Power and combinations of lenses

What the optician writes

An optician hands you a card that says $-2.50$ for the left eye. What does the number mean? It is the power of the lens, in a unit called the dioptre. The minus sign says the lens is concave. A strong lens has a short focal length and a large power. A weak lens has a long focal length and a small power.

The power of a lens is the reciprocal of its focal length in metre:

Its unit is the dioptre, symbol D. One dioptre is one reciprocal metre, $1\ \text{D} = 1\ \text{m}^{-1}$. A convex lens has a positive power and a concave lens has a negative power.

• A lens with $f = +25\ \text{cm} = +0.25\ \text{m}$ has $P = \dfrac{1}{0.25\ \text{m}} = +4.0\ \text{D}$.
• A lens with $f = -40\ \text{cm}$ has $P = \dfrac{1}{-0.40\ \text{m}} = -2.5\ \text{D}$.
• A lens of $+2.0\ \text{D}$ has $f = \dfrac{1}{2.0\ \text{D}} = 0.50\ \text{m} = 50\ \text{cm}$.

Always change the focal length to metre first. A common slip is to put it in centimetre.

Why the unit is useful

The power of a lens tells how much it bends the light. A lens that bends rays twice as strongly has twice the power. It also tells us how powers combine. We find this out next.

An optician's card has a few more terms. Sphere is the power of the lens as we use it here. A reading addition such as $+1.50$ is extra convex power for the lower part of a lens, which helps near work. A cylinder value is for an eye that is more curved in one direction than in another. It uses a lens with a different power along two directions, which the course on the eye takes up.

Two lenses in contact

Put two thin lenses against each other. Light from a point object goes through the first lens, which forms an image. That image is the object for the second lens. For a distant object we can follow the rays.

For the first lens, with object distance $u$ and image distance $v_1$:

The lenses touch, so the image at $v_1$ is at the distance $v_1$ from the second lens too. For the second lens it is the object, and its image is at $v$:

Add the two equations. The term $\dfrac{1}{v_1}$ cancels:

The pair behaves like one lens whose focal length $f$ obeys $\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$. In terms of power:

The magnification is the product of the two, $m = m_1m_2$, because the second lens enlarges the image made by the first.

Worked example 1: a convex and a concave lens

A convex lens of $+5.0\ \text{D}$ is pressed against a concave lens of $-2.0\ \text{D}$. Find the power and the focal length of the pair.

1. The power. $P = +5.0\ \text{D} + (-2.0\ \text{D}) = +3.0\ \text{D}$.
2. The focal length. We want $f$, so $f = \dfrac{1}{P} = \dfrac{1}{3.0\ \text{D}} = 0.333\ \text{m} = 33.3\ \text{cm}$.

The pair is converging, because the convex lens is stronger. If the two powers were equal and opposite, the pair would have zero power. Parallel light would pass straight through, like a plain sheet of glass.

Worked example 2: object and image through a pair

A convex lens of $f_1 = +20\ \text{cm}$ is in contact with a concave lens of $f_2 = -30\ \text{cm}$. An object stands $90\ \text{cm}$ in front. Find the image and the magnification.

1. The pair. $\dfrac{1}{f} = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{30\ \text{cm}} = \dfrac{1}{60\ \text{cm}}$, so $f = +60\ \text{cm}$.
2. The image. $u = -90\ \text{cm}$. We want $v$, so $\dfrac{1}{v} = \dfrac{1}{u} + \dfrac{1}{f} = -\dfrac{1}{90\ \text{cm}} + \dfrac{1}{60\ \text{cm}} = \dfrac{1}{180\ \text{cm}}$. Then $v = +180\ \text{cm}$.
3. The magnification. $m = \dfrac{v}{u} = \dfrac{180\ \text{cm}}{-90\ \text{cm}} = -2.0$.

Check lens by lens. The first lens gives $v_1 = \dfrac{(-90)(20)}{-90 + 20}\ \text{cm} = 25.7\ \text{cm}$ with $m_1 = -0.286$. The second lens has a virtual object $25.7\ \text{cm}$ behind it, so $u_2 = +25.7\ \text{cm}$ and $v_2 = \dfrac{(25.7)(-30)}{25.7 - 30}\ \text{cm} = 180\ \text{cm}$, with $m_2 = 7.0$. Then $m_1m_2 = -2.0$. It agrees.

Two lenses with a gap

Lenses in a camera or a telescope are not in contact. They are a distance $d$ apart. A ray that goes through both is bent twice, and the second bending happens at a different height. We follow a ray to find out how the powers combine.

Send a ray parallel to the axis at the height $h$ into the first lens $L_1$. After $L_1$ it heads for the focus of $L_1$, so it has the slope $-\dfrac{h}{f_1}$. It reaches the second lens $L_2$ at the height

A thin lens of focal length $f_2$ adds the slope $-\dfrac{h_2}{f_2}$ to a ray at height $h_2$. So after $L_2$ the ray has a slope of size

The equivalent focal length $F$ is the height at which the ray came in, divided by this slope. So

with $d$ in metre. For $d = 0$ this is the rule for lenses in contact.

There is one more point to notice. The ray leaves $L_2$ and crosses the axis at a distance $\dfrac{h_2}{\text{slope}}$ behind $L_2$. This is the back focal distance, $F\left(1 - \dfrac{d}{f_1}\right)$. It is not equal to $F$. The focal length $F$ is measured from a plane of the equivalent lens, which lies inside the pair, as the picture shows.

Worked example 3: the numbers behind the picture

Two convex lenses have $f_1 = 30\ \text{cm}$ and $f_2 = 20\ \text{cm}$, and they are $10\ \text{cm}$ apart.

