Chapter 1 · Light – Reflection and Refraction

1.4 Spherical Lenses, the Lens Formula and Power

A lens refracts light twice, at two curved surfaces. Two rays find the image by drawing; the lens formula $\frac1v-\frac1u=\frac1f$, the magnification $m=\frac vu$ and the power $P=\frac1f$ in dioptres find it by calculation.

Hold a magnifying glass over a page and the print swells up; hold it at arm's length towards a window and an upside-down picture of the window appears on the wall. The lens in a phone camera throws a sharp image of the whole room onto a sensor the size of a fingernail, and the lenses in a pair of spectacles shift images just far enough for the eye to focus them. Every one of these uses a lens: a piece of transparent material whose curved surfaces refract light, by exactly the laws of Section 1.3, so that the rays meet — or seem to spread from — a single point.

This section works like Section 1.2 did for mirrors. We name the parts of a lens, draw images with two standard rays, then calculate with the lens formula and the New Cartesian sign convention. Finally we meet the quantity that opticians write on every prescription — the power of a lens, measured in dioptres — and see why powers simply add when lenses are placed together.

Convex and Concave Lenses

A lens bounded by two spherical surfaces is a spherical lens. A lens that is thicker at the middle than at the edges is convex (double-convex if both faces bulge); it bends parallel rays inwards to meet at a point, so it is also called a converging lens. A lens that is thinner at the middle is concave; it spreads parallel rays out as though they came from a point, so it is a diverging lens. Think of a ray at the top of a convex lens: it meets a slanting face, bends towards the normal on entering the glass and away from it on leaving, and both bends turn it towards the axis — like a ray passing through a thin prism.

📘 Definition 1.7 — Parts of a spherical lens
  • Centres of curvature $C_1$, $C_2$: the centres of the two spheres of which the lens faces are parts. A double-convex or double-concave lens has two, one for each face.
  • Principal axis: the straight line through both centres of curvature.
  • Optical centre $O$: the central point of the lens on the principal axis. A ray through $O$ passes undeviated.
  • Aperture: the effective diameter of the circular outline of the lens. We consider thin lenses of small aperture.
  • Principal foci $F_1$, $F_2$: rays parallel to the principal axis converge to a point $F_2$ on the far side of a convex lens, or appear to diverge from a point $F_1$ on the near side of a concave lens. Light can pass through a lens either way, so every lens has two principal foci, one on each side, at equal distances from $O$.
  • Focal length $f$: the distance from $O$ to a principal focus. Points at distance $2f$ on each side are labelled $2F_1$ and $2F_2$.
🧭 The standard rays for thin lenses
  1. A ray parallel to the principal axis: after refraction it passes through $F_2$ (convex), or appears to come from $F_1$ on the same side (concave).
  2. A ray through the optical centre $O$: goes straight on, undeviated.
  3. A ray through $F_1$ of a convex lens (or aimed at $F_2$ of a concave lens): emerges parallel to the principal axis.

Any two of these from the top of the object meet at the top of the image. If the emerging rays spread apart, trace them back (dashed) to where they appear to meet — a virtual image on the same side as the object.

Images Formed by Lenses

Moving an object in from far away towards a convex lens gives the six cases below, with $2F_1$ and $F_1$ playing the parts that $C$ and $F$ played for a concave mirror. Try each in Interactive 1.

Position of objectPosition of imageSize of imageNature
At infinityAt $F_2$Highly diminished, point-sizedReal, inverted
Beyond $2F_1$Between $F_2$ and $2F_2$DiminishedReal, inverted
At $2F_1$At $2F_2$Same sizeReal, inverted
Between $F_1$ and $2F_1$Beyond $2F_2$EnlargedReal, inverted
At $F_1$At infinityHighly enlargedReal, inverted
Between $F_1$ and $O$On the same side as the objectEnlargedVirtual, erect

A concave lens, like a convex mirror, always gives a virtual, erect, diminished image on the same side as the object:

Position of objectPosition of imageSize of imageNature
At infinityAt $F_1$Highly diminished, point-sizedVirtual, erect
Between infinity and $O$Between $F_1$ and $O$DiminishedVirtual, erect

The last row of the convex-lens table is the magnifying glass: hold the page closer to the lens than its focal length and you see an enlarged, erect image. The rows above it are the camera and the eye (distant object, small real image near $F_2$) and the projector (object just outside $F_1$, large real image far away on a screen).

