Thin lenses and the lens maker's formula

A glass that can start a fire

Hold a magnifying glass in bright sunshine over a dry leaf. The sunlight gathers into a small, very bright spot. The leaf warms, and then it may smoke. How can a piece of clear glass do this? It bends the light that passes through it, and it bends all the rays the same way, towards one point. This is a lens.

In this lesson we learn the rules that tell us where a lens makes an image. Then we ask a second question: why does one lens have a long focal length and another a short one? The answer is the shape of the glass and the kind of glass.

Two kinds of lens

A lens is a piece of transparent material with two curved faces, or one curved and one flat. A lens that is thicker in the middle than at the edge is convex. It makes parallel rays come together, so it is also called a converging lens. A lens that is thinner in the middle is concave. It makes parallel rays spread out, so it is a diverging lens.

This is true when the glass is denser than the air around it. In a liquid of higher refractive index than the glass, the two kinds swap their jobs. We come back to this at the end of the lesson.

A lens is thin when its thickness is very small against the radii of its faces and against the distances we use. Then we can measure every distance from one point in the middle, the optical centre $O$. A ray that goes through $O$ is not bent. The straight line through $O$ and the centres of the two faces is the principal axis.

The focus and the focal length

Send a beam of rays parallel to the principal axis through a convex lens. The rays meet at one point on the axis, the second principal focus $F_2$. Its distance from the lens is the focal length $f$. A lens has a second focus on the other side too, the first principal focus $F_1$, at the same distance. A beam that leaves the lens parallel to the axis came from there.

For a concave lens the rays after the lens spread out. They seem to come from a point on the side of the incoming light. That point is its focus, and by the sign convention below its focal length is negative.

Three rays that tell the whole story

To find the image of the top of an object, we need only two rays. We draw three, and the third is a check.

1. A ray parallel to the axis goes, after the lens, through $F_2$. For a concave lens it spreads as if it came from $F_2$.
2. A ray through the optical centre goes straight on, because the thin lens is like a flat sheet at its middle.
3. A ray aimed at $F_1$ leaves the lens parallel to the axis.

Where the rays meet, or where their backward extensions meet, is the top of the image.

The picture is drawn to scale. The object is $40\ \text{cm}$ in front of a lens of $f = 15\ \text{cm}$. The three rays meet $24\ \text{cm}$ behind the lens. The image is on the other side from the object, so it is real. It is upside down and smaller.

The sign convention

To use a formula, we must say which distances are positive. We use the New Cartesian convention, which NCERT uses.

• The light travels from left to right. The principal axis is the $x$ axis.
• All distances are measured from the optical centre $O$.
• A distance in the direction of the light, to the right, is positive. A distance against the light, to the left, is negative.
• Heights above the axis are positive. Heights below it are negative.

A real object is on the left, so its distance $u$ is negative. A real image is on the right, so $v$ is positive. A convex lens has a positive $f$ and a concave lens has a negative $f$.

A-Level, IB and AP books often use another rule, real is positive. A real object and a real image have positive distances $u$ and $v$, and a converging lens has a positive $f$. The formula then looks different, and the answer is the same.

• New Cartesian (NCERT): $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$, with $u < 0$ for a real object and $m = \dfrac{v}{u}$.
• Real is positive (A-Level, IB, AP): $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$, with $u > 0$ for a real object and $m = -\dfrac{v}{u}$.

Replace $u$ by $-u$ in the first form and you get the second. In this lesson we use the New Cartesian form, and we say so whenever the other one could confuse.

The thin lens formula

For a thin lens the object distance $u$, the image distance $v$ and the focal length $f$ are linked by

The linear magnification $m$ is the height of the image divided by the height of the object. For a thin lens it equals the ratio of the distances:

A negative $m$ means an inverted image. A positive $m$ means an upright one. If $|m| > 1$ the image is larger than the object.

Worked example 1: a real image

An object $4.0\ \text{cm}$ tall stands $40\ \text{cm}$ in front of a convex lens of $f = 15\ \text{cm}$. Find the image distance, the magnification and the height of the image.

1. Signs. $u = -40\ \text{cm}$ and $f = +15\ \text{cm}$.
2. The distance. We want $v$, so $v = \dfrac{uf}{u + f} = \dfrac{(-40\ \text{cm})(15\ \text{cm})}{-40\ \text{cm} + 15\ \text{cm}} = 24\ \text{cm}$.
3. The magnification. $m = \dfrac{v}{u} = \dfrac{24\ \text{cm}}{-40\ \text{cm}} = -0.60$.
4. The height. $h' = mh = -0.60 \times 4.0\ \text{cm} = -2.4\ \text{cm}$.

