Lenses: Lens Maker's Formula & the Thin-Lens Equation

A Lens Is Two Surfaces (The Lens Maker's Formula)

A thin lens = two spherical refracting surfaces back-to-back. Apply Section 5's formula at each:

• Surface 1 (radius $R_1$): forms image $I_1$ of object O.
• Surface 2 (radius $R_2$): treats $I_1$ as its (possibly virtual) object and forms the final image I.

Adding the two surface equations for a thin lens (thickness $\to 0$), and putting the object at infinity (so the image lands at the focus f), gives the lens maker's formula:

$$\boxed{\frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$

• $n_{21}$ = lens material's index relative to the surroundings — a lens behaves differently in water than in air!
• Signs: a double-convex lens has $R_1 > 0$, $R_2 < 0$, giving $f > 0$ (converging). A double-concave lens reverses both: $f < 0$ (diverging).
• The formula designs lenses: choose radii to manufacture any desired f.

The Thin-Lens Equation and Image Tracing

Keeping the object finite, the same two-surface addition yields the thin lens formula:

$$\boxed{\frac{1}{v} - \frac{1}{u} = \frac{1}{f}}$$

(note the minus, unlike the mirror's plus!) — valid for convex and concave lenses, real and virtual images. A lens has two foci F and F', equidistant from the optical centre.

The three standard rays (any two locate the image):

1. Parallel to the axis → emerges through F' (convex) or appears to come from F (concave).
2. Through the optical centre → passes undeviated.
3. Through the first focus F (convex) → emerges parallel to the axis.

Magnification: $m = \dfrac{h'}{h} = \dfrac{v}{u}$ (no minus sign in the lens version!). $m < 0$: real and inverted; $m > 0$: virtual and erect.

The case map (convex lens): beyond 2F → image between F' and 2F' (real, inverted, diminished); at 2F → at 2F' (same size); between 2F and F → beyond 2F' (magnified); inside F → virtual, erect, magnified (the magnifying glass!). Concave lens: always virtual, erect, diminished, on the object's side.

[NEET Important] Mirror: $m = -v/u$, equation with +. Lens: $m = +v/u$, equation with −. Examiners live in this asymmetry.

Solved Examples

Example 1: The vanishing lens

A magician makes a glass lens (n = 1.47) disappear in a trough of liquid. What is the liquid's refractive index? Could it be water?

Solution:

1. The lens vanishes when it stops bending light: $\frac{1}{f} = (n_{21} - 1)(...) = 0$ requires $n_{21} = 1$, i.e. liquid index = 1.47.
2. Then $f \to \infty$: the lens acts as a plain glass sheet — invisible in the liquid.
3. Water (n = 1.33) won't do; glycerine (n about 1.47) would.

Example 2: Designing a lens [JEE Numerical]

A double-convex lens (n = 1.5) has radii 10 cm and 15 cm. Find its focal length.

Solution:

1. $R_1 = +10$ cm, $R_2 = -15$ cm.
2. $\frac{1}{f} = 0.5\left(\frac{1}{10} + \frac{1}{15}\right) = 0.5 \times \frac{1}{6} = \frac{1}{12}$.
3. $f = +12$ cm — converging, as the shape promised.

Example 3: Image hunt, convex lens [NEET Numerical]

An object stands 30 cm from a convex lens of f = +10 cm. Locate and describe the image.

Solution:

1. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$.
2. $v = +15$ cm (real, beyond F' on the far side).
3. $m = v/u = \frac{15}{-30} = -0.5$: real, inverted, half-size — the object is beyond 2F.

Example 4: Image hunt, concave lens [NEET Numerical]

The same object (30 cm away) faces a concave lens of f = -10 cm.

Solution:

1. $\frac{1}{v} = -\frac{1}{10} - \frac{1}{30} = -\frac{4}{30}$.
2. $v = -7.5$ cm: on the object's side.
3. $m = \frac{-7.5}{-30} = +0.25$: virtual, erect, quarter-size — the invariable concave-lens verdict.

Example 5: The magnifying-glass position [JEE Numerical]

Where must a stamp sit before a convex lens (f = +10 cm) to appear erect and 5x enlarged?

Solution:

1. Erect and magnified: virtual image, $m = +5 = v/u \Rightarrow v = 5u$.
2. $\frac{1}{5u} - \frac{1}{u} = \frac{1}{10} \Rightarrow -\frac{4}{5u} = \frac{1}{10} \Rightarrow u = -8$ cm.
3. Inside the focus (8 < 10) — exactly the case map's magnifier slot. Image at $v = -40$ cm.

Example 6: Plano-convex design [JEE Numerical]

What radius must the curved face of a plano-convex lens (n = 1.5) have for f = +20 cm?

Solution:

1. Flat face: $R_2 \to \infty$. $\frac{1}{f} = (n-1)\frac{1}{R_1}$.
2. $R_1 = (n-1)f = 0.5 \times 20 = 10$ cm.
3. One curved face does all the work; for n = 1.5, R is always half… rather, $R = f/2$. Equivalently an equiconvex lens (|R| both sides) with n = 1.5 has $f = R$.

Example 7: The lens in water [JEE Numerical]

A convex lens of f = +20 cm in air (n_glass = 1.5) is dipped in water ($n_w = 1.33$). Find its new focal length.

Solution:

1. In air: $\frac{1}{20} = (1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, so $\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = \frac{1}{10}$.
2. In water the relative index is $\frac{1.5}{1.33}$: $\frac{1}{f_w} = \left(\frac{1.5}{1.33}-1\right)\frac{1}{10} = \frac{0.17}{1.33}\times\frac{1}{10}$.
3. $f_w \approx +78.2$ cm — the lens converges nearly four times more weakly under water. (Why swimmers see blurry: their eye-lenses lose power the same way.)

Example 8: Same lens, both faces swapped

Does turning a lens around change its focal length?

Solution:

1. Reversing the lens swaps $R_1 \leftrightarrow -R_2$ (each radius flips sign and role).
2. $\frac{1}{R_1} - \frac{1}{R_2}$ is unchanged by that swap.
3. No — focal length (and power) is the same from either side; the two foci stay equidistant from the optical centre.

Example 9: Finding f from one conjugate pair [NEET Numerical]

A lens forms a real image at 60 cm of an object 20 cm away (on the other side). Find f and the lens type.

Solution:

1. Real image opposite side: $u = -20$ cm, $v = +60$ cm.
2. $\frac{1}{f} = \frac{1}{60} + \frac{1}{20} = \frac{4}{60}$, so $f = +15$ cm.
3. Positive f: convex lens; $m = 60/(-20) = -3$ — inverted, tripled.

Example 10: Why m has no minus for lenses

Reconcile $m_{mirror} = -v/u$ with $m_{lens} = +v/u$.

Solution:

1. Both come from the same similar-triangles construction; the sign difference is bookkeeping from the conventions (reflection folds distances back; refraction continues forward).
2. The decoder is identical: negative m = inverted (real), positive m = erect (virtual).
3. Trust the formulas with signed u, v — never re-derive signs midway through an exam.