Interactive 1 — The Lens Ray-Diagram Lab

Drag the tip of the green object (or use the slider). Two standard rays from its tip — parallel to the axis, and through the optical centre — refract at the lens and meet at the image, which is placed by the lens formula. Real images are solid red, on the far side; virtual images are dashed, on the object's side, with the rays traced back. Tick the third ray to see the ray through the focus. Play sweeps the object from far away up to the lens.

4015
$u$—
$v$—
$m=v/u$—
$P=1/f$—
image—
object is—

Sign Convention and the Lens Formula

We use the same New Cartesian sign convention as for mirrors (Definition 1.4), with one change: distances are measured from the optical centre $O$. The object is on the left, light travels left to right, so $u$ is always negative. The difference from mirrors is what happens to the light afterwards: it carries on through the lens, so a real image forms on the right and has $v$ positive, while a virtual image is on the left, with $v$ negative. The focus $F_2$ of a convex lens is on the right, so $f$ is positive for a convex lens and negative for a concave lens.

📐 Law 1.6 — The lens formula and magnification

For a thin spherical lens of small aperture, with $u$, $v$ and $f$ measured from the optical centre using the New Cartesian sign convention,

$$\frac1v-\frac1u=\frac1f .$$

The magnification produced by the lens is

$$m=\frac{h'}{h}=\frac vu .$$

As for mirrors: $m$ negative ⟹ real and inverted; $m$ positive ⟹ virtual and erect; $|m|\gt1$ enlarged, $|m|\lt1$ diminished. Note the differences from the mirror formulas — a minus sign between $\tfrac1v$ and $\tfrac1u$, and no minus sign in $m$.

🧭 Problem-solving strategy — lens numericals
  1. Write the convention: object on the left, distances from $O$, light left to right.
  2. Put signs on the data: $u$ negative; $f$ positive for convex, negative for concave. Same unit for every length.
  3. Rearrange: $\dfrac1v=\dfrac1f+\dfrac1u$, substitute with brackets round negative numbers, and combine the fractions.
  4. Find $m=\dfrac vu$ and $h'=mh$.
  5. Interpret: $v\gt0$ real, on the far side; $v\lt0$ virtual, on the object's side. Check against the tables above.
✏️ Example 1 — A real, enlarged image
A $2.0\ \text{cm}$ tall object is placed $15\ \text{cm}$ from a convex lens of focal length $10\ \text{cm}$. Find the position, size and nature of the image.
Signs New Cartesian convention: $u=-15\ \text{cm}$, $f=+10\ \text{cm}$ (convex), $h=+2.0\ \text{cm}$.
Formula $$\frac1v=\frac1f+\frac1u=\frac1{10}+\frac1{(-15)}=\frac{3-2}{30}=\frac1{30},\qquad v=+30\ \text{cm}.$$
Size $m=\dfrac vu=\dfrac{+30}{-15}=-2.0$, so $h'=mh=-4.0\ \text{cm}$.
Meaning The image is $30\ \text{cm}$ from the lens on the far side, real, inverted and twice the size ($4.0\ \text{cm}$ tall). Check: the object is between $F_1$ ($10\ \text{cm}$) and $2F_1$ ($20\ \text{cm}$), so the table predicts an enlarged real image beyond $2F_2$. ✓
✏️ Example 2 — The magnifying glass
A stamp is held $6.0\ \text{cm}$ from a convex lens of focal length $10\ \text{cm}$. Where is the image, and how much is the stamp magnified?
Signs $u=-6.0\ \text{cm}$, $f=+10\ \text{cm}$.
Formula $$\frac1v=\frac1{10}+\frac1{(-6)}=\frac{3-5}{30}=-\frac1{15},\qquad v=-15\ \text{cm}.$$
Size $m=\dfrac{-15}{-6.0}=+2.5$.
Meaning $v$ is negative: the image is $15\ \text{cm}$ from the lens on the same side as the stamp — virtual, erect and $2.5$ times as large. The stamp is inside $F_1$, as the magnifier row of the table requires. ✓
✏️ Example 3 — A diverging lens
A $3.0\ \text{cm}$ tall object is $10\ \text{cm}$ from a concave lens of focal length $15\ \text{cm}$. Find the image position and height.
Signs Concave: $f=-15\ \text{cm}$; $u=-10\ \text{cm}$; $h=+3.0\ \text{cm}$.
Formula $$\frac1v=\frac1{(-15)}+\frac1{(-10)}=-\frac{2+3}{30}=-\frac16,\qquad v=-6.0\ \text{cm}.$$
Size $m=\dfrac{-6.0}{-10}=+0.60$, $h'=+1.8\ \text{cm}$.
Meaning A virtual, erect image $1.8\ \text{cm}$ tall, $6.0\ \text{cm}$ from the lens on the object's side — between $F_1$ and $O$, exactly as the concave-lens table says.
✏️ Example 4 — Finding the focal length with a screen
A candle is $20\ \text{cm}$ from a lens. A sharp image of the flame is caught on a screen $60\ \text{cm}$ from the lens on the other side. Find the focal length, the type of lens and the magnification.
Signs The image is on a screen, so it is real and on the far side: $u=-20\ \text{cm}$, $v=+60\ \text{cm}$.
Formula $$\frac1f=\frac1v-\frac1u=\frac1{60}-\frac1{(-20)}=\frac{1+3}{60}=\frac1{15},\qquad f=+15\ \text{cm}.$$
Meaning $f$ is positive: a convex lens of focal length $15\ \text{cm}$. $m=\dfrac{60}{-20}=-3$: the flame's image is inverted and three times as tall.

