Derive and apply for a double convex lens.

Lens Maker's Equation — 1/f = (n-1)(1/R₁ - 1/R₂)

Question

Derive the Lens Maker’s Equation: $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Then apply it: A double convex lens has radii of curvature $R_1 = 20\ \text{cm}$ and $R_2 = 30\ \text{cm}$ , and refractive index $n = 1.5$ . Find its focal length.

Solution — Step by Step

A double convex lens has two refracting surfaces. We treat each surface separately and use the New Cartesian Sign Convention — distances measured from the optical centre, left is negative, right is positive.

For a double convex lens: $R_1 > 0$ (centre of curvature on the right) and $R_2 < 0$ (centre of curvature on the left). This is the step where most students lose marks — the sign of $R_2$ .

Light travels from air ( $n_1 = 1$ ) into glass ( $n_2 = n$ ). Using the refraction formula at a spherical surface:

\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}

Object is at infinity ( $u \to \infty$ ), so the $n_1/u$ term vanishes:

\frac{n}{v_1} = \frac{n - 1}{R_1}

The image formed by Surface 1 ( $v_1$ ) acts as the virtual object for Surface 2.

Now light goes from glass ( $n_2 = n$ ) back into air ( $n_1 = 1$ ). The object distance for Surface 2 is $v_1$ :

\frac{1}{v} - \frac{n}{v_1} = \frac{1 - n}{R_2}

Adding the two equations from Steps 2 and 3 — the $n/v_1$ terms cancel:

\frac{1}{v} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

Since the object was at infinity, the final image forms at the focal point, so $v = f$ :

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

This is the Lens Maker’s Equation. The name makes sense — it tells the lens manufacturer what radii to grind to get a desired focal length.

$R_1 = +20\ \text{cm}$ , $R_2 = -30\ \text{cm}$ , $n = 1.5$ :

\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-30}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{30}\right)

= 0.5 \times \frac{3 + 2}{60} = 0.5 \times \frac{5}{60} = \frac{1}{24}

Focal length $f = 24\ \text{cm}$ (positive — converging lens, as expected).

Why This Works

The derivation works by treating a lens as two back-to-back refracting surfaces and applying superposition. Each surface bends the wavefront independently, and the combined effect gives us the net focal power.

The $(n-1)$ factor is the optical “punch” of the material — a denser medium bends light more, giving a shorter focal length. The $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ factor encodes the geometry — sharper curvature means stronger focusing.

This equation is fundamental in JEE and NEET. It also explains why a lens submerged in water (where $n_{\text{eff}} = n_{\text{glass}}/n_{\text{water}}$ is close to 1) becomes nearly powerless — the $(n-1)$ term shrinks dramatically.

Alternative Method — Using Lens Power

For quick MCQs, once you know the Lens Maker’s Equation, you can work directly in terms of power ( $P = 1/f$ in dioptres, with lengths in metres):

P = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

With $R_1 = 0.20\ \text{m}$ , $R_2 = -0.30\ \text{m}$ :

P = 0.5 \times \left(5 + 3.33\right) = 0.5 \times 8.33 \approx 4.17\ \text{D}

f = \frac{1}{P} \approx 0.24\ \text{m} = 24\ \text{cm}\ ✓

In JEE Main 2024, a variation asked for the new focal length when the same lens is placed in water ( $n_w = 4/3$ ). Use $(n_{\text{rel}} - 1)$ where $n_{\text{rel}} = n_{\text{glass}}/n_{\text{water}} = 1.5/1.333 = 1.125$ . The focal length jumps to about $96\ \text{cm}$ — four times larger. Always check whether the surrounding medium is air or something else.

Common Mistake

The most frequent error: taking $R_2 = +30\ \text{cm}$ for a double convex lens. Students see “double convex” and assume both radii are positive. Wrong — $R_2$ is negative because its centre of curvature lies to the left of the surface (in the incident medium). If you use $+30$ , you get $\frac{1}{f} = 0.5 \times \left(\frac{1}{20} - \frac{1}{30}\right) = 0.5 \times \frac{1}{60} = \frac{1}{120}$ , giving $f = 120\ \text{cm}$ — a wildly wrong answer. Always draw the lens, mark both centres of curvature, and assign signs before substituting.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Lens Power

Common Mistake

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