PHY132H1/Unit 8

Unit 8 · Introduction to Physics II

Geometric Optics

When obstacles are much larger than the wavelength, light can be treated as straight rays. That approximation covers mirrors, lenses, eyes and telescopes — everything optical you use, right up to the point where Unit 9 shows why it must eventually fail.

By the end you can

• Apply the laws of reflection and refraction.
• Compute critical angles and explain total internal reflection.
• Use the mirror and lens equations with correct sign conventions.
• Distinguish real from virtual images.

Why it matters & where it comes from

Why it exists

Tracing rays is far simpler than solving wave equations, and it gives the right answer whenever the wavelength is negligible. Knowing when the approximation holds is as important as the approximation itself.

In the real world

Spectacles, cameras, microscopes, telescopes. Optical fibre carries essentially all long-distance internet traffic by total internal reflection, trapping light inside a glass thread for thousands of kilometres.

Story & open questions

Snell found the refraction law in 1621; Ibn Sahl had described it in Baghdad six centuries earlier. Fermat later showed it follows from light taking the path of least time — a reformulation that turned out to anticipate how modern physics states its laws.

Reflection and refraction

Rule Reflection

The angle of reflection equals the angle of incidence, both measured from the normal — the perpendicular to the surface, not the surface itself.

Rule Refraction: Snell's law

$n_1\sin\theta_1 = n_2\sin\theta_2$

The refractive index is $n = c/v$, so a larger $n$ means slower light. Air is about $1.00$, water $1.33$, glass around $1.5$.

Entering a slower medium, the ray bends toward the normal.

Entering a faster medium, it bends away.

Example Why a straw looks bent

Light from the submerged part leaves the water and bends away from the normal.

Your eye traces the rays back in straight lines, so the submerged part appears shallower and displaced.

The same effect makes a pool look shallower than it is, by a factor of about $1.33$ — which is a genuine safety issue.

Rule Total internal reflection

Going from slower to faster, the refracted ray bends away from the normal. Beyond a certain incidence angle it would need to exceed $90$ degrees, which is impossible — so none escapes and all the light reflects.

$\sin\theta_c = \dfrac{n_2}{n_1}$, valid only when $n_1 > n_2$.

For water to air, $\theta_c = \arcsin(1/1.33) = 48.8$ degrees.

This is the only kind of reflection that is perfectly lossless, which is exactly why optical fibre works over long distances.

Mirrors and lenses

Rule One equation for both

$\dfrac{1}{d_o} + \dfrac{1}{d_i} = \dfrac{1}{f}$

Magnification $m = -\dfrac{d_i}{d_o}$. Negative $m$ means inverted.

Sign conventions, which is where the marks are:

$f > 0$ for converging — a concave mirror or convex lens.

$f < 0$ for diverging — a convex mirror or concave lens.

$d_i > 0$ for a real image, formed where light actually converges. $d_i < 0$ for a virtual image.

Definition

A real image is where rays actually meet. It can be caught on a screen. Cameras and projectors form them.

A virtual image is where rays only appear to come from when traced backwards. It cannot be projected. A plane mirror gives one.

Example Converging lens, object outside the focus

An object $30$ cm from a lens of focal length $10$ cm.

$1/30 + 1/d_i = 1/10$, so $1/d_i = 1/10 - 1/30 = 1/15$ and $d_i = 15$ cm.

Positive, so the image is real and on the far side. $m = -15/30 = -0.5$: inverted and half size.

This is a camera. Moving the object closer to the focus pushes the image further away and makes it larger.

Example Same lens, object inside the focus

Now $5$ cm from the same lens.

$1/5 + 1/d_i = 1/10$, so $1/d_i = 1/10 - 1/5 = -1/10$ and $d_i = -10$ cm.

Negative, so the image is virtual, on the same side as the object. $m = -(-10)/5 = +2$: upright and double size.

This is a magnifying glass. The negative sign is not an error — it is the answer telling you the image is virtual.