Check. The image is real ($v > 0$), inverted ($m < 0$) and smaller. This agrees with the picture. A check with the formula: $\dfrac{1}{24} - \dfrac{1}{-40} = 0.0667\ \text{cm}^{-1} = \dfrac{1}{15\ \text{cm}}$.

What happens as the object comes closer

Keep a convex lens of focal length $f$ and move the object from far away towards the lens.

• Very far away. The image is at $F_2$. It is real, inverted and tiny.
• Beyond $2F_1$. The image is between $F_2$ and $2F_2$. It is real, inverted and smaller than the object.
• At $2F_1$. The image is at $2F_2$, real, inverted and the same size.
• Between $F_1$ and $2F_1$. The image is beyond $2F_2$. It is real, inverted and larger.
• At $F_1$. The rays leave parallel. The image is at infinity.
• Between $F_1$ and the lens. The image is on the same side as the object. It is virtual, upright and larger. This is how a magnifying glass is used.

Start with the defaults. Is the image real? Now move the object to $d = f$. What happens to the image distance? Then bring the object inside the focal length and watch the image move to the other side.

Practise with new numbers

Worked example 2: a virtual image

An insect $2.0\ \text{cm}$ long is $9.0\ \text{cm}$ from a convex lens of $f = 15\ \text{cm}$. Where is the image, and how big is it?

1. Signs. $u = -9.0\ \text{cm}$ and $f = +15\ \text{cm}$.
2. The distance. $v = \dfrac{uf}{u + f} = \dfrac{(-9.0\ \text{cm})(15\ \text{cm})}{-9.0\ \text{cm} + 15\ \text{cm}} = -22.5\ \text{cm}$.
3. The magnification. $m = \dfrac{v}{u} = \dfrac{-22.5\ \text{cm}}{-9.0\ \text{cm}} = +2.5$.
4. The size. $h' = mh = 2.5 \times 2.0\ \text{cm} = 5.0\ \text{cm}$.

The negative $v$ puts the image on the object side, and the positive $m$ says it is upright. It is virtual, because no light really goes through it. The rays only seem to come from there. A screen would show nothing, but an eye can see it.

Worked example 3: a concave lens

A concave lens has $f = -20\ \text{cm}$. An object $4.0\ \text{cm}$ tall is $30\ \text{cm}$ in front of it. Find the image.

1. Signs. $u = -30\ \text{cm}$ and $f = -20\ \text{cm}$.
2. The distance. $v = \dfrac{uf}{u + f} = \dfrac{(-30\ \text{cm})(-20\ \text{cm})}{-30\ \text{cm} - 20\ \text{cm}} = -12\ \text{cm}$.
3. The magnification. $m = \dfrac{-12\ \text{cm}}{-30\ \text{cm}} = +0.40$, so $h' = +1.6\ \text{cm}$.

The image is virtual, upright and smaller, and it sits between the lens and the object. This is true for every real object and every concave lens. Both $u$ and $f$ are negative, so $\dfrac{1}{v} = \dfrac{1}{u} + \dfrac{1}{f}$ is negative, and $v$ is negative.

Where does the focal length come from?

Two lenses of the same size can have different focal lengths. A fat, strongly curved lens has a short focal length. A flatter lens has a long one. A lens of glass with a higher refractive index bends the light more. These two facts are in one formula, the lens maker's formula.

Each face of the lens is a part of a sphere. The radius of curvature is the radius of that sphere. We use the same sign rule as before. The light comes from the left. If the centre of a face is on the right, its radius is positive. If the centre is on the left, its radius is negative. The first face, met first by the light, has radius $R_1$. The second face has radius $R_2$. For a biconvex lens, $R_1$ is positive and $R_2$ is negative. A flat face has an infinite radius.

The lens maker's formula

We build it from two refractions. In the first course of this chapter we found that for a single spherical surface between media of indices $n_1$ and $n_2$,

Take a lens of glass of index $n$ in air. A point object is at distance $u$ from the lens.

First face. Light goes from air to glass. The image $I'$ forms at the distance $v'$:

Second face. Light goes from glass to air. The image $I'$ is the object for this face, at the distance $v'$. The lens is thin, so we measure both distances from the same point. The final image is at $v$:

Add the two equations. The term $\dfrac{n}{v'}$ cancels:

Now let the object go very far away, $u \to -\infty$. Then $v = f$, and

This is the lens maker's formula. It tells us three things. The glass matters through $(n - 1)$. The shape matters through the two curvatures. And the same lens has the same $f$ from either side, because turning it over swaps $R_1$ and $R_2$ and changes their signs.

