Physics · Secondary School · Geometric Optics

Optics

A shadow on the ground, a reflection in a shop window, a spoon that looks 'broken' in a glass, a pair of spectacles and a rainbow are things you probably see every week, and we can explain all of them by following the path of light as straight-line rays. On top of that we only need three rules, namely that light reflects at equal angles, that it refracts when it changes medium and that each colour bends a slightly different amount.

1Light travels in straight lines
2Reflection and plane mirrors
3Curved (spherical) mirrors
4Refraction
5Total internal reflection
6Spherical lenses
7The human eye
8Dispersion and colour
✓Challenges

STEP 1

Light travels in a straight line, and that is why shadows exist

We only see an object when light coming from it reaches our eye. Some objects, which we call primary sources, make their own light (the Sun, a light bulb, a candle), while secondary sources only reflect the light they receive (the Moon, this page, you).

Light passes through a transparent medium (air, clear glass) in an orderly way. A translucent medium (frosted glass, tracing paper) also lets it through, though scattered, and an opaque one (wood, a wall) does not let it pass at all.

In a uniform medium light travels in a straight line, so an opaque object in its way blocks the rays and a shadow appears behind it. With a point source the edge of the shadow is sharp. An extended source also produces a penumbra, a region that receives light from only part of the source. Eclipses are exactly this on an astronomical scale.

In a pinhole camera, each point of the object sends one ray through the tiny hole, and because the rays cross there the image comes out upside down, with a size we can find by similar triangles.

\(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\)\(\dfrac{\text{shadow}}{\text{object}} = \dfrac{L}{a}\) (point source)o = object height · i = image height · p = distance from object to pinhole · p′ = depth of the camera · L and a = distances from the source to the screen and to the object

Let's discuss

Click 'Point source' and check whether there is a penumbra, then slowly increase the size of the source and watch the edge of the shadow.
Move the object closer to the source. Does the shadow get bigger or smaller, and what happens to the penumbra?
Click 'Annular eclipse', where the source is bigger than the object, and look for the shadow. Where did it go?
In the pinhole camera, move the candle farther away and see what happens to the size of the image, then check \(o/i = p/p^{\prime}\) with the numbers in the readouts.
Why does the image in the pinhole camera come out upside down? Following a single ray from the top of the candle may help.

edge rays

Source size

Object size

Source–object distance

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Classify each as a primary source (makes its own light) or a secondary source (only reflects light): the Sun, the Moon, a lit candle, a mobile phone screen that is on, and the page of this notebook.

Show solution
A primary source produces its own light; a secondary source only sends back (reflects) the light it receives.
The Sun, the lit candle and the screen that is on emit their own light.
The Moon and the page only reflect light from the Sun or from a lamp.
Primary: Sun, candle, screen that is on. Secondary: Moon and page.
basic
Classify each medium as transparent, translucent or opaque: clean air, clear window glass, frosted shower-door glass, tracing paper, a brick wall.

Show solution
In a transparent medium light passes through in an orderly way and we see clearly through it.
In a translucent one light also passes through, though scattered, so we cannot see sharp shapes.
In an opaque one light does not pass through.
Transparent: air and clear glass. Translucent: frosted glass and tracing paper. Opaque: brick.
basic
A building \(6\ \text{m}\) tall is \(30\ \text{m}\) from a pinhole camera that is \(20\ \text{cm}\) deep. How tall is the image on the back of the camera?

Show solution
By similar triangles, \(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\)
\(i = \dfrac{o \cdot p^{\prime}}{p} = \dfrac{6 \cdot 0.20}{30}\)
\(i = 0.04\ \text{m}\)
\(i = 4\ \text{cm}\) (and upside down)
basic
At the same moment, a \(3\ \text{m}\) lamppost casts a \(4\ \text{m}\) shadow on the ground. How long is the shadow of a \(1.8\ \text{m}\) person standing next to it?

Show solution
Sunlight arrives in parallel rays, so the post–shadow and person–shadow triangles are similar.
\(\dfrac{3}{4} = \dfrac{1.8}{s}\)
\(s = \dfrac{1.8 \cdot 4}{3} = 2.4\ \text{m}\)
\(s = 2.4\ \text{m}\)
intermediate
A pinhole camera is \(25\ \text{cm}\) deep. The image of a \(1.75\ \text{m}\) person is \(3.5\ \text{cm}\) tall. How far from the camera is the person?

Show solution
\(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\) \(\Rightarrow p = \dfrac{o \cdot p^{\prime}}{i}\)
With everything in metres, \(p = \dfrac{1.75 \cdot 0.25}{0.035}\)
\(p = \dfrac{0.4375}{0.035} = 12.5\ \text{m}\)
\(p = 12.5\ \text{m}\)
intermediate
In a solar eclipse, why is the total eclipse seen only along a narrow strip of the Earth, while the partial eclipse is seen over a much larger region? In what order are the Sun, the Moon and the Earth?

Show solution
The Sun is an extended source, so the Moon casts a shadow (umbra) and a penumbra.
Anyone in the umbra sees no part of the Sun, which is a total eclipse, and the umbra that reaches the Earth is small.
Anyone in the penumbra sees only part of the Sun, which is a partial eclipse, and the penumbra is much wider.
Order: Sun – Moon – Earth. Total in the umbra (narrow strip); partial in the penumbra (wide region).
intermediate
To measure the height of a building, a student pushes a \(1\ \text{m}\) stick into the ground. The stick's shadow is \(0.6\ \text{m}\) long and the building's is \(15\ \text{m}\). How tall is the building?

Show solution
Since sunlight arrives in parallel rays, the triangles are similar.
\(\dfrac{H}{15} = \dfrac{1}{0.6}\)
\(H = \dfrac{15}{0.6} = 25\ \text{m}\)
\(H = 25\ \text{m}\)
intermediate
A small (point) lamp is \(1\ \text{m}\) from an opaque disk of radius \(10\ \text{cm}\), parallel to a wall that is \(3\ \text{m}\) from the lamp. What is the radius of the shadow on the wall? How many times larger is the area of the shadow than the area of the disk?

Show solution
With a point source there is only a shadow, with no penumbra.
By similar triangles, \(\dfrac{R}{r} = \dfrac{3}{1}\) \(\Rightarrow R = 3 \cdot 10 = 30\ \text{cm}\)
The area grows with the square of that ratio, \(\left(\dfrac{30}{10}\right)^2 = 9\)
\(R = 30\ \text{cm}\); the shadow's area is 9 times the disk's.
challenge
A pinhole camera \(20\ \text{cm}\) deep, placed \(10\ \text{m}\) from a tree, shows an image \(6\ \text{cm}\) tall. (a) How tall is the tree? (b) If the camera is moved to \(6\ \text{m}\) from the tree, how tall will the new image be?

Show solution
(a) \(o = \dfrac{i \cdot p}{p^{\prime}} = \dfrac{0.06 \cdot 10}{0.20} = 3\ \text{m}\)
(b) \(i = \dfrac{o \cdot p^{\prime}}{p} = \dfrac{3 \cdot 0.20}{6} = 0.10\ \text{m}\)
Moving closer makes the image bigger, since \(i\) is inversely proportional to \(p\).
(a) \(3\ \text{m}\); (b) \(10\ \text{cm}\)
challenge
A lamp \(10\ \text{cm}\) in diameter (extended source) is \(1\ \text{m}\) from an opaque disk \(20\ \text{cm}\) in diameter. The wall is \(3\ \text{m}\) from the lamp. Find the diameter of the shadow (umbra) and the outer diameter of the penumbra.

Show solution
The crossed edge rays give the edge of the shadow, and the same-side rays give the outer edge of the penumbra. By similar triangles, with \(\dfrac{L}{a} = \dfrac{3}{1} = 3\):
For the shadow, \(D_s = d \cdot 3 - F \cdot (3 - 1) = 20 \cdot 3 - 10 \cdot 2 = 40\ \text{cm}\)
For the penumbra, \(D_p = d \cdot 3 + F \cdot (3 - 1) = 60 + 20 = 80\ \text{cm}\)
A shadow \(40\ \text{cm}\) in diameter, surrounded by penumbra out to \(80\ \text{cm}\).

STEP 2

In a mirror, light bounces back at the same angle

When light hits a polished surface, it reflects, and we measure the angles from the normal, the line perpendicular to the mirror at the point where the ray arrives.

