Waves and optics

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

When does a propagating refracted ray cease to exist?

Move forward one step at a time. Change the values and observe what happens.

A ray reaches the boundaryn₁ = 1,5n₂ = 1θ₁ = 35°
Drawing description

The yellow ray is incident at 35 degrees to the normal, coming from medium n₁=1,5 to the interface with n₂=1.

Step 1 of 6

Measure from the perpendicular

Compare θ₁=0° and θ₁=60° with the vertical dashed line.

Light arrives from the upper medium, with index n₁. Both media are transparent and nonmagnetic. Angle θ₁ is between the incident ray and the vertical normal, not between the ray and horizontal boundary.

Incidence
35°
Indices
1,5 → 1

What if you change this?

0 °80 °
11,8
11,8

Before moving on

For n₁=1,5 and n₂=1, what is the approximate critical angle?

Compare with the explanation

θc≈41,81°. At incidence 50°, total reflection occurs.

You decide when to mark this lesson as studied.

A ray reaches the boundary

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Relate speed, frequency and wavelength.
  • Apply reflection, Snell’s law and total internal reflection with the correct angles.
  • Analyse thin lenses and interference while respecting assumptions and signs.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Review a wave and choose a description

A periodic wave has frequency f, period T=1/f and wavelength λ. Propagation speed satisfies v=λf. Frequency counts oscillations per second, in hertz; λ is the distance between in-phase points and is measured in metres. Propagation speed is not a medium particle’s velocity. In vacuum, light propagates at c, approximately 3,00×10⁸ m/s.

Rays represent the local direction of energy propagation in a geometric approximation. They are useful for mirrors and lenses when relevant apertures and distances allow diffraction effects to be neglected. They are not material threads and do not describe the field’s entire structure. Passing through an aperture comparable to λ spreads propagation, and the description using isolated straight paths can fail. No model should be used outside the scale justifying its approximations.

Reflection and refraction

In specular reflection from a smooth surface, the incident ray, normal and reflected ray lie in the same plane; incidence and reflection angles measured from the normal are equal. A rough surface may distribute reflected directions, although each small region obeys the local relationship. The normal is the straight line perpendicular to the surface at the meeting point.

In refraction, speed changes between media. Refractive index is n=c/v for the model considered. Snell’s law is n₁ sinθ₁=n₂ sinθ₂. If n₂>n₁, the refracted ray moves towards the normal. Frequency remains determined by the source across a stationary boundary; wavelength changes, λ=v/f. Thus a lower speed in the second medium implies lower λ, not necessarily lower frequency. Indices may depend on frequency, explaining dispersion.

Total internal reflection: two conditions

For total internal reflection, light must start in a higher-index medium and enter a lower-index one: n₁>n₂. As θ₁ increases, the refracted angle reaches 90°. At this limit, Snell gives sinθc=n₂/n₁. Angle θc is the critical angle. For incidence greater than θc, there is no propagating refracted ray in the second medium in the ideal geometric model.

It is not enough to say the angle is large: the index order must be checked and the angle compared with θc. Going from air into ordinary glass, n₂/n₁ would exceed 1 and no real critical angle exists. Optical fibres use an index structure to guide light by internal reflection. The full explanation of fields near the boundary is wave-based; rays provide a useful approximation for introducing the mechanism.

Thin lenses and a sign convention

A converging lens brings approximately axis-parallel rays together at a focus; a diverging lens makes rays spread as if coming from a focus. The thin-lens equation is 1/f=1/dₒ+1/dᵢ in the paraxial approximation: a thin lens and rays at small angles to the axis. It is not an exact description of every real lens or large aperture.

We use f>0 for a converging lens and f<0 for a diverging one; dₒ>0 for a real object on the entrance side; dᵢ>0 for a real image on the exit side and dᵢ<0 for a virtual image on the object side. Transverse magnification is m=hᵢ/hₒ=−dᵢ/dₒ. Negative m indicates inversion; |m| indicates the size ratio. A real image can be captured on a screen placed at its position; a virtual one is located by extending rays.

Rays and equations must agree

For a converging lens, a ray parallel to the axis leaves towards the focus on the other side; a ray near the optical centre is approximated as undeviated in the thin-lens model. Their intersection locates the image. If they do not meet on the exit side, extend them backwards to locate a virtual image. The drawing helps interpret the sign found in the equation.

When the object is between the focus and a converging lens, the image is virtual, upright and enlarged. At the focus, outgoing rays are parallel and no image forms at a finite distance in the model. For an object beyond the focus, the image is real and inverted. These are not rules to memorise without geometry: they arise as the meeting of rays changes. Use all distances in the same unit and translate the final sign into a location.