Power of a Lens

A lens with a short focal length bends rays sharply; one with a long focal length bends them gently. Opticians therefore describe a lens not by $f$ but by its reciprocal — the bigger the number, the stronger the lens.

📘 Definition 1.8 — Power of a lens

The power $P$ of a lens is the reciprocal of its focal length in metres:

$$P=\frac1f .$$

The SI unit of power is the dioptre (D): $1\ \text{D}=1\ \text{m}^{-1}$, the power of a lens of focal length $1\ \text{m}$. A convex lens has positive power and a concave lens negative power. A prescription of "$-2.0\ \text{D}$" means a concave lens with $f=-0.50\ \text{m}$.

📐 Result 1.7 — Thin lenses in contact

When thin lenses of powers $P_1, P_2, P_3,\dots$ are placed in contact (on a common principal axis), the combination behaves as a single lens of power

$$P=P_1+P_2+P_3+\cdots$$

Opticians use this every day: a trial frame holds several lenses together and their powers are simply added until the patient sees clearly.

✏️ Example 5 — Powers of two lenses
Find the power of (a) a convex lens of focal length $25\ \text{cm}$; (b) a concave lens of focal length $40\ \text{cm}$. (c) A lens has power $-2.0\ \text{D}$. What are its type and focal length?
(a) $f=+25\ \text{cm}=+0.25\ \text{m}$, so $P=\dfrac{1}{0.25}=+4.0\ \text{D}$.
(b) $f=-40\ \text{cm}=-0.40\ \text{m}$, so $P=\dfrac{1}{-0.40}=-2.5\ \text{D}$.
(c) $f=\dfrac1P=\dfrac{1}{-2.0}=-0.50\ \text{m}=-50\ \text{cm}$: a concave (diverging) lens of focal length $50\ \text{cm}$.
✏️ Example 6 — Lenses in contact
A convex lens of power $+5.0\ \text{D}$ is placed in contact with a concave lens of power $-2.0\ \text{D}$. Find the power and focal length of the combination. Is it converging or diverging?
Add $P=P_1+P_2=(+5.0)+(-2.0)=+3.0\ \text{D}$.
Focal length $f=\dfrac1P=\dfrac{1}{3.0}=+0.33\ \text{m}\approx33\ \text{cm}$.
Meaning Positive power: the pair acts as a single converging lens of focal length about $33\ \text{cm}$. Check: focal lengths do not add — $20\ \text{cm}+(-50\ \text{cm})=-30\ \text{cm}$ would be wrong. Try it in Interactive 2.
Interactive 2 — Power and Lenses in Contact

A parallel beam meets one lens, or two thin lenses in contact. Set each power in dioptres. The beam converges to (or diverges from) the focus at $f=\tfrac1P$, with $P=P_1+P_2$; the pulses show the light travelling. The scale along the axis is in centimetres. Press Play to sweep the second lens from strongly diverging to strongly converging.