The 1st law of reflection says that the incident ray, the normal and the reflected ray lie in the same plane, and the 2nd law says that the angle of reflection is equal to the angle of incidence.

In a plane mirror, the reflected rays seem to come from a point behind the mirror. We call the image there virtual, because the rays don't actually go there, and it is also upright, the same size as the object and symmetric, as far behind the mirror as the object is in front of it, with left and right swapped.

A mirror half your height is enough to see your whole body, at any distance, something that tends to surprise people the first time they check it. When two mirrors stand at an angle, they make several images, each one reflecting another.

\(i = r\)\(d_{\text{image}} = d_{\text{object}}\)\(\ell_{\min} = \dfrac{H}{2}\)\(N = \dfrac{360^\circ}{\alpha} - 1\)i and r measured from the normal · N holds when 360°/α is a whole number (and, if it is odd, with the object on the bisector)

Let's discuss

In 'Ray' mode, drag the laser and check whether the reflected angle always matches the incident one.
In 'Object and image', drag the eye. Where do the rays reaching it seem to come from, and are there positions from which the eye cannot see the image?
In 'Full body', move the person away from the mirror. Many people expect that stepping back shows more of the body, so check whether the part of the mirror being used changes size.
In 'Two mirrors', close the angle from 90° to 60° and then to 45°, count the images each time and check the count with \(N = 360^\circ/\alpha - 1\).

incident ray

reflected ray

normal / extension

Angle of incidence

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A ray of light makes \(30^\circ\) with the surface of a plane mirror. What are the angle of incidence and the angle of reflection? What is the angle between the incident ray and the reflected ray?

Show solution
Angles are measured from the normal, not from the mirror.
\(i = 90^\circ - 30^\circ = 60^\circ\) and, by the law of reflection, \(r = i = 60^\circ\).
The angle between the rays is \(i + r = 120^\circ\).
\(i = r = 60^\circ\); the rays are \(120^\circ\) apart.
basic
You are \(2\ \text{m}\) from a plane mirror. How far from you is your image? And if you take half a metre towards the mirror?

Show solution
The image is behind the mirror, as far behind as the object is in front.
The distance from you to the image is \(2 + 2 = 4\ \text{m}\).
After walking \(0.5\ \text{m}\), you are \(1.5\ \text{m}\) from the mirror and so is the image, which gives \(1.5 + 1.5 = 3\ \text{m}\).
\(4\ \text{m}\); afterwards, \(3\ \text{m}\).
basic
Why is the word AMBULANCE painted 'backwards' on the front of ambulances?

Show solution
The image in a plane mirror is symmetric, swapping left and right (it is a mirror image).
The driver ahead sees the ambulance in the rear-view mirror, which flips the text again.
Flipped twice, the word reads correctly in the rear-view mirror.
basic
How many images of a candle do you see between two plane mirrors at \(90^\circ\)? And at \(60^\circ\)?

Show solution
\(N = \dfrac{360^\circ}{\alpha} - 1\)
\(\alpha = 90^\circ\): \(N = 4 - 1 = 3\)
\(\alpha = 60^\circ\): \(N = 6 - 1 = 5\)
3 images at \(90^\circ\); 5 images at \(60^\circ\).
intermediate
A person \(1.70\ \text{m}\) tall, with eyes \(1.60\ \text{m}\) above the floor, wants to see their whole body in a vertical plane mirror. What is the smallest mirror size, and how high above the floor must its bottom edge be? Does this depend on the distance to the mirror?

Show solution
The ray from the feet reflects halfway between the feet and the eyes, at \(\dfrac{1.60}{2} = 0.80\ \text{m}\).
The ray from the top of the head reflects halfway between the head and the eyes, at \(\dfrac{1.70 + 1.60}{2} = 1.65\ \text{m}\).
The size is then \(1.65 - 0.80 = 0.85\ \text{m}\), half the person's height.
\(0.85\ \text{m}\), with the bottom edge \(0.80\ \text{m}\) above the floor; it does not depend on the distance.
intermediate
A person walks towards a fixed plane mirror at \(1.5\ \text{m/s}\). What is the speed of the image relative to the ground? And relative to the person?

Show solution
Since the image is always symmetric, if the person gets \(1.5\ \text{m}\) closer each second, the image also gets \(1.5\ \text{m}\) closer to the mirror each second, in the opposite direction.
Relative to the ground, the image moves at \(1.5\ \text{m/s}\).
Relative to the person, since the two move in opposite directions, the speeds add up to \(1.5 + 1.5 = 3\ \text{m/s}\).
\(1.5\ \text{m/s}\) relative to the ground and \(3\ \text{m/s}\) relative to the person.
intermediate
A ray strikes a plane mirror. If the mirror is rotated by \(15^\circ\) (with the incident ray fixed), by how much does the reflected ray rotate?

Show solution
Rotating the mirror rotates the normal by \(15^\circ\), so the angle of incidence changes by \(15^\circ\).
The angle of reflection also changes by \(15^\circ\), and the two effects add up for the reflected ray.
Rotation of the reflected ray \(= 2 \cdot 15^\circ = 30^\circ\).
The reflected ray rotates by \(30^\circ\).
intermediate
Two plane mirrors at an angle form 7 images of an object placed between them. What is the angle between the mirrors?

Show solution
\(N = \dfrac{360^\circ}{\alpha} - 1\) \(\Rightarrow 7 + 1 = \dfrac{360^\circ}{\alpha}\)
\(\alpha = \dfrac{360^\circ}{8} = 45^\circ\)
\(\alpha = 45^\circ\)
challenge
Two plane mirrors form \(90^\circ\). A ray hits the first one with an angle of incidence of \(40^\circ\). At what angle does it strike the second mirror, and in what direction does it leave after the two reflections?

Show solution
In the triangle formed by the corner and the two reflection points, the ray reflected by the first mirror makes \(90^\circ - 40^\circ = 50^\circ\) with that mirror.
Since the mirrors make \(90^\circ\), this ray makes \(180^\circ - 90^\circ - 50^\circ = 40^\circ\) with the second mirror, that is, an angle of incidence of \(90^\circ - 40^\circ = 50^\circ\).
Each reflection reverses one component of the motion, and together they reverse the ray's direction.
It strikes the second mirror at \(50^\circ\) and leaves parallel to the original ray, going the opposite way (a retroreflector).
challenge
You are \(2\ \text{m}\) from a plane mirror \(1\ \text{m}\) wide, with your eyes in front of its middle. Behind you, \(5\ \text{m}\) from the mirror, there is a wall. How wide a section of that wall can you see in the mirror?

Show solution
The field of view is bounded by the lines that start at the image of the eye (\(2\ \text{m}\) behind the mirror) and pass through the edges of the mirror.
From the eye's image to the mirror there are \(2\ \text{m}\), and to the wall \(2 + 5 = 7\ \text{m}\).
By similar triangles, \(\dfrac{x}{1} = \dfrac{7}{2}\) \(\Rightarrow x = 3.5\ \text{m}\)
You see \(3.5\ \text{m}\) of wall.

STEP 3

Curved mirrors: magnify, shrink, flip

A spherical mirror is a piece of a mirrored sphere. In a concave mirror the reflecting side is the inside (like the inside of a spoon), while in a convex mirror it is the outside. We call C the centre of curvature, V the vertex and F the focus, which lies halfway between them.

To find the image we follow a few principal rays. A ray arriving parallel to the axis reflects through the focus (in a convex mirror, as if it came from it), and a ray through the focus comes back parallel. A ray through C comes back on itself, while one hitting the vertex reflects symmetrically about the axis. The image is where the reflected rays cross.

In a concave mirror, the image depends on where we place the object: beyond C → real, inverted, smaller; at C → real, inverted, same size; between C and F → real, inverted, larger; at F → at infinity (the rays leave parallel: a headlight); between F and V → virtual, upright, larger (a make-up mirror). A convex mirror always gives a virtual, upright, smaller image (wing mirror, car park mirror).