Take the reciprocal at the end

The commonest arithmetic error is solving for $1/d_i$ and reporting it as $d_i$. Check the sign as well: it tells you real or virtual, which is usually what the question is actually asking.

The eye and corrective lenses

Example Two common defects

Myopia, short sight: the eye focuses in front of the retina, so distant objects blur. Corrected with a diverging lens, which pushes the focus back.

Hyperopia, long sight: the focus falls behind the retina and near objects blur. Corrected with a converging lens.

Prescription strength is in dioptres, $1/f$ with $f$ in metres. A $-2.5$ D prescription is a diverging lens of focal length $-0.4$ m, and the sign alone tells you which defect is being corrected.

Common pitfalls

• Measuring angles from the surface rather than the normal.
• Applying the critical-angle formula going into a denser medium. Total internal reflection only happens dense to sparse.
• Forgetting to invert after solving for $1/d_i$.
• Getting the focal-length sign wrong for a diverging element.
• Treating a negative image distance as a mistake rather than as "virtual".

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Snell

Light hits water at $40$ degrees from the normal. Find the refraction angle.

Answer

$\sin\theta_2 = \sin 40/1.33 = 0.483$, so $\theta_2 = 28.9$ degrees — bent toward the normal, as entering a slower medium requires.

Practice 2 · Critical angle

Find the critical angle for glass of $n = 1.5$ to air.

Answer

$\sin\theta_c = 1/1.5$, so $\theta_c = 41.8$ degrees.

Practice 3 · Which direction

Does light bend toward or away from the normal entering glass from air?

Answer

Toward. Glass is slower, so the ray bends toward the normal.

Practice 4 · Lens equation

An object $20$ cm from a $15$ cm converging lens. Find the image.

Answer

$1/d_i = 1/15 - 1/20 = 1/60$, so $d_i = 60$ cm. Real, inverted, $m = -3$.

Practice 5 · Virtual image

How do you know an image is virtual from the arithmetic?

Answer

$d_i$ comes out negative. The image is on the same side as the object and cannot be projected onto a screen.

Practice 6 · Fibre optics

Why does light stay in an optical fibre?

Solution

The core has a higher refractive index than the cladding, so light striking the boundary beyond the critical angle is totally internally reflected — losslessly, and repeatedly along the whole length.

Practice 7 · Prescription

What does a $+2.0$ D lens do, and which defect does it correct?

Answer

Converging, with $f = 0.5$ m. Positive means converging, which corrects long sight.

Practice 8 · Plane mirror

Where is the image in a flat mirror, and is it real?

Answer

As far behind the mirror as the object is in front, upright and the same size. Virtual — no light actually goes behind the mirror, so it cannot be caught on a screen.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY132H1 topic map. It should cover the following items before you move on:

• Reflection and refraction. Angles from the normal, and Snell's law.
• Total internal reflection. The critical angle and its one-way condition.
• Mirrors and lenses. The equation, the sign conventions, and magnification.
• Images. Real versus virtual, and reading the answer off the sign.

Textbook/source alignment

The calendar names "optics". Check the syllabus on Quercus for assigned problems; ray diagrams and the sign conventions are standard laboratory and exam material.

Core idea map

Light changes speed between media, which bends it. Curved surfaces bend rays to a focus, and one equation with careful signs handles every mirror and lens. Unit 9 takes over where the ray picture fails.

Study routine. Sketch the ray diagram before computing. It tells you whether to expect a real or virtual image, so the sign of your answer becomes a check rather than a surprise.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Measuring angles

From what are optical angles measured?

Guide

The normal, the perpendicular to the surface — never the surface itself.

Extension 2 · TIR condition

When can total internal reflection occur?

Guide

Only going from a higher index to a lower one, at an angle beyond the critical.

Extension 3 · Reading the sign

What does a negative image distance mean?

Guide

The image is virtual and on the same side as the object.

Extension 4 · When rays fail

When does geometric optics break down?

Guide

When apertures or obstacles are comparable to the wavelength. Then diffraction matters and Unit 9's wave treatment is required.

← Unit 07Electromagnetic waves & light Unit 09 →Wave optics