Lens experiments and lenses in other media

Where is the centre of a lens?

You want to measure the focal length of a lens on an optical bench. The thin lens formula needs the distance from the object to the lens, and from the lens to the screen. But a real lens has a thickness. Its optical centre is somewhere inside the glass, and you cannot see where. Any ruler reading of $u$ and $v$ then carries a small error.

There is a neat way round the problem. Fix the object and the screen. Slide only the lens between them. The distance the lens slides can be read exactly, even if you do not know where its centre is. This is the displacement method, also called Bessel's method.

Two positions, one sharp image

In this derivation we use the real-is-positive convention of A-Level and IB. The object distance $x$ and the image distance $y$ are both positive for a real image, and $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{f}$. The object and the screen are a fixed distance $D$ apart, so $y = D - x$. Put this in the lens formula:

Clear the fractions. This gives a quadratic equation for the position of the lens:

There are two roots. So there are two lens positions that give a sharp image. Call the distance between them $d$. It is the difference of the two roots:

We want $f$, so we square this and solve: $d^2 = D^2 - 4Df$ gives

This is the result of the displacement method. Notice what we need: only $D$ and $d$. Both are distances between marks on the bench, so the unknown position of the lens centre does not matter.

The two positions are $x_1 = \dfrac{D - d}{2}$ and $x_2 = \dfrac{D + d}{2}$. Look at the picture. In the first position the object is at $x_1$ and the image is at $D - x_1 = x_2$. In the second position the two distances swap. Because they swap, the two magnifications are also linked:

One image is enlarged and the other is reduced, and the product of the magnifications is exactly $1$. If the two image heights are $h_1$ and $h_2$ and the object height is $h$, then $h_1 = m_1h$ and $h_2 = m_2h$. So $h_1h_2 = h^2$ and

What if $D$ is less than $4f$? Then $D^2 - 4Df$ is negative and the square root does not exist. There is no position of the lens that gives a sharp image on the screen. At $D = 4f$ the two positions meet, $d = 0$, and the image is the same size as the object. So the condition for the experiment is $D > 4f$.

A worked example: the bench

An object and a screen are $90\ \text{cm}$ apart. A lens gives a sharp image at two positions, $30\ \text{cm}$ apart. Find $f$, the two positions and the two magnifications.

1. The focal length. $f = \dfrac{D^2 - d^2}{4D} = \dfrac{(90\ \text{cm})^2 - (30\ \text{cm})^2}{4 \times 90\ \text{cm}} = \dfrac{7200\ \text{cm}^2}{360\ \text{cm}} = 20\ \text{cm}$.
2. The positions. $x_1 = \dfrac{90\ \text{cm} - 30\ \text{cm}}{2} = 30\ \text{cm}$ and $x_2 = \dfrac{90\ \text{cm} + 30\ \text{cm}}{2} = 60\ \text{cm}$ from the object.
3. The magnifications. $m_1 = \dfrac{60\ \text{cm}}{30\ \text{cm}} = 2$ and $m_2 = \dfrac{30\ \text{cm}}{60\ \text{cm}} = 0.5$. Their product is $1$.

Check with the lens formula. At the first position $\dfrac{1}{30\ \text{cm}} + \dfrac{1}{60\ \text{cm}} = \dfrac{3}{60\ \text{cm}} = \dfrac{1}{20\ \text{cm}}$. It agrees.

A second example: the slit

A slit is lit and its images are caught on a screen. The two sharp images are $9.0\ \text{mm}$ and $4.0\ \text{mm}$ tall. How tall is the slit? The two magnifications are $m$ and $1/m$, so $h = \sqrt{9.0\ \text{mm} \times 4.0\ \text{mm}} = \sqrt{36}\ \text{mm} = 6.0\ \text{mm}$. The slit is $6.0\ \text{mm}$ tall. The larger image is $1.5$ times the slit, and the smaller image is $\dfrac{1}{1.5} = 0.67$ of it.

A third example finds how far the lens must move. Take $D = 120\ \text{cm}$ and a lens of $f = 25\ \text{cm}$. We want $d$, so $d = \sqrt{D^2 - 4Df} = \sqrt{14400 - 12000}\ \text{cm} = 49.0\ \text{cm}$. The positions are $35.5\ \text{cm}$ and $84.5\ \text{cm}$. Check: $\dfrac{1}{35.5} + \dfrac{1}{84.5} = 0.0400\ \text{cm}^{-1} = \dfrac{1}{25\ \text{cm}}$.