\(f = \dfrac{R}{2}\)\(\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{p^{\prime}}\)\(A = \dfrac{i}{o} = -\dfrac{p^{\prime}}{p}\)Signs: p > 0 (real object) · p′ > 0 real image, in front; p′ < 0 virtual, behind · f > 0 concave, f < 0 convex · A > 0 upright, A < 0 inverted

Let's discuss

Drag the object from far away to close to the concave mirror and find the point where the image 'switches' from real to virtual.
Place the object at C and check in the readout that \(A = -1\).
Put the object at F. Where do the reflected rays go, and why might headlights use this arrangement?
Switch to convex and drag the object anywhere. Does the type of image change at any point?
Choose p and f, calculate \(p^{\prime}\) in your notebook with the thin lens / mirror equation (Gauss) and compare with the readout.

parallel → focus

focus → parallel

through the vertex

through C

Focal length |f|

Object distance p

Image position p′

Magnification A

Height (10 cm object)

Image

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A concave mirror has a radius of curvature \(R = 40\ \text{cm}\). Where is the focus?

Show solution
\(f = \dfrac{R}{2} = \dfrac{40}{2}\)
\(f = 20\ \text{cm}\) from the vertex, halfway between the vertex and the centre of curvature.
basic
An object is \(30\ \text{cm}\) from a concave mirror with a focal length of \(10\ \text{cm}\). Where does the image form and what are its characteristics?

Show solution
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{f} - \dfrac{1}{p} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{2}{30}\)
\(p^{\prime} = 15\ \text{cm}\) (positive: real, in front of the mirror)
\(A = -\dfrac{p^{\prime}}{p} = -\dfrac{15}{30} = -0.5\)
Image at \(15\ \text{cm}\): real, inverted and half the size.
basic
Why is the passenger-side wing mirror of a car convex, and why does it say 'objects in mirror are closer than they appear'?

Show solution
A convex mirror always forms a virtual, upright and smaller image.
Because the image is smaller, more fits in the mirror, and the field of view is wider.
Since the image is small, the brain interprets the car as being farther away.
Convex = wide field of view; the reduced image is misleading about distance.
basic
In car headlights, the bulb sits at the focus of a concave mirror. Why?

Show solution
Rays leaving the focus and hitting the concave mirror come back parallel to the axis (a principal ray).
So the light leaves in a narrow beam that travels far without spreading out.
With the object at the focus, the image is at infinity.
Bulb at the focus → parallel, concentrated beam.
intermediate
In a concave make-up mirror with a focal length of \(20\ \text{cm}\), the face is \(10\ \text{cm}\) away. Where is the image and how much is it magnified? A \(3\ \text{mm}\) freckle appears how big?

Show solution
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{20} - \dfrac{1}{10} = -\dfrac{1}{20}\) \(\Rightarrow p^{\prime} = -20\ \text{cm}\) (virtual, behind the mirror)
\(A = -\dfrac{-20}{10} = 2\) (upright, 2 times larger)
\(i = 2 \cdot 3 = 6\ \text{mm}\)
Virtual image \(20\ \text{cm}\) behind the mirror, upright, magnified 2 times: \(6\ \text{mm}\).
intermediate
A \(9\ \text{cm}\) object is \(30\ \text{cm}\) from a convex mirror with a focal length of \(15\ \text{cm}\) (use \(f = -15\ \text{cm}\)). Find the position and height of the image.

Show solution
\(\dfrac{1}{p^{\prime}} = -\dfrac{1}{15} - \dfrac{1}{30} = -\dfrac{3}{30}\) \(\Rightarrow p^{\prime} = -10\ \text{cm}\)
\(A = -\dfrac{-10}{30} = \dfrac{1}{3}\)
\(i = \dfrac{9}{3} = 3\ \text{cm}\)
Virtual image \(10\ \text{cm}\) behind the mirror, upright, \(3\ \text{cm}\) tall.
intermediate
An object sits exactly at the centre of curvature of a concave mirror with \(R = 30\ \text{cm}\). Where is the image and what is the magnification?

Show solution
\(f = 15\ \text{cm}\) and \(p = 30\ \text{cm}\)
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{15} - \dfrac{1}{30} = \dfrac{1}{30}\) \(\Rightarrow p^{\prime} = 30\ \text{cm}\)
\(A = -\dfrac{30}{30} = -1\)
The image is also at C: real, inverted and the same size.
intermediate
A concave mirror projects onto a screen, \(60\ \text{cm}\) away from it, a real image of a candle 3 times larger. How far away is the candle and what is the mirror's focal length?

Show solution
Since the image is real and projected, it is inverted, so \(A = -3 = -\dfrac{p^{\prime}}{p}\)
\(p = \dfrac{60}{3} = 20\ \text{cm}\)
\(\dfrac{1}{f} = \dfrac{1}{20} + \dfrac{1}{60} = \dfrac{4}{60}\) \(\Rightarrow f = 15\ \text{cm}\)
Candle at \(20\ \text{cm}\); \(f = 15\ \text{cm}\).
challenge
You want a make-up mirror that, with your face \(20\ \text{cm}\) away, shows an upright image 3 times larger. What type of mirror, with what focal length and what radius of curvature?

Show solution
An upright, larger image only happens in a concave mirror, with the object between F and V.
\(A = 3 = -\dfrac{p^{\prime}}{20}\) \(\Rightarrow p^{\prime} = -60\ \text{cm}\)
\(\dfrac{1}{f} = \dfrac{1}{20} - \dfrac{1}{60} = \dfrac{2}{60}\) \(\Rightarrow f = 30\ \text{cm}\)
\(R = 2f = 60\ \text{cm}\)
Concave, with \(f = 30\ \text{cm}\) and \(R = 60\ \text{cm}\).
challenge
A car's convex side mirror has a radius of curvature of \(2\ \text{m}\). A car \(1.5\ \text{m}\) tall is \(9\ \text{m}\) from it. How tall is the image and where does it form?

Show solution
\(f = -\dfrac{R}{2} = -1\ \text{m}\) (convex)
\(\dfrac{1}{p^{\prime}} = -1 - \dfrac{1}{9} = -\dfrac{10}{9}\) \(\Rightarrow p^{\prime} = -0.9\ \text{m}\)
\(A = -\dfrac{-0.9}{9} = 0.1\) \(\Rightarrow i = 0.1 \cdot 1.5 = 0.15\ \text{m}\)
Virtual, upright image, \(15\ \text{cm}\) tall, \(0.9\ \text{m}\) behind the mirror.

STEP 4

New medium, new speed, new direction

Light is slower in water and glass than in air, and the refractive index \(n = c/v\) tells us how many times slower it gets. Some typical values are air ≈ 1.00, water ≈ 1.33, glass ≈ 1.50 and diamond ≈ 2.42.

When light passes from one medium into another at an oblique angle, the ray changes direction, and we call this refraction. Entering an optically denser medium (higher n), it bends towards the normal, and leaving it, it bends away. A small part of the light is always reflected.

That is why a pool looks shallower than it is and a pencil looks 'broken' in a glass, since the eye extends in a straight line rays that actually bent at the surface.

\(n = \dfrac{c}{v}\)\(n_1\sin\theta_1 = n_2\sin\theta_2\)\(h^{\prime} \approx h\,\dfrac{n_{\text{obs}}}{n_{\text{obj}}}\)c = 3·10⁸ m/s · θ measured from the normal · the frequency (the colour) does not change in refraction; the speed and the wavelength do · h′ holds when looking almost straight down

Let's discuss

Send light from air into water at 40°. Does the ray bend towards or away from the normal, and what changes when it goes from water into air?
Set the angle to zero and check whether the ray bends and whether the speed changes.
Turn on the wavefronts and think about why they are closer together in the slower medium.
Now try air → diamond and compare θ₂ with the water case.
In 'Coin in the pool' mode, tilt your line of sight. Does the coin seem to rise or sink as the view gets more slanted?

Medium 1 (top)

Medium 2 (bottom)

Angle of incidence θ₁

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Light travels through glass with index \(n = 1.5\). What is its speed in the glass? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
\(n = \dfrac{c}{v}\) \(\Rightarrow v = \dfrac{c}{n}\)
\(v = \dfrac{3 \cdot 10^8}{1.5}\)
\(v = 2 \cdot 10^8\ \text{m/s}\)
basic
In diamond, light travels at \(1.25 \cdot 10^8\ \text{m/s}\). What is the refractive index of diamond? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
\(n = \dfrac{c}{v} = \dfrac{3 \cdot 10^8}{1.25 \cdot 10^8}\)
\(n = 2.4\)
basic
A ray passes from air into water. Does it bend towards or away from the normal? And if it arrives perpendicular to the surface, what happens?

