Refraction at a spherical surface

Why does a glass ball turn the world upside down?

Hold a clear glass ball in front of a window. Inside it you see the whole garden, small and upside down. A fish in a round bowl looks bigger than it is. A raindrop on a leaf makes a tiny picture of the sky. All of them use one idea: refraction at a curved surface between two materials. The same idea, used twice, gives a lens in the next course. Here we learn it for one surface.

The setting and the signs

A spherical surface of radius $R$ separates a material of index $n_1$ on the left from one of index $n_2$ on the right. The light comes from the left. The point where the axis meets the surface is the pole $P$. The centre of curvature is $C$.

We keep the New Cartesian convention of lesson 1. Distances are measured from $P$, and positive to the right. The object distance $u$ is negative for a real object. The image distance $v$ is positive when the image is on the right, which is the side the light goes to. The radius $R$ is positive when the centre of curvature is on the right, on the side of the outgoing light. A surface that bulges toward the incoming light, a convex surface, has a positive $R$.

The formula

Take an object $O$ on the axis, and a ray from it that meets the surface at a point $M$ close to the axis. Let $N$ be the foot of the perpendicular from $M$ to the axis, so that $N$ is very close to $P$. The normal at $M$ is the line $CM$. We name three small angles: $\alpha$ at $O$, $\beta$ at $C$ and $\gamma$ at the image $I$.

1. The angle of incidence $i$ is an exterior angle of the triangle $OMC$, so $i = \alpha + \beta$.
2. The angle of refraction $r$ comes from the triangle $MCI$, where $\beta$ is an exterior angle at $C$, so $r = \beta - \gamma$.
3. For small angles Snell's law is $n_1 i = n_2 r$, so $n_1(\alpha + \beta) = n_2(\beta - \gamma)$.
4. Each small angle is about its tangent, which is $MN$ divided by a distance on the axis. The distances are $OP$, $PC$ and $PI$.
5. Put them in, and $MN$ cancels. We get $\dfrac{n_1}{OP} + \dfrac{n_2}{PI} = \dfrac{n_2 - n_1}{PC}$.
6. With the signs, $OP = -u$, $PI = v$ and $PC = R$. So

The magnification of a refracting surface is

This formula holds for rays close to the axis, and for a convex surface, a concave surface and a flat one alike. Only the sign of $R$ changes. Other books may print the formula with different signs, so check their rule for $R$ before you use it.

Example 1: the rounded end of a glass rod. The end of a glass rod ($n_2 = 1.50$) in air ($n_1 = 1.00$) is ground into a convex surface with $R = +10\ \text{cm}$. A small bright object is on the axis, $40\ \text{cm}$ in front of the surface, so $u = -40\ \text{cm}$.

1. We want $v$. The formula gives $\dfrac{n_2}{v} = \dfrac{n_2 - n_1}{R} + \dfrac{n_1}{u} = \dfrac{0.50}{10\ \text{cm}} - \dfrac{1.00}{40\ \text{cm}} = 0.025\ \text{cm}^{-1}$.
2. So $v = \dfrac{1.50}{0.025\ \text{cm}^{-1}} = +60\ \text{cm}$.
3. The magnification is $m = \dfrac{n_1 v}{n_2 u} = \dfrac{1.00 \times 60\ \text{cm}}{1.50 \times (-40\ \text{cm})} = -1.0$.

The image is real, $60\ \text{cm}$ inside the glass, upside down and the same size as the object. Check. A ray from the object that goes along the axis does not bend, so the image must lie on the axis. It does.

The flat surface

A flat surface has $R \to \infty$, so the right side of the formula is $0$. Then $\dfrac{n_2}{v} = \dfrac{n_1}{u}$, so

The magnification is $m = \dfrac{n_1 v}{n_2 u} = 1$, so the image is as large as the object.

Example 2: the apparent depth again. A fish is $1.60\ \text{m}$ below the surface of a pond, so $u = -1.60\ \text{m}$. The light goes from water ($n_1 = 1.33$) into air ($n_2 = 1.00$). Then $v = \dfrac{1.00}{1.33} \times (-1.60\ \text{m}) = -1.20\ \text{m}$. The fish seems to be $1.20\ \text{m}$ below the surface. This is the apparent depth formula of lesson 2, now found from the surface formula.

The focal lengths of one surface

If the object is very far away, $u \to -\infty$ and the rays come in parallel. The image is then at the second focal point, at the distance $f_2$:

If instead the image is at infinity, the rays leave parallel. The object is then at the first focal point, at the distance $f_1$:

A single surface has two focal lengths, unless $n_1 = n_2$. They are in the ratio $\dfrac{f_2}{f_1} = -\dfrac{n_2}{n_1}$.

For the glass rod above, $f_2 = \dfrac{1.50 \times 10\ \text{cm}}{0.50} = 30\ \text{cm}$ and $f_1 = -\dfrac{1.00 \times 10\ \text{cm}}{0.50} = -20\ \text{cm}$. The program later in this lesson gives the same $30\ \text{cm}$.