Start with $D = 100\ \text{cm}$ and $f = 18\ \text{cm}$. How far does the lens move? Now make $f$ larger, up to $20\ \text{cm}$. Why do the two positions come closer? Then make $D$ larger and watch them part.

Practise with new numbers

A lens in a liquid

The lens maker's formula you know is for a lens in air: $\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$. Why does it have $n - 1$? Because the bending of light at a surface depends on the difference of the two indices, and the lens is glass on one side and air on the other.

Put the same lens in water. Now the glass is next to water, which is a much weaker contrast. We expect the lens to bend light less. To see exactly how much, take a thin lens of index $n_l$ in a liquid of index $n_m$. Each surface of a lens gives the term $\dfrac{n_2 - n_1}{R}$ for $\dfrac{n_2}{v} - \dfrac{n_1}{u}$. Add the two surfaces, with the liquid on both sides:

Divide by $n_m$:

Compare this with the same lens in air. The ratio of the two focal lengths is

There are three cases, and they are worth remembering.

• If $n_m < n_l$, the bracket is positive. A convex lens stays converging, but its focal length is longer than in air.
• If $n_m = n_l$, the bracket is zero. The focal length is infinite. The lens does nothing. A glass rod in a liquid of the same index seems to disappear.
• If $n_m > n_l$, the bracket is negative. A convex lens now spreads the light out. It behaves as a diverging lens.

An example. A glass lens of $n_l = 1.50$ has $f_{air} = 20\ \text{cm}$. In water, $n_m = 1.33$, so $\dfrac{n_l}{n_m} - 1 = \dfrac{1.50}{1.33} - 1 = 0.1278$. The focal length is $f_m = 20\ \text{cm} \times \dfrac{0.50}{0.1278} = 78.2\ \text{cm}$. The lens is about four times weaker. In a dense oil of $n_m = 1.62$ the bracket is $\dfrac{1.50}{1.62} - 1 = -0.0741$, so $f_m = 20\ \text{cm} \times \dfrac{0.50}{-0.0741} = -135\ \text{cm}$. The lens now diverges.

An air bubble that acts as a lens. Inside a block of glass there is a cavity of air in the shape of a biconvex lens. Its two faces have radii of $6.0\ \text{cm}$. Here the "lens" is air and the medium is glass, so $n_l = 1.00$ and $n_m = 1.50$. The first bracket is $\dfrac{1.00}{1.50} - 1 = -\dfrac{1}{3}$. The second bracket is $\dfrac{1}{6.0\ \text{cm}} + \dfrac{1}{6.0\ \text{cm}} = \dfrac{1}{3.0\ \text{cm}}$. So $\dfrac{1}{f} = -\dfrac{1}{3} \times \dfrac{1}{3.0\ \text{cm}}$ and $f = -9.0\ \text{cm}$. A cavity that looks like a convex lens spreads light out.

What about a mirror? The focal length of a mirror is $f = R/2$ in air, in water and in any other medium. It depends only on the shape of the surface, because light is reflected and never crosses the surface. Only a lens changes its focal length when the medium changes.

Put $n_l = 1.50$ and move the red point from air to water. Where does the power reach zero? What happens beyond it? Then make $n_l$ larger and see where the zero moves.

Different media on the two sides

Sometimes the two sides of the lens have different media. The window of a fish tank is an example. The eye is another: air is in front and a watery fluid is behind. The surface formula still works for each face, $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$. Take a thin lens of glass index $n_g$ with the index $n_1$ on its left and $n_3$ on its right. The first surface makes an image at $v'$, which is the object for the second surface:

The lens is thin, so $v'$ is the same distance for both. Add the two equations and $v'$ cancels:

All distances use the New Cartesian convention. The quantity $P$ is the power of the lens between these two media. The two focal lengths are now different. A beam coming from far on the left has $u = -\infty$, so it focuses at $f_2 = \dfrac{n_3}{P}$. For the beam to leave parallel we need $\dfrac{n_3}{v} = 0$, so the object sits at $f_1 = -\dfrac{n_1}{P}$. The ratio of the two focal lengths is the ratio of the two indices:

The magnification also changes. The ray through the centre of the lens crosses two nearly parallel faces. For small angles, $n\theta$ stays the same, so $n_1\theta_1 = n_3\theta_3$. With $\theta_1 = \dfrac{h}{|u|}$ and $\theta_3 = \dfrac{|h'|}{v}$ this gives

An example. A thin biconvex lens of $n_g = 1.50$ has $R_1 = +20\ \text{cm}$ and $R_2 = -20\ \text{cm}$. Air is on its left and water of $n_3 = 1.33$ is on its right. An object is $60\ \text{cm}$ in front of it, in the air.