Show solution
Water is optically denser than air (higher \(n\)), so light slows down and the ray bends towards the normal.
At perpendicular incidence (\(\theta_1 = 0^\circ\)), \(\sin\theta_2 = 0\), so the ray changes speed but not direction.
It bends towards the normal; at perpendicular incidence it goes straight through.
basic
A ray comes from air (\(n_1 = 1\)) at an angle of \(30^\circ\) and, in the other medium, travels with \(\sin\theta_2 = 0.25\). What is the refractive index of that medium? Use \(\sin 30^\circ = 0.5\).

Show solution
\(n_1\sin\theta_1 = n_2\sin\theta_2\)
\(1 \cdot 0.5 = n_2 \cdot 0.25\)
\(n_2 = 2\)
intermediate
A ray comes from air with an angle of incidence of \(60^\circ\) and enters glass with index \(\sqrt{3}\). What is the angle of refraction? Use \(\sin 60^\circ = \dfrac{\sqrt 3}{2}\).

Show solution
\(1 \cdot \sin 60^\circ = \sqrt 3 \cdot \sin\theta_2\)
\(\sin\theta_2 = \dfrac{\sqrt3/2}{\sqrt 3} = \dfrac{1}{2}\)
\(\theta_2 = 30^\circ\)
intermediate
A swimming pool is \(2\ \text{m}\) deep. Looking almost straight down from outside, how deep does it appear? Use \(n_{\text{water}} = \dfrac{4}{3}\).

Show solution
For nearly vertical viewing, \(h^{\prime} = h \cdot \dfrac{n_{\text{observer}}}{n_{\text{object}}}\)
\(h^{\prime} = 2 \cdot \dfrac{1}{4/3} = 2 \cdot \dfrac{3}{4}\)
\(h^{\prime} = 1.5\ \text{m}\): the pool looks shallower.
intermediate
How many times faster is light in water (\(n = \dfrac{4}{3}\)) than in glass (\(n = 1.5\))?

Show solution
\(v = \dfrac{c}{n}\), so \(\dfrac{v_{\text{water}}}{v_{\text{glass}}} = \dfrac{n_{\text{glass}}}{n_{\text{water}}}\)
\(\dfrac{1.5}{4/3} = \dfrac{1.5 \cdot 3}{4} = 1.125\)
\(1.125\) times (about 12.5% faster).
intermediate
A spearfisher sees a fish in the water. Should they aim exactly where they see the fish, above it or below it? Explain using refraction.

Show solution
Light from the fish leaves the water and bends away from the normal as it enters the air.
The eye extends the ray in a straight line, and that extension passes above the real position.
That is why the fish seems shallower (like the pool and the 'broken' pencil in a glass).
They should aim below where they see the fish.
challenge
A ray passes from water (\(n = \dfrac{4}{3}\)) into glass (\(n = 1.5\)) with an angle of incidence of \(45^\circ\). What is the angle of refraction? Use \(\sin 45^\circ \approx 0.707\).

Show solution
\(\dfrac{4}{3} \cdot 0.707 = 1.5 \cdot \sin\theta_2\)
\(\sin\theta_2 = \dfrac{0.943}{1.5} \approx 0.629\)
\(\theta_2 \approx 38.9^\circ\), smaller than \(45^\circ\) because glass is optically denser than water.
\(\theta_2 \approx 39^\circ\)
challenge
A diver underwater looks up and sees a bird hovering \(3\ \text{m}\) above the surface. Looking almost straight up, how high does the bird appear to be? Use \(n_{\text{water}} = \dfrac{4}{3}\).

Show solution
Now the observer is in the denser medium, and \(h^{\prime} = h \cdot \dfrac{n_{\text{observer}}}{n_{\text{object}}}\)
\(h^{\prime} = 3 \cdot \dfrac{4/3}{1} = 4\ \text{m}\)
For someone in the water, things outside look farther away.
The bird seems to be \(4\ \text{m}\) above the surface.

STEP 5

When light can't get out

Going from an optically denser medium to a less dense one (from water into air, for example), the ray bends away from the normal. As we increase the angle of incidence, the refracted ray tilts more and more until it skims along the surface (\(\theta_2 = 90^\circ\)), and we call that angle of incidence the critical angle L.

Above L there is no refraction and all the light comes back, which is total internal reflection, more efficient than any mirror. It only happens going from the denser to the less dense medium.

We find this effect in several places. An optical fibre traps light by successive total internal reflections and carries internet over many kilometres. A diamond (L ≈ 24°) sparkles because light gets trapped and exits through the top face, and a mirage on hot tarmac is light from the sky that curves in the hot air near the ground.

\(\sin L = \dfrac{n_{\text{lower}}}{n_{\text{higher}}}\)total internal reflection if \(\theta > L\)water–air: L ≈ 48.6° (with n = 4/3; with n = 1.33 it is 48.8°) · glass–air: L ≈ 41.8° · diamond–air: L ≈ 24.4°

Let's discuss

With the source in the water, which rays get out and which come back, and where is the boundary between them?
Bring the highlighted ray up to the critical angle and look at how the refracted ray leaves.
With diamond instead of water, does the 'cone' of light that gets out become wider or narrower?
In fibre mode, increase the entry angle until light starts to escape, and then switch the cladding to air to see what changes.
Why is there no total internal reflection when light goes from air into water?

ray that gets out

total internal reflection

critical angle

Highlighted ray θ

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the critical angle for light going from water (\(n = \dfrac{4}{3}\)) into air?

Show solution
\(\sin L = \dfrac{n_{\text{lower}}}{n_{\text{higher}}} = \dfrac{1}{4/3} = 0.75\)
\(L \approx 48.6^\circ\)
basic
What are the two conditions for total internal reflection to happen?

Show solution
1. Light must go from the more optically dense medium to the less dense one (for example, from water into air).
2. The angle of incidence must be greater than the critical angle \(L\).
From higher to lower n, with \(\theta > L\).
basic
The glass–air critical angle is about \(42^\circ\) (\(n_{\text{glass}} = 1.5\)). A ray inside the glass reaches the surface at \(45^\circ\). Does it leave the glass?

Show solution
\(\sin L = \dfrac{1}{1.5} \approx 0.667\) \(\Rightarrow L \approx 41.8^\circ\)
Since \(45^\circ > 41.8^\circ\), the ray cannot get out.
No: it undergoes total internal reflection and goes back into the glass.
basic
How can an optical fibre carry light along a cable full of bends?

Show solution
The fibre's core has a higher index than the cladding around it.
Light hits the wall at an angle greater than the critical angle and undergoes total internal reflection, with almost no loss.
Bouncing like this, it follows the path of the fibre, even around gentle bends.
By successive total internal reflections in the core.
intermediate
Diamond has \(n = 2.42\). What is its critical angle with air? Why does this make diamond sparkle so much?

Show solution
\(\sin L = \dfrac{1}{2.42} \approx 0.413\) \(\Rightarrow L \approx 24.4^\circ\)
With such a small critical angle, almost all light that enters hits the inner faces above \(L\) and undergoes total internal reflection several times.
The cut of the stone makes the light come out concentrated through the top face.
\(L \approx 24^\circ\): many total internal reflections send the light back to the viewer.
intermediate
By measuring, a student finds that the critical angle of a material with air is \(30^\circ\). What is the material's refractive index?

Show solution
\(\sin L = \dfrac{1}{n}\) \(\Rightarrow n = \dfrac{1}{\sin 30^\circ}\)
\(n = \dfrac{1}{0.5}\)
\(n = 2\)
intermediate
Can there be total internal reflection for light going from glass (\(n = 1.5\)) into water (\(n = \dfrac{4}{3}\))? If so, what is the critical angle?

Show solution
Yes, because glass is optically denser than water.
\(\sin L = \dfrac{4/3}{1.5} = \dfrac{8}{9} \approx 0.889\)
\(L \approx 62.7^\circ\) (larger than glass–air, because the indices are closer).
intermediate
On hot days, the tarmac seems to have a 'puddle of water' that disappears as you get closer. Explain this mirage.

Show solution
The air near the hot tarmac is less dense and has a slightly lower refractive index.
Light from the sky, coming down at a slant, passes through layers with smaller and smaller \(n\), keeps bending away from the normal and ends up curving (total internal reflection) upward.
The eye sees an image of the sky on the ground, and the brain interprets it as water reflecting the sky.
It is an image of the sky, bent by the hot air near the ground.
challenge
A torch at the bottom of a \(2\ \text{m}\) deep pool shines light upward in all directions. At the surface, light only gets out through a circle. Find the radius of that circle. Use \(n_{\text{water}} = \dfrac{4}{3}\).