Looking into a glass sphere

Example 3. A small coin is at the centre of a solid glass sphere ($n = 1.50$) of radius $6.0\ \text{cm}$. You look at it from outside. The light leaves the glass, so $n_1 = 1.50$ and $n_2 = 1.00$. The object distance is $u = -6.0\ \text{cm}$. The centre of curvature is where the coin is, on the left, so $R = -6.0\ \text{cm}$.

1. $\dfrac{n_2}{v} = \dfrac{n_2 - n_1}{R} + \dfrac{n_1}{u} = \dfrac{-0.50}{-6.0\ \text{cm}} + \dfrac{1.50}{-6.0\ \text{cm}} = -0.167\ \text{cm}^{-1}$.
2. So $v = -6.0\ \text{cm}$. The image is exactly where the coin is.
3. The magnification is $m = \dfrac{1.50 \times (-6.0\ \text{cm})}{1.00 \times (-6.0\ \text{cm})} = +1.5$.

The coin seems to be in the same place, but one and a half times as large. A ray through the centre of a sphere meets the surface along the normal, so it does not bend. This is why the image stays on the axis.

The ball lens

Two surfaces one after the other make a lens. A glass ball is the simplest one. We follow parallel light through a ball of radius $R = 10\ \text{cm}$ and $n = 1.50$.

1. First surface. Here $u = -\infty$, so $\dfrac{n}{v_1} = \dfrac{n - 1}{R}$ and $v_1 = \dfrac{nR}{n - 1} = \dfrac{1.50 \times 10\ \text{cm}}{0.50} = 30\ \text{cm}$. The light would meet at $30\ \text{cm}$ from the front, which is beyond the back of the ball.
2. Second surface. The back surface is $20\ \text{cm}$ from the front, so the rays head toward a point $30\ \text{cm} - 20\ \text{cm} = +10\ \text{cm}$ beyond it. This is a virtual object for the second surface: $u_2 = +10\ \text{cm}$. The light goes from glass to air, so $n_1 = 1.50$, $n_2 = 1.00$ and $R_2 = -10\ \text{cm}$.
3. $\dfrac{1.00}{v} = \dfrac{-0.50}{-10\ \text{cm}} + \dfrac{1.50}{10\ \text{cm}} = 0.20\ \text{cm}^{-1}$, so $v = 5.0\ \text{cm}$.

The light comes to a focus $5.0\ \text{cm}$ behind the ball, which is $15\ \text{cm}$ from its centre. For any ball, the focus is at $\dfrac{nR}{2(n - 1)}$ from the centre. A glass ball left on a sunny window sill can start a fire, because it brings sunlight to a bright spot. A ball with $n = 2.00$ focuses on its own back surface. Ball lenses are used to couple light into optical fibres.

Start with the defaults, and write down $v$ and $m$. Now make $R$ smaller. Does the image come closer? Then put the object at $15\ \text{cm}$ and look at the sign of $v$.

Practise with new numbers

Try it in Python

The program uses the surface formula for the glass rod of Example 1 and moves the object closer, step by step. It prints the image distance and the magnification.

Common mistakes

• Writing $u$ as a positive number. A real object has a negative $u$ in this convention.
• Taking $R$ as always positive. The sign of $R$ depends on which side the centre of curvature is. For light that leaves a glass sphere, $R$ is negative.
• Mixing up $n_1$ and $n_2$. $n_1$ is the material the light starts in, and $n_2$ is the material it goes into.
• Using the mirror magnification $-\dfrac{v}{u}$ for a refracting surface. For a surface, $m = \dfrac{n_1 v}{n_2 u}$.
• Measuring the focus of a ball from its centre when the question asks from its back. Subtract the radius.

Remember these

• Refraction at a spherical surface: $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$, with the light coming from the left.
• $R > 0$ when the centre of curvature is on the side of the outgoing light. The magnification is $m = \dfrac{n_1 v}{n_2 u}$.
• A flat surface: $v = \dfrac{n_2}{n_1}u$ and $m = 1$.
• One surface has $f_2 = \dfrac{n_2 R}{n_2 - n_1}$ and $f_1 = -\dfrac{n_1 R}{n_2 - n_1}$.
• A ball lens focuses parallel light at $\dfrac{nR}{2(n - 1)}$ from its centre.

Check yourself

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

1. An object is $48\ \text{cm}$ in front of a convex surface of radius $12\ \text{cm}$ between air and glass ($n = 1.50$). Where is the image?

2. Find the two focal lengths of the surface of Question 1.

3. What is the sign of $R$ for the back surface of a glass sphere, when light goes through the sphere from left to right?

4. A flat window of water is viewed from the air side. A fish is $0.90\ \text{m}$ below the surface. Use the formula for a flat surface to find the apparent depth. Take $n = 1.33$.

5. A glass ball has a radius of $4.0\ \text{cm}$ and $n = 1.50$. Sunlight falls on it. How far behind the ball is the hot spot?

References

• NCERT Physics Part II (Class 12), Chapter 9, the section on refraction at spherical surfaces.
• OpenStax University Physics Volume 3, Chapter 2, the section on images formed by refraction.