+5.0−2.0
$f_1$—
$f_2$—
$P=P_1+P_2$—
$f=1/P$—
combination—
⚠️ Common mistakes

1. Using the mirror formula. For lenses it is $\dfrac1v-\dfrac1u=\dfrac1f$ and $m=+\dfrac vu$. Mixing these up with $\dfrac1v+\dfrac1u=\dfrac1f$ and $m=-\dfrac vu$ is the commonest slip in board papers.

2. Power in centimetres. $P=\dfrac1f$ needs $f$ in metres. A lens of $f=20\ \text{cm}$ has $P=\dfrac{1}{0.20}=+5.0\ \text{D}$, not $\dfrac1{20}=0.05\ \text{D}$. Shortcut: $P=\dfrac{100}{f\ (\text{in cm})}$.

3. Wrong sign of $f$. Convex lens: $f$ positive. Concave lens: $f$ negative — the reverse of mirrors, where the concave one has negative $f$.

4. Adding focal lengths. For lenses in contact add powers, then take the reciprocal.

🚀 Beyond the board syllabus (JEE Advanced) — optional: two lenses a distance apart

In JEE Advanced 2026 ("combinations of mirrors and thin lenses"); JEE Main and NEET 2026 list only lenses in contact, and it is not in the CBSE Class 10 syllabus.

When two lenses are separated, use the lens formula twice: the image formed by the first lens acts as the object for the second. For example, an object is $30\ \text{cm}$ in front of a convex lens $L_1$ ($f_1=+15\ \text{cm}$); a second convex lens $L_2$ ($f_2=+10\ \text{cm}$) is $50\ \text{cm}$ beyond $L_1$. For $L_1$: $\tfrac1{v_1}=\tfrac1{15}-\tfrac1{30}=\tfrac1{30}$, so $v_1=+30\ \text{cm}$ and $m_1=-1$. That image is $50-30=20\ \text{cm}$ in front of $L_2$, so $u_2=-20\ \text{cm}$: $\tfrac1{v_2}=\tfrac1{10}-\tfrac1{20}=\tfrac1{20}$, $v_2=+20\ \text{cm}$, $m_2=-1$. The final image is $20\ \text{cm}$ beyond $L_2$, and the total magnification is $m_1m_2=+1$ — erect, real and the same size. Microscopes and telescopes (Class 12 Physics, Ray Optics) are built this way. The rule $P=P_1+P_2$ is the special case of zero separation.