Show solution
Outside the circle, the rays arrive with \(\theta > L\) and undergo total internal reflection.
\(\sin L = \dfrac{3}{4}\) \(\Rightarrow \cos L = \dfrac{\sqrt 7}{4}\) \(\Rightarrow \tan L = \dfrac{3}{\sqrt 7} \approx 1.134\)
\(r = h \cdot \tan L = 2 \cdot 1.134 \approx 2.27\ \text{m}\)
\(r \approx 2.3\ \text{m}\) (an area of about \(16\ \text{m}^2\)).
challenge
An optical fibre has a core with \(n_1 = 1.50\) and cladding with \(n_2 = 1.40\). (a) What is the core–cladding critical angle? (b) What is the largest angle the ray can make with the fibre's axis and still stay trapped?

Show solution
(a) \(\sin L = \dfrac{1.40}{1.50} \approx 0.933\) \(\Rightarrow L \approx 69.0^\circ\)
(b) The angle of incidence on the wall is measured from the normal, which is perpendicular to the axis. If the ray makes \(\alpha\) with the axis, it strikes the wall at \(90^\circ - \alpha\).
The ray stays trapped if \(90^\circ - \alpha > 69.0^\circ\) \(\Rightarrow \alpha < 21.0^\circ\).
(a) \(L \approx 69^\circ\); (b) up to about \(21^\circ\) with the axis.

STEP 6

Lenses: refraction working twice

A lens is a piece of glass with curved faces, and light refracts both on the way in and on the way out; in air, lenses with thin edges are converging, bringing parallel rays together at a real focus. Lenses with thick edges are diverging and spread the rays out, so that they seem to come from a virtual focus.

We can again trace principal rays to find the image. A ray arriving parallel to the axis leaves through the focus (or as if it came from it, in a diverging lens), a ray through the optical centre goes straight on, and a ray through the focus on the incoming side leaves parallel.

In a converging lens, an object far away (beyond 2f) gives a real, inverted, smaller image, which is what happens in a camera and the eye. Between 2f and f the image is real, inverted and larger, as in a projector, and between f and the lens it becomes virtual, upright and larger, as in a magnifying glass. A diverging lens always gives a virtual, upright, smaller image.

\(\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{p^{\prime}}\)\(A = -\dfrac{p^{\prime}}{p}\)\(V = \dfrac{1}{f}\)V (optical power) in dioptres (D) with f in metres · f > 0 converging, f < 0 diverging · p′ > 0: real image, on the other side of the lens · p′ < 0: virtual, on the object's side

Let's discuss

Drag the object from far away to close to the converging lens and note when the image moves to the other side and becomes virtual.
Click 'Projector'. Why does the slide have to go in upside down?
Click 'Magnifier' and compare with the make-up mirror from step 3.
Switch to diverging and look for a position that gives a real image. Is there one?
Decrease f and watch whether the optical power V goes up or down. Does a 'stronger' lens, then, have a shorter or a longer focal length?

parallel → focus

focus → parallel

through the centre

Focal length |f|

Object distance p

Image position p′

Magnification A

Optical power V

Image

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the optical power of a converging lens with a focal length of \(50\ \text{cm}\)? And of a diverging lens with a focal length of \(25\ \text{cm}\)?

Show solution
\(V = \dfrac{1}{f}\), with \(f\) in metres (positive for converging, negative for diverging).
\(V_1 = \dfrac{1}{0.5} = +2\ \text{D}\)
\(V_2 = \dfrac{1}{-0.25} = -4\ \text{D}\)
\(+2\ \text{D}\) and \(-4\ \text{D}\).
basic
An object is \(30\ \text{cm}\) from a converging lens with a focal length of \(10\ \text{cm}\). Where does the image form and what is the magnification?

Show solution
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{2}{30}\) \(\Rightarrow p^{\prime} = 15\ \text{cm}\)
\(A = -\dfrac{15}{30} = -0.5\)
Real image \(15\ \text{cm}\) away on the other side of the lens, inverted, half the size.
basic
To use a converging lens as a magnifying glass, where must the object be? What is the image like?

Show solution
The object must be between the focus and the lens (\(p < f\)).
In that case \(p^{\prime} < 0\), and the rays leave spreading apart and the eye sees their extensions.
Virtual, upright and larger image, on the same side as the object.
basic
What does a diverging lens do to a beam of rays parallel to the axis? What kind of image does it form of a real object?

Show solution
The rays leave spreading out, as if they came from the focus on the side the light comes from (a virtual focus).
For any position of a real object, the image is virtual, upright and smaller.
It spreads the beam; the image is always virtual, upright and smaller.
intermediate
A projector has a lens with a focal length of \(10\ \text{cm}\) and the screen is \(5\ \text{m}\) from the lens. How far from the lens must the slide be, and how many times larger than the slide is the image?

Show solution
With \(p^{\prime} = 500\ \text{cm}\), \(\dfrac{1}{p} = \dfrac{1}{10} - \dfrac{1}{500} = \dfrac{49}{500}\)
\(p = \dfrac{500}{49} \approx 10.2\ \text{cm}\) (just beyond the focus)
\(A = -\dfrac{500}{500/49} = -49\)
Slide at \(\approx 10.2\ \text{cm}\); image 49 times larger (and inverted: that is why the slide goes in upside down).
intermediate
A magnifying glass with a focal length of \(5\ \text{cm}\) is \(4\ \text{cm}\) from a printed letter. Where does the image form and what is the magnification?

Show solution
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{5} - \dfrac{1}{4} = -\dfrac{1}{20}\) \(\Rightarrow p^{\prime} = -20\ \text{cm}\)
\(A = -\dfrac{-20}{4} = 5\)
Virtual image \(20\ \text{cm}\) from the lens, on the letter's side, upright and 5 times larger.
intermediate
An object is \(20\ \text{cm}\) from a diverging lens with a focal length of \(20\ \text{cm}\) (\(f = -20\ \text{cm}\)). Find the image.

Show solution
\(\dfrac{1}{p^{\prime}} = -\dfrac{1}{20} - \dfrac{1}{20} = -\dfrac{1}{10}\) \(\Rightarrow p^{\prime} = -10\ \text{cm}\)
\(A = -\dfrac{-10}{20} = 0.5\)
Virtual, \(10\ \text{cm}\) from the lens, upright, half the size.
intermediate
A camera has a lens with a focal length of \(50\ \text{mm}\). It photographs a \(1.80\ \text{m}\) person \(5\ \text{m}\) away. How far from the lens is the sensor and how tall is the image?

Show solution
With \(p = 5000\ \text{mm}\), \(\dfrac{1}{p^{\prime}} = \dfrac{1}{50} - \dfrac{1}{5000} = \dfrac{99}{5000}\)
\(p^{\prime} = \dfrac{5000}{99} \approx 50.5\ \text{mm}\)
\(|A| = \dfrac{50.5}{5000} \approx 0.0101\) \(\Rightarrow i \approx 0.0101 \cdot 1800 \approx 18.2\ \text{mm}\)
Sensor at \(\approx 50.5\ \text{mm}\); real, inverted image about \(18\ \text{mm}\) tall.
challenge
A lens forms an image on a screen that is 4 times larger than the object. The distance between object and screen is \(125\ \text{cm}\). Find \(p\), \(p^{\prime}\), the focal length and the optical power of the lens.

Show solution
Since the image falls on a screen, it is real and inverted, so \(p^{\prime} = 4p\).
\(p + 4p = 125\) \(\Rightarrow p = 25\ \text{cm}\) and \(p^{\prime} = 100\ \text{cm}\)
\(\dfrac{1}{f} = \dfrac{1}{25} + \dfrac{1}{100} = \dfrac{5}{100}\) \(\Rightarrow f = 20\ \text{cm}\)
\(V = \dfrac{1}{0.20} = 5\ \text{D}\)
\(p = 25\ \text{cm}\), \(p^{\prime} = 100\ \text{cm}\), \(f = 20\ \text{cm}\), \(V = 5\ \text{D}\).
challenge
An object and a screen are fixed \(90\ \text{cm}\) apart. A converging lens with a focal length of \(20\ \text{cm}\) is slid between them. At what positions is the image sharp on the screen?