Worked example 4: six lenses of one glass

Glass has $n = 1.5$, so $n - 1 = 0.5$. Radii are in centimetre.

• Biconvex, $R_1 = +20$ and $R_2 = -20$. $\dfrac{1}{f} = 0.5\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = 0.050\ \text{cm}^{-1}$, so $f = 20\ \text{cm}$.
• Plano-convex, $R_1 = +12$ and $R_2 = \infty$. $\dfrac{1}{f} = 0.5 \times \dfrac{1}{12}$, so $f = 24\ \text{cm}$.
• Convex meniscus, $R_1 = +10$ and $R_2 = +25$. $\dfrac{1}{f} = 0.5\left(\dfrac{1}{10} - \dfrac{1}{25}\right) = 0.030\ \text{cm}^{-1}$, so $f = 33.3\ \text{cm}$.
• Biconcave, $R_1 = -20$ and $R_2 = +20$. $f = -20\ \text{cm}$.
• Plano-concave, $R_1 = \infty$ and $R_2 = +15$. $\dfrac{1}{f} = 0.5\left(0 - \dfrac{1}{15}\right)$, so $f = -30\ \text{cm}$.
• Concave meniscus, $R_1 = +25$ and $R_2 = +10$. $f = -33.3\ \text{cm}$.

Look at the two meniscus lenses. They have the same two radii in a different order, and the focal length changes sign. A meniscus is a convex lens when its first face, the one that meets the light first, is the more strongly curved one. The sure test is the sign of $f$.

Start with the defaults, a biconvex lens. Read the power and the focal length. Make $1/R_2$ equal to $0$ for a plano-convex lens, and watch the power halve. Then set $1/R_1$ and $1/R_2$ equal, and see that the power is zero.

Practise the lens maker's formula

A lens in another medium

A lens of glass of index $n_g$ in a medium of index $n_m$ works with the ratio of the two:

A glass lens with $f = 20\ \text{cm}$ in air ($n_g = 1.5$) has $\dfrac{1}{R_1} - \dfrac{1}{R_2} = 0.10\ \text{cm}^{-1}$. Put it in water of $n_m = 1.33$. Then $\dfrac{n_g}{n_m} - 1 = 0.128$, so $\dfrac{1}{f} = 0.0128\ \text{cm}^{-1}$ and $f = 78\ \text{cm}$. The lens is about four times weaker in water, because glass and water differ less in index. If the medium has a larger index than the glass, the sign changes. A convex shape then acts as a diverging lens, as an air bubble in water does. The paid course of this chapter takes this further.

Try it in Python

The program uses the thin lens formula for a convex lens of $15\ \text{cm}$, with the object at five distances. It prints the image distance, the magnification and whether the image is real.

Common mistakes

• Using $u$ as a positive number in the New Cartesian formula. A real object has a negative $u$. If you write $u = +40$, the answer is wrong.
• Reading $m$ without its sign. A negative $m$ means the image is inverted. Its size is $|m|$.
• Forgetting the sign of $R_2$ for a biconvex lens. The second face has its centre on the left, so $R_2$ is negative and the two terms add.
• Calling a virtual image "not there". A virtual image can be seen. It cannot be caught on a screen.
• Measuring from the wrong point. In a thin lens every distance is from the optical centre, not from a focus.
• Mixing the two conventions. Do not use $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ with a negative $u$.

Remember these

• A convex lens converges light ($f > 0$). A concave lens diverges it ($f < 0$).
• Three rays: parallel to the axis, through the optical centre, and aimed at the first focus.
• Thin lens formula, New Cartesian: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ and $m = \dfrac{v}{u}$.
• Real is positive form: $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ and $m = -\dfrac{v}{u}$.
• Lens maker's formula: $\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$.
• A concave lens gives a virtual, upright, smaller image of a real object, whatever the distance.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A convex lens has $f = 16\ \text{cm}$. An object is $24\ \text{cm}$ from it. Find the image distance and the magnification, and say what kind of image it is.

2. A concave lens of focal length $25\ \text{cm}$ is used with an object $50\ \text{cm}$ away. Find the image distance and say if the image is real or virtual.

3. A plano-convex lens of glass with $n = 1.5$ has a curved face of radius $30\ \text{cm}$. Find its focal length. Does the answer change if the curved face is turned to face the light?

4. Why can a concave lens never form a real image of a real object?

5. A convex glass lens works well in air. Will it still converge light if it is placed in a liquid that has a larger refractive index than the glass?

References

• NCERT Physics Part II (Class 12), Chapter 9, the sections on refraction by lenses and the lens maker's formula.
• OpenStax University Physics Volume 3, Chapter 2, the section on thin lenses.