The lens formula and magnification

From a picture to a number

A ray diagram shows the idea. For an exact answer, we use the lens formula. It works for convex and for concave lenses, with the same sign convention that we used for mirrors.

The sign convention

The optical centre $O$ is the origin. The object is always on the left, so the light travels from left to right.

• All distances are measured from the optical centre.
• A distance in the direction of the incident light (to the right) is positive. A distance against it (to the left) is negative.
• A height above the principal axis is positive. A height below is negative.

So:

• The object distance $u$ is always negative.
• A convex lens has a positive focal length $f$. A concave lens has a negative $f$.
• A real image is on the right of the lens, so $v$ is positive. A virtual image is on the left, so $v$ is negative.

Be careful: for a lens, a real image has a positive $v$. For a mirror, it is the opposite, because the light turns back.

The lens formula

When the unknown is $v$, we write $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$ (remember that $u$ has its own sign).

Magnification

• $|m| > 1$: the image is larger. $|m| < 1$: it is smaller.
• $m$ negative: the image is real and inverted. $m$ positive: the image is virtual and upright.

You should find the table of the last lesson in the curve. At $u = 2f$ the point sits on $v = 2f$: the same-size image. Inside $f$, $v$ turns negative and the image is virtual.

Worked example 1: a convex lens

An object $6\ \text{cm}$ tall is $30\ \text{cm}$ in front of a convex lens of focal length $12\ \text{cm}$. Find the image.

Given: $f = +12\ \text{cm}$, $u = -30\ \text{cm}$, $h = 6\ \text{cm}$.

Step 1. We want $v$, so $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{12\ \text{cm}} - \dfrac{1}{30\ \text{cm}} = \dfrac{5 - 2}{60\ \text{cm}} = \dfrac{1}{20\ \text{cm}}$. So $v = +20\ \text{cm}$.

Step 2. $m = \dfrac{v}{u} = \dfrac{20\ \text{cm}}{-30\ \text{cm}} = -0.67$.

Step 3. $h' = m \times h = -0.67 \times 6\ \text{cm} = -4\ \text{cm}$.

The image is $20\ \text{cm}$ from the lens on the other side. It is real and inverted, and it is $4\ \text{cm}$ tall, two thirds of the object. The object is beyond $2F$ ($24\ \text{cm}$), so this is what the ray diagram shows.

Worked example 2: a concave lens

An object is $10\ \text{cm}$ in front of a concave lens of focal length $15\ \text{cm}$. Find the image.

Given: $f = -15\ \text{cm}$, $u = -10\ \text{cm}$.

Step 1. $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = -\dfrac{1}{15\ \text{cm}} - \dfrac{1}{10\ \text{cm}} = -\dfrac{2 + 3}{30\ \text{cm}} = -\dfrac{1}{6\ \text{cm}}$. So $v = -6\ \text{cm}$.

Step 2. $m = \dfrac{v}{u} = \dfrac{-6\ \text{cm}}{-10\ \text{cm}} = +0.6$.

The image is $6\ \text{cm}$ from the lens, on the same side as the object. It is virtual, upright and smaller ($0.6$ of the object). This is what a concave lens always does.

Practise with new numbers

Try it in Python

The program uses the lens formula and says what the image is like.

Common mistakes

• Using the mirror formula for a lens. The signs are different: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ for a lens, and $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ for a mirror.
• Writing $u$ as positive. The object distance is always negative.
• Forgetting that a concave lens has a negative $f$.

Remember these

• Lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. Magnification: $m = \dfrac{h'}{h} = \dfrac{v}{u}$.
• $u$ is negative. $f$ is positive for a convex lens and negative for a concave lens.
• A real image has a positive $v$ and a negative $m$. A virtual image has a negative $v$ and a positive $m$.

Check yourself

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

1. What is the sign of $f$ for a concave lens?

2. An object is $20\ \text{cm}$ in front of a convex lens of $f = 10\ \text{cm}$. Where is the image?

3. A lens gives $m = -0.25$. What do you know about the image?

4. An object is $20\ \text{cm}$ from a concave lens of focal length $20\ \text{cm}$. Find $v$ and $m$.