Show solution
\(p + p^{\prime} = 90\) and \(\dfrac{1}{p} + \dfrac{1}{p^{\prime}} = \dfrac{1}{20}\) \(\Rightarrow p \cdot p^{\prime} = 20 \cdot 90 = 1800\)
\(p\) and \(p^{\prime}\) are the roots of \(x^2 - 90x + 1800 = 0\), which gives \(x = \dfrac{90 \pm \sqrt{8100 - 7200}}{2} = \dfrac{90 \pm 30}{2}\)
\(x = 60\) or \(x = 30\)
With the lens \(30\ \text{cm}\) from the object (image 2 times larger) or \(60\ \text{cm}\) from it (image half the size).

STEP 7

The eye is a camera with a lens that refocuses

The cornea and the crystalline lens work together as a converging lens that forms a real, inverted, smaller image on the retina. Since the distance to the retina does not change, the eye focuses by changing the lens itself, which muscles make more curved when we look at something up close. We call this accommodation.

In myopia (nearsightedness) the eye converges too much (or is too long), so the image of a distant object forms in front of the retina, and we correct it with a diverging lens. In hyperopia (farsightedness) the eye converges too little (or is too short), the image of a close object would form behind the retina and the correction is a converging lens. With presbyopia ('tired eyes' with age), the lens stiffens and loses accommodation, and people use converging reading glasses.

The strength of a glasses prescription is the lens's optical power in dioptres, so −2 D means a diverging lens with a 50 cm focal length.

\(V = \dfrac{1}{f}\)myopia: \(f = -d_{\text{far}}\)\(V_{\text{eye}} = \dfrac{1}{p} + \dfrac{1}{p^{\prime}_{\text{retina}}}\)Normal eye: near point ≈ 25 cm, far point at infinity · the eye has about 60 D · in the simulation the glasses lens is treated as touching the eye and the focusing error is exaggerated so it is visible

Let's discuss

With the normal eye, switch the object from far to near and see how much accommodation the lens needed.
Take the myopic eye with a distant object. Where does the image form, and what changes when we put on the glasses?
Can the hyperopic eye see far away, and how much effort (accommodation) does it use for that? Then try an object up close.
For presbyopia, put on the reading glasses and then look far away. What happens to the image?
Why do glasses for myopia have negative optical power?

light rays

where the image forms

Accommodation used

The image forms

Glasses

Sees sharply from… to…

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
In the eye, which part works as an adjustable-focus lens, and where must the image form? What is that image like?

Show solution
The crystalline lens (together with the cornea) is a converging lens; muscles change its curvature (accommodation).
The image must form on the retina.
It is real, inverted and smaller, and the brain 'flips' the image back.
The crystalline lens; a real, inverted, smaller image on the retina.
basic
A glasses prescription says '\(-2.0\) D'. Is the lens converging or diverging? What is its focal length? What vision problem does it correct?

Show solution
A negative optical power means a diverging lens.
\(f = \dfrac{1}{V} = \dfrac{1}{-2} = -0.5\ \text{m}\)
A diverging lens corrects myopia.
Diverging, \(f = -50\ \text{cm}\), for myopia.
basic
In hyperopia, where does the image of a nearby object tend to form? Which lens corrects it?

Show solution
A hyperopic eye is 'weak' (or too short), so the rays have not yet converged when they reach the retina.
The image tends to form behind the retina.
A converging lens makes the rays converge sooner.
Behind the retina; corrected with a converging lens.
basic
Why do many people over 40 hold their mobile phone farther away to read? What is this condition called and how is it corrected?

Show solution
With age, the crystalline lens stiffens and loses its ability to accommodate.
The near point moves away, and the eye can no longer focus on nearby objects.
Presbyopia; it is corrected with converging reading glasses.
intermediate
A myopic person sees sharply only up to \(50\ \text{cm}\) (far point). Which lens corrects this? Give its optical power.

Show solution
The lens must make an object at infinity form a virtual image at the far point, which means \(f = -50\ \text{cm} = -0.5\ \text{m}\).
\(V = \dfrac{1}{-0.5} = -2\ \text{D}\)
Diverging, \(-2\ \text{D}\) (a −2 D prescription for myopia).
intermediate
A person wears \(-4\ \text{D}\) glasses for myopia. Up to what distance can they see clearly without glasses?

Show solution
For the correction, \(|f|\) equals the far point.
\(f = \dfrac{1}{-4} = -0.25\ \text{m}\)
Far point at \(25\ \text{cm}\): without glasses, they only see sharply up to \(25\ \text{cm}\).
intermediate
A hyperopic person has a near point at \(1\ \text{m}\). Which lens lets them read at \(25\ \text{cm}\)?

Show solution
The lens must make the book (\(p = 25\ \text{cm}\)) form a virtual image at \(100\ \text{cm}\) (\(p^{\prime} = -100\ \text{cm}\)).
\(\dfrac{1}{f} = \dfrac{1}{25} - \dfrac{1}{100} = \dfrac{3}{100}\) \(\Rightarrow f \approx 33.3\ \text{cm}\)
\(V = \dfrac{1}{0.333} = 3\ \text{D}\)
Converging, \(+3\ \text{D}\).
intermediate
In a simple model, the retina is \(2.0\ \text{cm}\) from the crystalline lens. What is the eye's optical power to see a very distant object? And to see an object at \(25\ \text{cm}\)? How much did the eye have to 'accommodate'?

Show solution
For a distant object (\(p \to \infty\)), \(V = \dfrac{1}{p^{\prime}} = \dfrac{1}{0.02} = 50\ \text{D}\)
Up close, \(V = \dfrac{1}{0.25} + \dfrac{1}{0.02} = 4 + 50 = 54\ \text{D}\)
The accommodation is \(54 - 50 = 4\ \text{D}\)
\(50\ \text{D}\) and \(54\ \text{D}\); an accommodation of \(4\ \text{D}\).
challenge
A person with presbyopia has a near point at \(50\ \text{cm}\). What optical power do the glasses need to read at \(25\ \text{cm}\)? With these glasses, why do they see poorly at a distance?

Show solution
\(\dfrac{1}{f} = \dfrac{1}{25} - \dfrac{1}{50} = \dfrac{1}{50}\) \(\Rightarrow f = 50\ \text{cm}\) \(\Rightarrow V = +2\ \text{D}\)
For distant objects, the eye is already relaxed and the extra converging lens makes the image form in front of the retina.
\(+2\ \text{D}\); that is why people use reading glasses or multifocal lenses.
challenge
A myopic person has a far point at \(50\ \text{cm}\) and a near point at \(10\ \text{cm}\). They wear \(-2\ \text{D}\) glasses (treat the lens as touching the eye). What is the new near point?

Show solution
With glasses, the closest visible object is the one whose virtual image falls at \(10\ \text{cm}\), so \(p^{\prime} = -10\ \text{cm}\), \(f = -50\ \text{cm}\).
\(\dfrac{1}{p} = \dfrac{1}{f} - \dfrac{1}{p^{\prime}} = -\dfrac{1}{50} + \dfrac{1}{10} = \dfrac{4}{50}\)
\(p = 12.5\ \text{cm}\)
The near point moves to \(12.5\ \text{cm}\).

STEP 8

White light is a mixed-up rainbow

Newton showed that white light is a mixture of colours. In glass, the refractive index depends on the colour, higher for violet and lower for red, so when light goes through a prism each colour bends by a different angle and the light fans out into a spectrum, an effect we call dispersion. A rainbow is the same effect in raindrops (with one reflection inside each drop), with the Sun behind the person looking.

The colour of an object depends on the light it reflects. A red apple, for instance, reflects red and absorbs the rest, so under green light it has no red to reflect and looks black.

Colours of light add up, as in screens and stage lights, and red, green and blue (RGB) together make white. Pigments work by subtraction, and cyan, magenta and yellow together absorb almost everything and make black.

\(n_{\text{violet}} > n_{\text{red}}\)\(v = \dfrac{c}{n}\)R + G = yellow · R + B = magenta · G + B = cyan · R + G + B = whiteIn ordinary glass: n ≈ 1.51 (red) to 1.53 (violet) · the colour (frequency) does not change when light changes medium

Let's discuss

Which colour comes out of the prism bent the most? Check each one's deviation in the readouts.
Increase the prism angle and see whether the fan of colours opens up or closes.
Turn off 'Exaggerate the separation'. Why is real dispersion so small, and yet we still see rainbows?
Light the objects with green light. Which ones turn black, and why?
In 'Light vs. paint', turn on only red and green and try to predict the colour in the middle before looking.

