Refraction by a lens
A double convex lens is bounded by two spherical surfaces. NCERT models the image formation as two sequential refractions at single spherical surfaces. The first surface forms an intermediate image $I_1$ of the object $O$; this image then acts as a virtual object for the second surface, which forms the final image $I$. Each refraction obeys the single-surface relation $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$.
For a thin lens, the two surfaces lie so close together that the intermediate image distance measured from either face is effectively the same. Adding the two single-surface equations then cancels the intermediate term and leaves a single relation governing the whole lens. This two-step construction is the foundation of both formulas derived below.
The first surface (toward the object) has radius $R_1$ with centre $C_1$; the second surface has radius $R_2$ with centre $C_2$. For a biconvex lens the sign convention makes $R_1$ positive and $R_2$ negative.
The lens maker's formula
Take the object at infinity, so the parallel incident beam converges to the focus and the final image distance equals the focal length $f$. Substituting $R_1 = +R_1$ and $R_2 = -R_2$ from the sign convention into the summed single-surface equations gives the result NCERT labels Eq. (9.21):
$$\frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Here $n_{21} = n_2/n_1$ is the refractive index of the lens material ($n_2$) relative to the surrounding medium ($n_1$). This is the lens maker's formula; it lets a maker design a lens of any desired focal length by choosing the material and the two radii. NCERT notes the formula is valid for a concave lens too, where $R_1$ is negative and $R_2$ positive, producing a negative $f$.
| Quantity | Symbol | Meaning | Sign for biconvex |
|---|---|---|---|
| Relative index | n21 = n2/n1 | Lens material relative to medium | > 1 (e.g. 1.5 in air) |
| First radius | R1 | Curvature of surface facing object | positive |
| Second radius | R2 | Curvature of far surface | negative |
| Focal length | f | Image distance for object at infinity | positive (converging) |
The medium and the radii both carry signs
Two errors recur. First, students drop the sign convention on $R_1$ and $R_2$: for a biconvex lens $R_1 > 0$ and $R_2 < 0$, so $\left(\tfrac{1}{R_1} - \tfrac{1}{R_2}\right)$ is a sum of two positive terms, not a difference. Second, $n_{21}$ depends on the surrounding medium, not just the glass. The focal length therefore changes when a lens is immersed in water, because $n_{21}$ falls from about $1.5$ in air to about $1.5/1.33 \approx 1.13$ in water, weakening the lens. NCERT Example 9.6 takes this to the limit: if the liquid matches the glass index, $n_{21} = 1$, so $1/f = 0$ and the lens disappears optically.
In air a glass lens of $n=1.5$ with $R_1 = -R_2 = R$ has $f = R$. Immerse it in a higher-index liquid and $f$ can grow, flip sign, or vanish.
Sign convention and focal length
All lens problems use the Cartesian sign convention: distances are measured from the optical centre, with those along the direction of incident light taken positive and those against it negative. A converging (convex) lens has a positive focal length; a diverging (concave) lens has a negative focal length. A lens has two foci, $F$ and $F'$, equidistant from the optical centre on either side.
| Lens type | R1 | R2 | Focal length f | Behaviour |
|---|---|---|---|---|
| Biconvex | positive | negative | positive | converging |
| Biconcave | negative | positive | negative | diverging |
| Plano-convex | positive (or ∞) | ∞ (or negative) | positive | converging |
| Plano-concave | ∞ (or negative) | positive (or ∞) | negative | diverging |
The reciprocal of focal length is the lens power, which adds simply when lenses touch. See Power of a Lens and Combination of Lenses.
The thin lens equation
When the object is at a finite distance, the same two-step refraction analysis, with the sign convention $BO = -u$ and $DI = +v$, yields NCERT Eq. (9.23), the familiar thin lens equation:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
NCERT stresses that although it is derived for a real image from a convex lens, this formula holds for both convex and concave lenses and for both real and virtual images. The minus sign before $1/u$ is what distinguishes it from the mirror equation, which carries a plus sign.
In every numerical, substitute object distance $u$, image distance $v$ and focal length $f$ with their signs from the convention before solving, then read off the sign of the answer to decide whether the image is real or virtual. A positive $v$ places the image on the far side of the lens and marks it as real; a negative $v$ places it on the same side as the object and marks it as virtual. The two foci $F$ and $F'$ being equidistant means the same equation describes light incident from either direction.