1. The equivalent focal length. $\dfrac{1}{F} = \dfrac{1}{30\ \text{cm}} + \dfrac{1}{20\ \text{cm}} - \dfrac{10\ \text{cm}}{(30\ \text{cm})(20\ \text{cm})} = 0.0667\ \text{cm}^{-1}$. So $F = 15\ \text{cm}$ and $P = 6.7\ \text{D}$.
2. The back focal distance. $F\left(1 - \dfrac{d}{f_1}\right) = 15\ \text{cm} \times \left(1 - \dfrac{10}{30}\right) = 10\ \text{cm}$ behind $L_2$.

Check without the formula. $L_1$ alone would focus parallel light $30\ \text{cm}$ behind it. That is $20\ \text{cm}$ behind $L_2$, so $L_2$ meets a virtual object, $u_2 = +20\ \text{cm}$. Then $\dfrac{1}{v_2} = \dfrac{1}{20\ \text{cm}} + \dfrac{1}{20\ \text{cm}}$, so $v_2 = 10\ \text{cm}$. It agrees.

Start with the defaults, $+3.0\ \text{D}$ and $+5.0\ \text{D}$. What is the power for $d = 0$ and for $d = 10\ \text{cm}$? Now make $d$ larger. At what gap does the power fall to zero? Then give $P_2$ a negative value, and see what a gap does to the pair.

Practise the separated pair

The direct method: one lens at a time

The formulas above give the equivalent lens. Often we want the image and the magnification for a real object. Then the simplest way is the direct method. We do not need any new formula.

1. Use the thin lens formula for $L_1$. Find $v_1$ and $m_1$.
2. The image of $L_1$ is the object of $L_2$. Its distance from $L_2$ is $u_2 = v_1 - d$. Keep the sign. If $v_1$ is larger than $d$, the object of $L_2$ is behind it. Then $u_2$ is positive, and the object is virtual.
3. Use the thin lens formula for $L_2$. Find $v_2$ and $m_2$.
4. The total magnification is $m = m_1m_2$.

Worked example 4: two convex lenses

An object $2.0\ \text{cm}$ tall is $30\ \text{cm}$ in front of $L_1$ with $f_1 = 20\ \text{cm}$. A second lens $L_2$ with $f_2 = 15\ \text{cm}$ is $30\ \text{cm}$ behind $L_1$. Find the final image.

1. Lens 1. $u_1 = -30\ \text{cm}$. $v_1 = \dfrac{(-30)(20)}{-30 + 20}\ \text{cm} = 60\ \text{cm}$ and $m_1 = \dfrac{60}{-30} = -2.0$. This image is $4.0\ \text{cm}$ tall and inverted.
2. The object of lens 2. $u_2 = v_1 - d = 60\ \text{cm} - 30\ \text{cm} = +30\ \text{cm}$. It is a virtual object, behind $L_2$.
3. Lens 2. $v_2 = \dfrac{(30)(15)}{30 + 15}\ \text{cm} = 10\ \text{cm}$ and $m_2 = \dfrac{10}{30} = +0.333$.
4. Total. $m = (-2.0)(+0.333) = -0.667$, so $h' = -0.667 \times 2.0\ \text{cm} = -1.33\ \text{cm}$.

The final image is real, $10\ \text{cm}$ behind $L_2$, inverted and smaller. The picture shows exactly this, drawn to scale.

Try it in Python

The program takes the two convex lenses $f_1 = 30\ \text{cm}$ and $f_2 = 20\ \text{cm}$ and finds the power and the equivalent focal length for six gaps.

Common mistakes

• Using centimetre in $P = \dfrac{1}{f}$. The power is in dioptre only if $f$ is in metre.
• Forgetting that a concave lens has a negative power. In contact, a concave lens takes away from a convex one.
• Adding powers for lenses that are apart. The simple sum is for lenses in contact. With a gap use $P_1 + P_2 - dP_1P_2$.
• Measuring $F$ from the second lens. The equivalent focal length is measured from the equivalent lens. The distance of the focus behind $L_2$ is the back focal distance.
• Dropping the sign of a virtual object. If the image of $L_1$ falls beyond $L_2$, then $u_2$ is positive.
• Adding magnifications. Magnifications multiply: $m = m_1m_2$.

Remember these

• $P = \dfrac{1}{f}$ with $f$ in metre. The unit is the dioptre, $1\ \text{D} = 1\ \text{m}^{-1}$.
• In contact: $P = P_1 + P_2$ and $m = m_1m_2$.
• With a gap $d$: $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1f_2}$, and the back focal distance is $F\left(1 - \dfrac{d}{f_1}\right)$.
• The direct method: image of $L_1$, then $u_2 = v_1 - d$, then $L_2$, and multiply $m_1$ by $m_2$.
• When $d = f_1 + f_2$ the pair is afocal.

Check yourself

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

1. A lens of a pair of spectacles has the power $-2.5$ D. What is its focal length, and what kind of lens is it?

2. A lens of $+10$ D is in contact with a lens of $-4.0$ D. What is the focal length of the pair?

3. Two identical convex lenses, each of $f = 20\ \text{cm}$, touch each other. What is the focal length of the pair?

4. A lens of $f_1 = 40\ \text{cm}$ and a lens of $f_2 = 10\ \text{cm}$ are $50\ \text{cm}$ apart. A wide parallel beam enters the first lens. What leaves the second?

5. Two lenses in contact have magnifications of $-2$ and $-3$ for the same object. What is the total magnification?

References

• NCERT Physics Part II (Class 12), Chapter 9, the section on power of a lens and combinations of thin lenses.
• OpenStax University Physics Volume 3, Chapter 2, the section on thin lenses.