Exercises

I
Images by drawing. A convex lens has $f=10\ \text{cm}$. Describe the position, size and nature of the image when the object is at the given distance from it, and draw a ray diagram for each.
  • 1 $30\ \text{cm}$
  • 2 $20\ \text{cm}$
  • 3 $15\ \text{cm}$
  • 4 $10\ \text{cm}$
  • 5 $5\ \text{cm}$
  • 6 Describe the image formed by a concave lens for an object anywhere in front of it.
  • 7 Which lens is used as a magnifying glass, and where must the object be placed?
  • 8 You are handed two lenses that look alike. Without measuring anything, how can you tell which is convex?
1 Beyond $2F_1$: image between $F_2$ and $2F_2$, real, inverted, diminished ($v=+15\ \text{cm}$, $m=-0.5$).
2 At $2F_1$: image at $2F_2$, real, inverted, same size ($v=+20\ \text{cm}$, $m=-1$).
3 Between $F_1$ and $2F_1$: image beyond $2F_2$, real, inverted, enlarged ($v=+30\ \text{cm}$, $m=-2$).
4 At $F_1$: the refracted rays are parallel; the image is at infinity, real, inverted, highly enlarged.
5 Between $F_1$ and $O$: image on the same side as the object, virtual, erect, enlarged ($v=-10\ \text{cm}$, $m=+2$).
6 Between $F_1$ and $O$ on the object's side; virtual, erect and diminished.
7 A convex lens, with the object between $O$ and $F_1$ (closer than the focal length).
8 Look at nearby print through each lens: the convex lens magnifies it (erect, enlarged); the concave lens makes it look smaller. Or feel them: the convex lens is thicker in the middle.
II
The lens formula. Use the New Cartesian sign convention throughout. Find $v$ and $m$ (or $f$ where asked) and state the nature of the image.
  • 1 Convex, $f=20\ \text{cm}$, object $30\ \text{cm}$ from the lens.
  • 2 Convex, $f=20\ \text{cm}$, object $60\ \text{cm}$ from the lens.
  • 3 Convex, $f=20\ \text{cm}$, object $12\ \text{cm}$ from the lens.
  • 4 Concave, $f=20\ \text{cm}$, object $20\ \text{cm}$ from the lens.
  • 5 Concave, $f=12\ \text{cm}$, object $24\ \text{cm}$ from the lens.
  • 6 An object $30\ \text{cm}$ from a lens gives a real image $20\ \text{cm}$ from it. Find $f$ and the type of lens.
  • 7 An object $20\ \text{cm}$ from a lens gives a virtual image $10\ \text{cm}$ from the lens. Find $f$ and the type of lens.
  • 8 An object $8.0\ \text{cm}$ from a lens gives a virtual image $24\ \text{cm}$ from the lens. Find $f$, the type of lens and $m$.
  • 9 An object is at the focus of a convex lens of focal length $15\ \text{cm}$. What does the formula say about $v$?
Check: type in $u$ and $f$ with their signs and compare $v$ and $m$ with your answer
$v$—
$m$—
image—
1 $u=-30$, $f=+20$: $\tfrac1v=\tfrac1{20}-\tfrac1{30}=\tfrac1{60}$, $v=+60\ \text{cm}$; $m=-2$. Real, inverted, enlarged, on the far side.
2 $\tfrac1v=\tfrac1{20}-\tfrac1{60}=\tfrac2{60}$, $v=+30\ \text{cm}$; $m=-0.5$. Real, inverted, diminished.
3 $\tfrac1v=\tfrac1{20}-\tfrac1{12}=\tfrac{3-5}{60}=-\tfrac1{30}$, $v=-30\ \text{cm}$; $m=+2.5$. Virtual, erect, enlarged, on the object's side.
4 $f=-20$: $\tfrac1v=-\tfrac1{20}-\tfrac1{20}=-\tfrac1{10}$, $v=-10\ \text{cm}$; $m=+0.5$. Virtual, erect, half size.
5 $f=-12$: $\tfrac1v=-\tfrac1{12}-\tfrac1{24}=-\tfrac3{24}$, $v=-8.0\ \text{cm}$; $m=+0.33$. Virtual, erect, diminished.
6 $u=-30$, $v=+20$: $\tfrac1f=\tfrac1{20}+\tfrac1{30}=\tfrac5{60}$, $f=+12\ \text{cm}$: convex.
7 $u=-20$, $v=-10$: $\tfrac1f=-\tfrac1{10}+\tfrac1{20}=-\tfrac1{20}$, $f=-20\ \text{cm}$: concave.
8 $u=-8.0$, $v=-24$: $\tfrac1f=-\tfrac1{24}+\tfrac18=\tfrac2{24}$, $f=+12\ \text{cm}$: convex (used as a magnifier); $m=\tfrac{-24}{-8.0}=+3.0$.
9 $\tfrac1v=\tfrac1{15}+\tfrac1{(-15)}=0$: no finite $v$ — the refracted rays are parallel and the image is at infinity.
III
Magnification and power.
  • 1 A $3.0\ \text{cm}$ object is $20\ \text{cm}$ from a convex lens of focal length $15\ \text{cm}$. Find the image distance and height.
  • 2 Find the power of a convex lens of focal length $50\ \text{cm}$.
