R_1 - 1/R_2$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the thin lens equation directly

Common Mistake

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