Lens magnification is m = v/u, not −v/u
Lenses and mirrors share the symbol $m$ but use different definitions. For a lens, $m = \dfrac{h'}{h} = \dfrac{v}{u}$ with no minus sign, whereas a mirror uses $m = -\dfrac{v}{u}$. With the lens convention, an erect (virtual) image gives positive $m$ and an inverted (real) image gives negative $m$. Likewise the lens equation reads $\tfrac{1}{v} - \tfrac{1}{u} = \tfrac{1}{f}$ against the mirror's $\tfrac{1}{v} + \tfrac{1}{u} = \tfrac{1}{f}$. Mixing the two sets of signs is a classic source of lost marks.
Lens: $m = v/u$, equation has $-1/u$. Mirror: $m = -v/u$, equation has $+1/u$.
Magnification and image nature
The magnification fixes both the size and the orientation of the image. For a real object, a convex lens forms a real, inverted image when the object lies beyond the focus, and a virtual, erect, magnified image when the object lies within the focus. A concave lens always forms a virtual, erect, diminished image of a real object. The ray diagram below contrasts the two converging cases.
Object beyond $F$ on the left: the parallel ray refracts through $F'$ and the central ray passes undeviated, meeting on the right to form a real, inverted image. Here $v > 0$, so $m = v/u < 0$ (inverted).
| Lens & object | Image distance v | Magnification m = v/u | Image nature |
|---|---|---|---|
| Convex, object beyond F | positive | negative | real, inverted |
| Convex, object within F | negative | positive (> 1) | virtual, erect, magnified |
| Concave, any real object | negative | positive (< 1) | virtual, erect, diminished |
Worked examples
A biconvex lens has both radii of curvature equal to $20\,\text{cm}$. The material has refractive index $1.5$. Find the focal length and power in air.
Apply the sign convention: $R_1 = +20\,\text{cm}$, $R_2 = -20\,\text{cm}$.
$\dfrac{1}{f} = (1.5 - 1)\left(\dfrac{1}{20} - \dfrac{1}{-20}\right) = 0.5 \times \dfrac{2}{20} = \dfrac{1}{20}\,\text{cm}^{-1}$.
So $f = 20\,\text{cm} = 0.20\,\text{m}$, and the power is $P = \dfrac{1}{f(\text{m})} = \dfrac{1}{0.20} = +5\,\text{D}$. This reproduces the NEET 2022 result for the same lens.
A magician makes a glass lens with $n = 1.47$ disappear in a trough of liquid. What is the refractive index of the liquid, and could it be water?
For the lens to vanish optically its power must be zero, i.e. $1/f = 0$. From the lens maker's formula this requires $n_{21} = 1$, so $n_2 = n_1$. The liquid must have refractive index $1.47$.
Water has $n \approx 1.33$, not $1.47$, so the liquid is not water; NCERT notes it could be glycerine.
An object is placed $60\,\text{cm}$ in front of a convex lens of focal length $30\,\text{cm}$. Find the image distance and magnification.
Using $u = -60\,\text{cm}$, $f = +30\,\text{cm}$ in $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$:
$\dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{-60} = \dfrac{2 - 1}{60} = \dfrac{1}{60}$, giving $v = +60\,\text{cm}$ (real image).
$m = \dfrac{v}{u} = \dfrac{60}{-60} = -1$: the image is real, inverted, and the same size. This matches the first refraction step of NEET 2021 Q.37.
Lens maker's formula and the thin lens equation in one glance
- Lens maker's formula: $\dfrac{1}{f} = (n_{21} - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$, valid for convex and concave lenses.
- $n_{21} = n_2/n_1$ depends on the surrounding medium; immersing a lens in a liquid changes $f$, and $n_{21}=1$ makes the lens vanish.
- Thin lens equation: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ — note the minus sign, unlike the mirror's plus.
- Convex lens: $f > 0$ (converging). Concave lens: $f < 0$ (diverging).
- Lens magnification $m = v/u$ (no minus). Negative $m$ = real inverted; positive $m$ = virtual erect.
- Power $P = 1/f$ in dioptres with $f$ in metres; a biconvex lens of $f=20\,\text{cm}$ has $P=+5\,\text{D}$.