1. The power. $P = \dfrac{1.50 - 1.00}{20\ \text{cm}} + \dfrac{1.33 - 1.50}{-20\ \text{cm}} = 0.0250\ \text{cm}^{-1} + 0.0085\ \text{cm}^{-1} = 0.0335\ \text{cm}^{-1}$.
2. The image. We want $v$, so $\dfrac{n_3}{v} = \dfrac{1}{u} + P = -\dfrac{1}{60\ \text{cm}} + 0.0335\ \text{cm}^{-1} = 0.01683\ \text{cm}^{-1}$. Then $v = \dfrac{1.33}{0.01683\ \text{cm}^{-1}} = 79.0\ \text{cm}$.
3. The magnification. $m = \dfrac{n_1 v}{n_3 u} = \dfrac{1.00 \times 79.0\ \text{cm}}{1.33 \times (-60\ \text{cm})} = -0.990$. The image is real and inverted, and almost the size of the object.

Check with the focal lengths. $f_1 = -\dfrac{1.00}{0.0335\ \text{cm}^{-1}} = -29.9\ \text{cm}$ and $f_2 = \dfrac{1.33}{0.0335\ \text{cm}^{-1}} = 39.7\ \text{cm}$. The ratio is $\dfrac{39.7\ \text{cm}}{29.9\ \text{cm}} = 1.33$. It is the ratio of the indices, as it must be.

Try it in Python

The program below takes four runs of the displacement method, each with a different screen distance. It finds the focal length from every run, and the two positions and magnifications.

The second program puts the lens of the worked example in five liquids and prints its focal length in each.

Common mistakes

• Using $D$ smaller than $4f$. The square root of a negative number means there is no sharp image at all. Check $D > 4f$ before you start.
• Averaging the two image heights. The object height is the square root of the product, $\sqrt{h_1h_2}$, because the two magnifications are $m$ and $1/m$.
• Using the air formula in a liquid. Replace $n_l$ by $\dfrac{n_l}{n_m}$. A lens that converges in air can diverge in a liquid of larger index.
• Changing a mirror's focal length in water. A mirror keeps $f = R/2$ in every medium. Only lenses change.
• Forgetting the $n_3$ in $\dfrac{n_3}{v}$. After finding the right side, multiply by $n_3$ to get $v$. The image distance $v$ is not just $1/(\ldots)$.

Remember these

• Displacement method: $D > 4f$, $d = \sqrt{D^2 - 4Df}$, $f = \dfrac{D^2 - d^2}{4D}$, $m_1m_2 = 1$ and $h = \sqrt{h_1h_2}$.
• Lens in a medium: $\dfrac{1}{f_m} = \left(\dfrac{n_l}{n_m} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$, so $\dfrac{f_m}{f_{air}} = \dfrac{n_l - 1}{n_l/n_m - 1}$.
• A lens with $n_m = n_l$ has no effect. With $n_m > n_l$ a convex lens diverges.
• Different media: $\dfrac{n_3}{v} - \dfrac{n_1}{u} = \dfrac{n_g - n_1}{R_1} + \dfrac{n_3 - n_g}{R_2}$, and $m = \dfrac{n_1 v}{n_3 u}$.
• The two focal lengths of such a lens are $f_1 = -\dfrac{n_1}{P}$ and $f_2 = \dfrac{n_3}{P}$.

Check yourself

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

1. A student has a lit slit and a screen $60\ \text{cm}$ apart. Her only lens has $f = 18\ \text{cm}$. Can she find a sharp image by sliding the lens?

2. With $D = 100\ \text{cm}$ a lens gives sharp images with its centre at $25\ \text{cm}$ and at $75\ \text{cm}$ from the slit. Find $f$ and the larger magnification.

3. The two sharp images of an arrow are $5.0\ \text{mm}$ and $20\ \text{mm}$ tall. How tall is the arrow?

4. A glass lens of $n = 1.50$ is dipped in a liquid of the same refractive index. Why can you hardly see the lens?

5. A cavity of air in a block of glass of $n = 1.60$ has two equal faces, each of radius $15\ \text{cm}$, and it is biconvex in shape. What is its focal length, and does it converge or diverge?

References

• NCERT Physics Part II (Class 12), Chapter 9, the sections on refraction at spherical surfaces and the lens maker's formula.
• OpenStax University Physics Volume 3, Chapter 2, the sections on refraction at a spherical interface and on thin lenses.
• OpenStax College Physics 2e, Chapter 25, the section on image formation by lenses.