  • 3 Find the power of a concave lens of focal length $20\ \text{cm}$.
  • 4 A lens has power $+2.5\ \text{D}$. Find its focal length and type.
  • 5 A lens has power $-4.0\ \text{D}$. Find its focal length and type.
  • 6 Lenses of power $+4.0\ \text{D}$ and $-1.5\ \text{D}$ are placed in contact. Find the power and focal length of the combination.
  • 7 Two convex lenses of focal lengths $20\ \text{cm}$ and $30\ \text{cm}$ are placed in contact. Find the power and focal length of the combination.
  • 8 A convex lens forms a real image twice the size of the object, $36\ \text{cm}$ from the lens. Find the object distance, the focal length and the power.
1 $\tfrac1v=\tfrac1{15}-\tfrac1{20}=\tfrac1{60}$, $v=+60\ \text{cm}$; $m=\tfrac{60}{-20}=-3.0$; $h'=-9.0\ \text{cm}$ (real, inverted, $9.0\ \text{cm}$ tall).
2 $P=\tfrac1{0.50}=+2.0\ \text{D}$.
3 $P=\tfrac{1}{-0.20}=-5.0\ \text{D}$.
4 $f=\tfrac1{2.5}=+0.40\ \text{m}=+40\ \text{cm}$: convex.
5 $f=\tfrac1{-4.0}=-0.25\ \text{m}=-25\ \text{cm}$: concave.
6 $P=4.0-1.5=+2.5\ \text{D}$; $f=+0.40\ \text{m}=40\ \text{cm}$ (converging).
7 $P=\tfrac1{0.20}+\tfrac1{0.30}=5.0+3.3=+8.3\ \text{D}$; $f=\tfrac1{8.33}=0.12\ \text{m}=12\ \text{cm}$.
8 Real, so $m=-2$: $v=+36\ \text{cm}$, $u=\tfrac vm=-18\ \text{cm}$. $\tfrac1f=\tfrac1{36}+\tfrac1{18}=\tfrac3{36}$, $f=+12\ \text{cm}$; $P=\tfrac1{0.12}=+8.3\ \text{D}$.
IV
Applications and challenges.
  • 1 Camera. A camera lens of focal length $5.0\ \text{cm}$ photographs a person $1.7\ \text{m}$ tall standing $2.0\ \text{m}$ away. How far behind the lens must the sensor be, and how tall is the image?
  • 2 Projector. A slide $2.5\ \text{cm}$ wide must fill a screen $1.0\ \text{m}$ wide, $5.0\ \text{m}$ from the lens. Find the magnification, the distance of the slide from the lens, and the focal length and power of the lens.
  • 3 Reading glasses. A prescription reads $+2.5\ \text{D}$. What kind of lens is it and what is its focal length?
  • 4 Trial frame. An optician places a $+3.0\ \text{D}$ lens and a $-1.0\ \text{D}$ lens together in a trial frame. What single lens would replace them?
  • 5 Challenge. A candle and a screen are fixed $90\ \text{cm}$ apart. A convex lens of focal length $20\ \text{cm}$ is moved between them. Show that a sharp image is formed for two positions of the lens, and find them.
  • 6 Challenge. Prove from the lens formula that a concave lens always forms its image between $O$ and $F_1$.
1 $u=-200\ \text{cm}$, $f=+5.0\ \text{cm}$: $\tfrac1v=\tfrac15-\tfrac1{200}=\tfrac{39}{200}$, $v=+5.13\ \text{cm}$ (just beyond $F_2$). $m=\tfrac{5.13}{-200}=-0.0256$, $h'=-0.0256\times170=-4.4\ \text{cm}$ (inverted, $4.4\ \text{cm}$ tall).
2 Real image: $m=-\tfrac{100}{2.5}=-40$; $v=+500\ \text{cm}$, $u=\tfrac vm=-12.5\ \text{cm}$. $\tfrac1f=\tfrac1{500}+\tfrac1{12.5}=0.082\ \text{cm}^{-1}$, $f=12\ \text{cm}$ ($12.2\ \text{cm}$); $P=\tfrac1{0.122}=+8.2\ \text{D}$.
3 Positive power: a convex (converging) lens, $f=\tfrac1{2.5}=0.40\ \text{m}=40\ \text{cm}$.
4 $P=3.0-1.0=+2.0\ \text{D}$: a single convex lens of focal length $50\ \text{cm}$.
5 Let the lens be $x$ cm from the candle: $u=-x$, $v=+(90-x)$. Then $\tfrac1{90-x}+\tfrac1x=\tfrac1{20}$, so $20\cdot90=x(90-x)$, i.e. $x^2-90x+1800=0$, $(x-30)(x-60)=0$. The lens can be $30\ \text{cm}$ or $60\ \text{cm}$ from the candle (image $60$ or $30\ \text{cm}$ beyond it; enlarged $m=-2$ or diminished $m=-0.5$).
6 With $u=-|u|$ and $f=-|f|$: $\tfrac1v=\tfrac1f+\tfrac1u=-\left(\tfrac1{|f|}+\tfrac1{|u|}\right)$. This is negative, so $v\lt0$ (object's side), and its size exceeds $\tfrac1{|f|}$, so $|v|\lt|f|$. The image lies between $O$ and $F_1$.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 3 marks An object $4.0\ \text{cm}$ high is placed $18\ \text{cm}$ from a convex lens of focal length $12\ \text{cm}$. Find the position, nature and size of the image, and the power of the lens.