Prism angle

Angle of incidence

Exercises for step 8 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What happens to white sunlight when it passes through a glass prism? Why?

Show solution
White light is a mixture of all colours.
The refractive index of glass is slightly different for each colour, so each colour bends by a different angle.
It splits into a spectrum (red, orange, yellow, green, blue, indigo, violet): this is dispersion.
basic
A red apple is lit only by green light in a dark room. What colour does it look?

Show solution
A red object reflects red light and absorbs the other colours.
Under green light there is no red to reflect, and the green is absorbed.
It looks black.
basic
With colours of light (RGB), what colour do you see when you add: (a) red + green; (b) red + blue; (c) red + green + blue?

Show solution
Colours of light add up (additive mixing, as in the pixels of a screen).
(a) red + green = yellow
(b) red + blue = magenta
(c) all three together = white
Yellow, magenta and white.
basic
In a prism, which colour is deviated the most, red or violet? What does this say about the refractive index for each colour?

Show solution
Violet is the most deviated colour; red is the least deviated.
The colour with the higher refractive index (the one 'slowed down' more in the glass) bends more.
Violet bends more: \(n_{\text{violet}} > n_{\text{red}}\).
intermediate
In a certain glass, \(n_{\text{red}} = 1.51\) and \(n_{\text{violet}} = 1.53\). Calculate the speed of each colour in the glass. Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
\(v = \dfrac{c}{n}\)
Red: \(v = \dfrac{3 \cdot 10^8}{1.51} \approx 1.99 \cdot 10^8\ \text{m/s}\)
Violet: \(v = \dfrac{3 \cdot 10^8}{1.53} \approx 1.96 \cdot 10^8\ \text{m/s}\)
Red travels slightly faster in the glass than violet.
intermediate
White light comes from air at an angle of incidence of \(60^\circ\) and enters the glass from the previous exercise (\(n_{\text{red}} = 1.51\), \(n_{\text{violet}} = 1.53\)). Calculate the angle of refraction for each colour. Use \(\sin 60^\circ \approx 0.866\).

Show solution
Red: \(\sin\theta = \dfrac{0.866}{1.51} \approx 0.574\) \(\Rightarrow \theta \approx 35.0^\circ\)
Violet: \(\sin\theta = \dfrac{0.866}{1.53} \approx 0.566\) \(\Rightarrow \theta \approx 34.5^\circ\)
The difference is only \(\approx 0.5^\circ\), yet it is enough to separate the colours along the way.
\(\approx 35.0^\circ\) (red) and \(\approx 34.5^\circ\) (violet).
intermediate
On a stage, a white shirt and a blue shirt are lit by a yellow spotlight (yellow = red + green). What colour does each one look?

Show solution
The white shirt reflects every colour it receives, so it reflects red and green → yellow.
The blue shirt reflects only blue, which is not in the spotlight, and it absorbs red and green → black.
The white shirt looks yellow; the blue one looks black.
intermediate
Mixing cyan paint with yellow paint gives green. Why doesn't mixing paints follow the same rule as adding colours of light?

Show solution
Paint emits no light of its own, and it absorbs (subtracts) colours from white light.
Cyan absorbs red; yellow absorbs blue.
Together, only green is left to be reflected.
Pigments subtract colours (subtractive mixing); lights add them (additive mixing).
challenge
White light strikes a glass plate at \(45^\circ\), with \(n_{\text{red}} = 1.50\) and \(n_{\text{violet}} = 1.54\). Calculate the two angles of refraction and say which colour ends up closer to the normal. Use \(\sin 45^\circ \approx 0.707\).

Show solution
Red: \(\sin\theta = \dfrac{0.707}{1.50} \approx 0.471\) \(\Rightarrow \theta \approx 28.1^\circ\)
Violet: \(\sin\theta = \dfrac{0.707}{1.54} \approx 0.459\) \(\Rightarrow \theta \approx 27.3^\circ\)
Violet, with the higher \(n\), ends up closer to the normal.
\(\approx 28.1^\circ\) and \(\approx 27.3^\circ\); violet bends closer to the normal.
challenge
The flag of Brazil (green, yellow, blue and white) is lit only with green light. Then only with red light. What colours appear in each case? (Treat yellow as red + green.)

Show solution
Each part reflects only the colours of the incoming light that it 'has'.
Green light: green → green; yellow (R+G) → green; blue → black; white → green.
Red light: green → black; yellow → red; blue → black; white → red.
Under green: green, green, black, green. Under red: black, red, black, red.

WRAP-UP

Challenges

Pinhole camera

A tree \(4.5\ \text{m}\) tall is \(15\ \text{m}\) from a pinhole camera \(20\ \text{cm}\) deep. How tall is the image?

Show solution

\(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\) \(\Rightarrow i = \dfrac{4.5 \cdot 0.20}{15}\)
\(i = 0.06\ \text{m} = 6\ \text{cm}\), upside down.

Plane mirror

A person \(1.80\ \text{m}\) tall, with eyes \(1.70\ \text{m}\) above the floor, wants to see their whole body in a wall mirror. What is the minimum size, and how high is the bottom edge?

Show solution

Bottom edge: \(\dfrac{1.70}{2} = 0.85\ \text{m}\)
Top edge: \(\dfrac{1.80 + 1.70}{2} = 1.75\ \text{m}\)
Size: \(1.75 - 0.85 = 0.90\ \text{m}\) (half the person's height).

Curved mirror

An object is \(18\ \text{cm}\) from a concave mirror with a focal length of \(12\ \text{cm}\). Where is the image and what is it like?

Show solution

\(\dfrac{1}{p^{\prime}} = \dfrac{1}{12} - \dfrac{1}{18} = \dfrac{1}{36}\) \(\Rightarrow p^{\prime} = 36\ \text{cm}\)
\(A = -\dfrac{36}{18} = -2\)
Real, inverted and 2 times larger (object between C and F).

Refraction

A ray comes from air with \(\sin\theta_1 = 0.8\) (about \(53^\circ\)) and enters water (\(n = \dfrac{4}{3}\)). What is the angle of refraction?

Show solution

\(1 \cdot 0.8 = \dfrac{4}{3} \cdot \sin\theta_2\)
\(\sin\theta_2 = 0.6\) \(\Rightarrow \theta_2 \approx 37^\circ\)
It bent towards the normal, because water is optically denser.

Total internal reflection

Inside a glass block (\(n = 1.5\)), a ray reaches the glass–air face making \(50^\circ\) with the normal. Does it leave the glass?

Show solution

\(\sin L = \dfrac{1}{1.5} \approx 0.667\) \(\Rightarrow L \approx 41.8^\circ\)
Since \(50^\circ > L\), there is total internal reflection and the ray does not get out.

Lenses

A \(+4\ \text{D}\) lens forms an image of an object placed \(50\ \text{cm}\) from it. Where is the image and what is the magnification?

Show solution

\(f = \dfrac{1}{4} = 0.25\ \text{m} = 25\ \text{cm}\)
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{25} - \dfrac{1}{50} = \dfrac{1}{50}\) \(\Rightarrow p^{\prime} = 50\ \text{cm}\)
\(A = -1\), so the image is real, inverted and the same size (object at \(2f\)).

Human eye

A myopic person sees sharply only up to \(2\ \text{m}\). What optical power do their glasses need? Converging or diverging?

Show solution

The lens must bring infinity to the far point, so \(f = -2\ \text{m}\)
\(V = \dfrac{1}{-2} = -0.5\ \text{D}\), a diverging lens (half a dioptre).

Think it through, no math

During a total lunar eclipse the Moon turns reddish instead of disappearing. Use the Earth's shadow and refraction in the atmosphere to explain why.

Show solution

In a lunar eclipse the order is Sun – Earth – Moon, and the Moon moves into the Earth's shadow.
The Earth's atmosphere refracts sunlight and bends it into the shadow.
Along the long path through the air, blue is scattered away and mostly red is left (the same effect as a sunset), and that red light illuminates the Moon.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with a short everyday text, a question and five options, only one of which is right. Further down we list real ENEM questions on the same topics, which may be useful practice once you have tried these.