Superposition, interference and diffraction

In the linear wave description, disturbances superpose. Two coherent waves with equal frequency and stable relative phase can reinforce or cancel locally. For two slits illuminated in phase by monochromatic light, maxima satisfy d sinθ=mλ and minima d sinθ=(m+1/2)λ, with integer m and slit separation d. Path difference, not merely total distance, determines the result.

On a distant screen L≫d and at small angles, neighbouring maxima are separated by approximately Δy=λL/d. Each slit’s diffraction influences the pattern’s overall intensity; two real slits do not produce equally intense maxima everywhere. Destructive interference at one point does not destroy energy globally: the spatial distribution changes, with reinforcement elsewhere. A phenomenon requiring waves does not invalidate rays at another scale; it shows we need the description appropriate to the question.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Refraction with frequency preserved

Light of 600 nm in air enters glass of index 1,50 at 30° incidence. Approximate air’s n as 1. Determine angle and wavelength in glass.

  1. Snell gives sinθ₂=(1/1,50)sin30°=1/3. Therefore θ₂≈19,47°.
  2. Frequency in air is approximately c/λ=3,00×10⁸/(600×10⁻⁹)=5,00×10¹⁴ Hz.
  3. In glass, v=c/1,50 and λ₂=λ₁/1,50=400 nm, retaining f.
Result

Refracted angle 19,47° and wavelength 400 nm.

Check the result

The ray bends towards the normal because it enters a higher index. Frequency does not change at the stationary boundary.

Example 2

A real image

A 3 cm object is 30 cm from a converging lens with f=10 cm. Locate and characterise the image.

  1. 1/dᵢ=1/10−1/30=1/15, so dᵢ=15 cm.
  2. Magnification is m=−15/30=−0,5.
  3. Signed height is hᵢ=−0,5·3=−1,5 cm. The negative sign indicates inversion.
Result

Real image, 15 cm on the exit side, inverted and 1,5 cm in size.

Check the result

1/10=1/30+1/15 confirms the equation. Since the object is beyond 2f, the reduced image lies between f and 2f.

Example 3

Fringe spacing

Two slits separated by 0,30 mm receive light of λ=600 nm. A screen is 2,0 m away. Estimate neighbouring-maxima spacing.

  1. Convert d=3,0×10⁻⁴ m and λ=6,0×10⁻⁷ m.
  2. Use Δy≈λL/d=(6,0×10⁻⁷·2,0)/(3,0×10⁻⁴).
  3. The result is 4,0×10⁻³ m. The ratio λ/d=0,002 is small, consistent with the small-angle approximation near the centre.
Result

Spacing approximately 4,0 mm.

Check the result

Doubling screen distance would double spacing; doubling slit separation would halve it.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

A wave enters a medium where speed halves. Does frequency change?
A hint
The source maintains frequency at a stationary boundary.
See the worked solution
Frequency stays the same; since λ=v/f, wavelength halves. Confusing temporal frequency with spatial spacing is the mistake to avoid.

Exercise 2

What is the critical angle from glass n=1,50 to air n=1,00?
A hint
Use lower index divided by higher index.
See the worked solution
θc=arcsin(1/1,50)≈41,81°. Total internal reflection requires incidence above this value and light leaving glass for air.

Exercise 3

Does the same critical angle exist when entering glass from air?
A hint
Check which medium has the greater index.
See the worked solution
No. The ray starts in the lower index; sinθc=1,50 would be impossible for a real angle. Partial reflection may occur, but not total internal reflection in this direction.

Exercise 4

A diverging lens has f=−12 cm and a real object at 24 cm. Calculate dᵢ.
A hint
Keep the negative focal-length sign.
See the worked solution
1/dᵢ=−1/12−1/24=−1/8, so dᵢ=−8 cm. The image is virtual on the object side; m=−(−8)/24=1/3, upright and reduced.

Exercise 5

For two in-phase slits, path difference is 3λ/2. Is there a maximum?
A hint
Compare integer and half-integer multiples.
See the worked solution
It is a half-integer multiple of λ, so corresponds to a minimum in the ideal description of two equal-amplitude waves. It is not a maximum; contributions arrive out of phase by half a cycle.

Mistakes worth avoiding

Snell angles are measured to the surface.
They are measured to the normal.
A virtual image corresponds to a meaningless distance.
The negative sign locates an image formed by extensions of rays.
Destructive interference removes energy from the system.
Energy is redistributed spatially; the global analysis includes regions of reinforcement.

The essentials to take with you

  • Check the scale when choosing rays or waves.
  • Refraction preserves the source frequency.
  • Total internal reflection requires higher initial index and angle above critical.
  • State the lens sign convention.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.