Convention New Cartesian: $u=-18\ \text{cm}$, $f=+12\ \text{cm}$, $h=+4.0\ \text{cm}$.
Working $\tfrac1v=\tfrac1{12}-\tfrac1{18}=\tfrac{3-2}{36}=\tfrac1{36}$, $v=+36\ \text{cm}$; $m=\tfrac{36}{-18}=-2.0$; $h'=-8.0\ \text{cm}$. $P=\tfrac1{0.12}=+8.3\ \text{D}$.
Answer $36\ \text{cm}$ beyond the lens; real, inverted, $8.0\ \text{cm}$ tall; $P=+8.3\ \text{D}$. Exam tip: convert $f$ to metres before finding $P$ — $\tfrac{1}{12}$ is not the power.

E2. Board · 5 marks (a) Draw a ray diagram showing how a convex lens forms a virtual image, and name a device that uses it. (b) An object is placed $30\ \text{cm}$ from a concave lens of focal length $20\ \text{cm}$. Find the position of the image and the magnification, and draw the ray diagram.

(a) Object between $F_1$ and $O$. A ray parallel to the axis refracts through $F_2$; a ray through $O$ goes straight. They diverge; produced backwards (dashed) they meet on the object's side, farther away than the object: an enlarged, erect, virtual image. Device: the magnifying glass (simple microscope).
(b) Working $u=-30\ \text{cm}$, $f=-20\ \text{cm}$: $\tfrac1v=-\tfrac1{20}-\tfrac1{30}=-\tfrac5{60}$, $v=-12\ \text{cm}$; $m=\tfrac{-12}{-30}=+0.40$.
(b) Diagram A ray parallel to the axis diverges as if from $F_1$ (dashed back to $F_1$); a ray through $O$ is undeviated. The dashed line and the central ray meet $12\ \text{cm}$ from the lens on the object's side.
Answer Image $12\ \text{cm}$ from the lens on the object's side; virtual, erect, $0.40$ times the size. Exam tip: mark $F_1$, $F_2$, $2F_1$, $2F_2$ on every lens diagram — they carry marks.

E3. Board · Assertion–Reason Assertion (A): The power of a concave lens is negative. Reason (R): In the New Cartesian sign convention the focal length of a concave lens is negative. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea $F_1$ of a concave lens is on the object's side, against the incident light, so $f\lt0$; and $P=\tfrac1f$ takes the same sign.
Answer (A). Exam tip: the sign of $P$ tells you the lens type at a glance — "+" converging, "−" diverging.

E4. Board · Case study A science club has three thin lenses of powers $+2.0\ \text{D}$, $+4.0\ \text{D}$ and $-1.0\ \text{D}$. (i) Find the focal length of each lens. [2 marks] (ii) Which two lenses, in contact, give a combination of focal length about $33\ \text{cm}$? [1 mark] (iii) What are the power and nature of all three in contact? [1 mark]

(i) $f=\tfrac1P$: $+0.50\ \text{m}$ ($50\ \text{cm}$, convex), $+0.25\ \text{m}$ ($25\ \text{cm}$, convex), $-1.0\ \text{m}$ ($100\ \text{cm}$, concave).
(ii) $f=0.33\ \text{m}$ needs $P=+3.0\ \text{D}$: the $+4.0\ \text{D}$ and $-1.0\ \text{D}$ lenses.
(iii) $P=2.0+4.0-1.0=+5.0\ \text{D}$: converging, $f=20\ \text{cm}$. Exam tip: in case studies, write $P=P_1+P_2+\dots$ before adding — the formula line earns a mark.