Pinhole camera · Step 1
For the science fair, Luana turned a tin \(25\ \text{cm}\) deep into a pinhole camera, with a tiny hole in one lid and tracing paper on the other. When she pointed the tin at a building \(40\ \text{m}\) away, she saw on the paper an inverted image of the building, \(15\ \text{cm}\) tall.

If we treat the hole as a point and neglect the thickness of the lids, roughly how tall is the building?
1. \(0.94\ \text{mm}\)
2. \(2.4\ \text{m}\)
3. \(6\ \text{m}\)
4. \(24\ \text{m}\)
5. \(160\ \text{m}\)
Show solution

Answer: D.
The rays crossing at the hole form two similar triangles, so \(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\).
With everything in metres, \(o = \dfrac{i \cdot p}{p^{\prime}} = \dfrac{0.15 \cdot 40}{0.25} = 24\ \text{m}\).
The other values come from common slips: \(6\ \text{m}\) forgets to divide by \(p^{\prime}\), \(2.4\ \text{m}\) uses \(2.5\ \text{m}\) instead of \(0.25\ \text{m}\), \(160\ \text{m}\) is just the ratio \(p/p^{\prime}\), and \(0.94\ \text{mm}\) swaps \(p\) and \(p^{\prime}\).
Spherical mirrors · Step 3
At the exit of a car park, a convex safety mirror with a \(4\ \text{m}\) radius of curvature shows drivers whoever is coming along the pavement. At one moment, a car \(1.6\ \text{m}\) tall is \(6\ \text{m}\) from the mirror.

Using the Gauss equation, with the sign convention we used in the lesson, what is the image of this car like?
1. Real and inverted, \(3\ \text{m}\) in front of the mirror, half the height of the car.
2. Virtual and upright, \(1.5\ \text{m}\) behind the mirror, \(0.4\ \text{m}\) tall.
3. Virtual and upright, \(2.4\ \text{m}\) behind the mirror, \(0.64\ \text{m}\) tall.
4. Virtual and upright, \(6\ \text{m}\) behind the mirror, \(1.6\ \text{m}\) tall.
5. Virtual and upright, \(1.5\ \text{m}\) behind the mirror, \(6.4\ \text{m}\) tall.
Show solution

Answer: B.
In a convex mirror the focal length is negative, \(f = -\dfrac{R}{2} = -2\ \text{m}\).
\(\dfrac{1}{p^{\prime}} = \dfrac{1}{f} - \dfrac{1}{p} = -\dfrac{1}{2} - \dfrac{1}{6}\)
\(\dfrac{1}{p^{\prime}} = -\dfrac{4}{6}\) \(\Rightarrow p^{\prime} = -1.5\ \text{m}\), a virtual image behind the mirror.
\(A = -\dfrac{p^{\prime}}{p} = \dfrac{1.5}{6} = 0.25\)
\(i = A \cdot o = 0.25 \cdot 1.6 = 0.4\ \text{m}\)
The small image also tends to make the driver think the car is further away than it is.
Using \(f = +2\ \text{m}\) leads to option A, mixing up \(f\) and \(R\) leads to C, D treats the convex mirror as a plane one, and E inverts the magnification.

Total internal reflection · Step 5

Binoculars and periscopes often use prisms instead of mirrors, because total internal reflection loses less light than a metal-coated surface. In these prisms, light reaches the glass–air face from inside at \(45^\circ\) to the normal. A team built pieces of this shape from the materials in the table.

Refractive index (yellow light)
Material	\(n\)
Ice	1.31
Water	1.33
Acrylic	1.49
Crown glass	1.52
Diamond	2.42

Taking \(\sin 45^\circ \approx 0.71\) and the refractive index of air as 1, in which pieces does the ray undergo total internal reflection at that face?

Only in the diamond one.
In all of them, because the light always goes from the material into air.
In the acrylic, glass and diamond ones.
Only in the ice and water ones.
In none of them, because total internal reflection requires light to go from air into the material.

Show solution

Answer: C.
Total internal reflection happens when the angle of incidence exceeds the critical angle, that is, when \(\sin 45^\circ > \sin L = \dfrac{1}{n}\).
This requires \(n > \dfrac{1}{0.71} \approx 1.41\).
Working out \(\sin L = \dfrac{1}{n}\), we get ice \(0.76\), water \(0.75\), acrylic \(0.67\), glass \(0.66\) and diamond \(0.41\), and only the last three fall below \(0.71\).
Option A is tempting because diamond is the famous case, B forgets the angle condition, D reverses the inequality, and E gets the direction the light must travel the wrong way round.

The human eye · Step 7
At his eye test, Rafael said he can only read the board from the front row. The test showed that he sees sharply only up to \(40\ \text{cm}\) from his eyes, his far point, and the optician prescribed glasses for distance.

If we treat the lens as sitting right against the eye, what should the optical power of Rafael's glasses be?
1. \(+2.5\ \text{D}\), with a converging lens.
2. \(-0.025\ \text{D}\), with a diverging lens.
3. \(+0.4\ \text{D}\), with a converging lens.
4. \(-4\ \text{D}\), with a diverging lens.
5. \(-2.5\ \text{D}\), with a diverging lens.
Show solution

Answer: E.
Rafael is myopic, and the lens has to bring the image of an object at infinity to his far point, so \(f = -0.40\ \text{m}\).
\(V = \dfrac{1}{f} = \dfrac{1}{-0.40} = -2.5\ \text{D}\)
Option A has the wrong sign and would correct hyperopia, B leaves \(f\) in centimetres, C mixes up \(f\) and \(V\), and D uses the \(25\ \text{cm}\) of the normal near point.
Colour · Step 8
At a theatre rehearsal, an actress wears a yellow blouse and a blue skirt, dyed with ordinary pigments. Under white light, the blouse reflects red and green, while the skirt reflects only blue. In the final scene, the stage is lit by a single red spotlight.

In that scene, how is the audience likely to see the blouse and the skirt?
1. A red blouse and a black skirt.
2. A yellow blouse and a blue skirt.
3. An orange blouse and a violet skirt.
4. A black blouse and a red skirt.
5. A black blouse and a black skirt.
Show solution

Answer: A.
An object can only reflect the light it receives, and under the spotlight only red arrives.
The blouse reflects the red it receives and looks red, while the skirt absorbs red, sends nothing back and looks black.
B assumes colour belongs to the object, C mixes the colours as if they were paint, D swaps the roles, and E forgets that the yellow blouse also reflects red.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, and we can find each question by its year, day, booklet colour and number. It may be worth starting with the 2015 one, which links the pinhole camera to the eye, as we did in Steps 1 and 7.

ENEM 2015, Day 1, blue booklet, question 85. It starts from the pinhole camera described by Alhazen and asks which part of the eye plays the role of the cloth where the inverted image forms.
ENEM 2018, Day 2, blue booklet, question 125. A beam of red, green and blue light passes through a prism, and we have to predict the order of the coloured spots, which we can do from how much each colour is deviated.
ENEM 2019, Day 2, blue booklet, question 132. It discusses why a people from the coast of Thailand see better underwater when they narrow their pupils, linking the effect to refraction inside the eye.
ENEM 2022, Day 2, blue booklet, question 135. It asks which optical phenomenon explains "bottled light", PET bottles of water fixed in a roof to light homes during the day.
ENEM 2024, Day 2, blue booklet, question 100. It deals with a toy made of two concave mirrors that produces a floating image, and asks for the nature of that image and its distance from the object.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Pinhole camera\(\dfrac{o}{i} = \dfrac{p}{p^{\prime}}\)

Reflection\(i = r\)

Plane mirror\(\ell_{\min} = H/2\)

Mirrors at an angle\(N = \dfrac{360^\circ}{\alpha} - 1\)

Curved mirror\(f = R/2\)

Thin lens / mirror equation (Gauss)\(\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{p^{\prime}}\)

Magnification\(A = \dfrac{i}{o} = -\dfrac{p^{\prime}}{p}\)

Refractive index\(n = c/v\)

Snell's law\(n_1\sin\theta_1 = n_2\sin\theta_2\)

Critical angle\(\sin L = n_{\text{lower}}/n_{\text{higher}}\)

Optical power\(V = 1/f\) (D, f in m)

Signsp′ > 0 real · f < 0 convex/diverging