E5. NEET A convex lens of focal length $20\ \text{cm}$ forms a real image of the same size as the object. The distance of the object from the lens is (A) $10\ \text{cm}$ (B) $20\ \text{cm}$ (C) $40\ \text{cm}$ (D) $80\ \text{cm}$

Working Real, same size: $m=-1$, so $v=-u$. Then $\tfrac1{-u}-\tfrac1u=-\tfrac2u=\tfrac1{20}$ gives $u=-40\ \text{cm}$ (the object is at $2F_1$).
Answer (C). Exam tip: "real and same size" always means object and image at $2F$ — no calculation needed.

E6. JEE Main Thin lenses of power $+3.0\ \text{D}$ and $-1.0\ \text{D}$ are in contact. An object is placed $1.0\ \text{m}$ in front of the combination. The image is formed at (A) $0.50\ \text{m}$ (B) $1.0\ \text{m}$ (C) $2.0\ \text{m}$ (D) $0.33\ \text{m}$ behind the lenses

Combine $P=+2.0\ \text{D}$, so $\tfrac1f=2.0\ \text{m}^{-1}$.
Working In metres, $u=-1.0$: $\tfrac1v=\tfrac1f+\tfrac1u=2.0-1.0=1.0\ \text{m}^{-1}$, $v=+1.0\ \text{m}$.
Answer (B). Exam tip: work in metres and dioptres — then $\tfrac1f$ is just $P$ and no conversion is needed.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How accurately can you measure the power of a lens with nothing more than a window, a sheet of paper and a ruler — and is that good enough to check a pair of reading glasses?

Why it matters. The World Health Organization estimates that around a billion people worldwide live with a vision problem that could have been prevented or simply has not been corrected — very often just the need for reading glasses. Low-cost "ready-made" readers, sold in steps of $0.5\ \text{D}$, and vision camps in rural India and Africa help close this gap, but someone has to check that each lens has the power printed on it.

What to do.

  1. Stand at the back of a room facing a bright window. Hold a convex lens (a magnifying glass or one lens of a pair of reading glasses) so that it throws a sharp, inverted image of the window onto a sheet of white paper. Measure the lens-to-paper distance. Explain why this is (very nearly) the focal length.
  2. Repeat five times and average. Calculate the power $P=\tfrac1f$ with $f$ in metres.
  3. Compare your answer with the power written on the glasses (or the box). By what percentage do they differ?
  4. Place two convex lenses in contact and measure the focal length of the pair. Does $P=P_1+P_2$ hold within your measurement error?

Tools: convex lenses or reading glasses of known power, white card, metre rule, calculator.

Hand in: A results table, your measured powers with an estimate of their uncertainty, and a short note on whether this method could be used at a vision camp to sort ready-made readers.

Going further. A window $5\ \text{m}$ away is not at infinity. Use the lens formula to find how much this makes your measured $f$ differ from the true focal length for a $+2.0\ \text{D}$ lens.

Instructor note: ready-made readers usually run from $+1.0$ to $+3.5\ \text{D}$, i.e. $f=100\ \text{cm}$ down to about $29\ \text{cm}$ ($+2.0\ \text{D}$ → $50\ \text{cm}$, $+2.5\ \text{D}$ → $40\ \text{cm}$, $+3.0\ \text{D}$ → $33\ \text{cm}$). For "going further": with $u=-5.0\ \text{m}$ and $P=+2.0\ \text{D}$, $\tfrac1v=2.0-0.20=1.8\ \text{m}^{-1}$, so $v=0.556\ \text{m}$ instead of $0.500\ \text{m}$ — the method over-estimates $f$ by about 11% and under-estimates $P$ ($1.8$ instead of $2.0\ \text{D}$). A window $10\ \text{m}$ away halves the error; students should notice that the error matters most for weak lenses.

🧪 Self-check quiz

0/5 correct

Q1. With the New Cartesian sign convention, the lens formula and magnification are

Q2. The power of a concave lens of focal length $25\ \text{cm}$ is

Q3. An object is placed at $2F_1$ of a convex lens. The image is

Q4. Lenses of power $+2.0\ \text{D}$ and $-0.5\ \text{D}$ are placed in contact. The focal length of the combination is about

Q5. Whatever the position of the object, a concave lens forms an image that is

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