Gr 11 · Optics · Deep Study

Understanding Light

Light travels in straight lines — until it hits something. This guide covers what happens when light bounces off a surface (reflection), bends at a boundary (refraction), gets trapped inside a medium (total internal reflection), and gets focused by mirrors and lenses.

💡 5 sections 🔍 Worked examples with steps ✅ Checkpoint questions 🧠 Intuition-first explanations
1
Reflection
Why angles are always measured from the normal, not the surface

When light hits a surface, it bounces off. The angle it arrives at equals the angle it leaves — both measured from the normal (a line perpendicular to the surface at the point of contact). This is the law of reflection, and it applies to every reflective surface from a mirror to a calm lake.

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Why the normal, not the surface?
Angles in optics are always measured from the normal because the normal is perpendicular to the surface at the exact point of reflection — it gives a consistent reference regardless of how the surface is tilted. If you measured from the surface, a ray coming in at 0° would be grazing along it, which is confusing. Measured from the normal, 0° means straight in, 90° means parallel to the surface.
Law of Reflection
θᵢ = θᵣ
Both angles measured from the normal · Plane mirror image: virtual, upright, same size, same distance behind
θᵢ θᵣ Incident ray Reflected ray Normal Mirror surface
Law of Reflection: θᵢ = θᵣ (angles from normal)
★ Easy
Finding the reflected ray angle
A ray hits a plane mirror at 35° to the surface. What is the angle of reflection?
Show solution
1
Find the angle of incidence (from normal)
Angle to surface = 35° → Angle to normal = 90° − 35° = 55°

The angle of incidence θᵢ = 55°

2
Apply the law of reflection
θᵣ = θᵢ = 55°
Angle of reflection: 55° from the normal
★★ Intermediate
Plane mirror image position
An object is 40 cm in front of a plane mirror. You walk 10 cm toward the mirror. How far are you from the image now?
Show solution
1
Initial setup
Object 40 cm in front → image 40 cm behind mirror Total object-to-image distance = 40 + 40 = 80 cm
2
After walking 10 cm toward mirror
New object distance = 40 − 10 = 30 cm Image is still 30 cm behind mirror (image moves with object) You-to-image = 30 + 30 = 60 cm
Distance from you to image: 60 cm (was 80 cm — both moved closer)
✅ Checkpoint 1
a) A ray hits a mirror at 50° to the normal. What is the angle of reflection?
b) An object is 60 cm from a plane mirror. Where is the image, and what are its characteristics?
c) Why does a mirror reverse left and right, but not up and down?

a) θᵣ = θᵢ = 50° from the normal.

b) Image is 60 cm behind the mirror. It is virtual, upright, same size, and laterally inverted.

c) A mirror doesn't really reverse left-right — it reverses front-to-back (depth). When you raise your right hand, the image raises its right hand too. The apparent left-right reversal happens because you mentally rotate the image to face you.

2
Refraction & Snell's Law
Why light bends when it crosses a boundary

When light passes from one medium into another, its speed changes. If it hits the boundary at an angle, one side of the wavefront slows down before the other — causing the ray to bend. This is refraction. The index of refraction n tells you how much a medium slows light down compared to a vacuum.

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The marching band analogy:
Imagine a row of soldiers marching in step across a field. When they hit a muddy patch at an angle, the soldiers on one end hit the mud first and slow down. The line pivots — bending toward the mud. Light does exactly this at a boundary: entering a denser medium (higher n, slower speed) bends toward the normal.
Index of refraction
n = c / v
Snell's Law
n₁ sinθ₁ = n₂ sinθ₂
Common n values
Air ≈ 1.00 · Water ≈ 1.33 · Glass ≈ 1.50
θ₁ θ₂ Air (n = 1.00) Water (n = 1.33) Incident ray Refracted ray Normal θ₁ > θ₂
Refraction — Light Bends Toward Normal Entering Denser Medium
★ Easy
Air to water refraction
Light travels from air (n=1.00) into water (n=1.33) at an angle of 40° to the normal. Find the refracted angle.
Show solution
1
Write Snell's Law
n₁ sinθ₁ = n₂ sinθ₂ 1.00 × sin 40° = 1.33 × sinθ₂
2
Solve for sinθ₂
sinθ₂ = sin 40° / 1.33 = 0.6428 / 1.33 = 0.4832
3
Find θ₂
θ₂ = arcsin(0.4832) ≈ 28.9°

The ray bends toward the normal (smaller angle) because water is denser.

Refracted angle: 28.9° — bends toward the normal entering water
★★ Intermediate
Glass to air + speed of light
Light travels in glass (n=1.50) and hits an air boundary at 25°. Find the refracted angle and the speed of light in glass.
Show solution
1
Apply Snell's Law (glass → air)
1.50 × sin 25° = 1.00 × sinθ₂ sinθ₂ = 1.50 × 0.4226 = 0.6339 θ₂ = arcsin(0.6339) ≈ 39.3°

Bends away from normal — going into less dense medium.

2
Find speed of light in glass
n = c / v → v = c / n = (3 × 10⁸) / 1.50 = 2.0 × 10⁸ m/s
Refracted angle: 39.3° · Speed in glass: 2.0 × 10⁸ m/s
✅ Checkpoint 2
a) A material slows light to 2.4 × 10⁸ m/s. What is its index of refraction?
b) Light travels from glass (n=1.60) to water (n=1.33) at θ₁=30°. Find θ₂.
c) Light enters water from air. Does it bend toward or away from the normal? Why?

a) n = c/v = (3 × 10⁸) / (2.4 × 10⁸) = 1.25

b) 1.60 × sin30° = 1.33 × sinθ₂ → sinθ₂ = (1.60 × 0.5)/1.33 = 0.601 → θ₂ = 36.9° (bends away from normal, going less dense)

c) Toward the normal. Water has higher n than air, so light slows down. The side of the wavefront hitting water first slows, bending the ray toward the normal.

3
Total Internal Reflection
When light can't escape — and how fibre optics exploit this

When light tries to leave a dense medium at a steep angle, Snell's Law would require sinθ₂ > 1 — which is impossible. Instead, all the light reflects back inside. This is total internal reflection (TIR), and it only occurs going from a denser medium to a less dense one.

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Critical angle: the smallest angle of incidence (measured from the normal) at which TIR occurs.
sin θc = n₂ / n₁  (where n₁ > n₂)
Any angle greater than θc → total internal reflection. Any angle less → light partially escapes (with partial reflection).
Critical Angle Formula
sin θc = n₂ / n₁
Only valid when going from dense (n₁) to less dense (n₂) · TIR occurs when θ > θc
Air (less dense) Glass (denser) θ < θc exits (refracted) θ = θc along surface θ > θc (TIR) no exit
Total Internal Reflection — Three Cases
★ Easy
Critical angle for glass-air
Find the critical angle for light going from glass (n=1.50) into air (n=1.00).
Show solution
1
Apply the critical angle formula
sin θc = n₂ / n₁ = 1.00 / 1.50 = 0.6667
2
Find θc
θc = arcsin(0.6667) ≈ 41.8°

Any ray hitting the glass-air boundary at more than 41.8° from the normal reflects entirely back into the glass.

Critical angle: 41.8°
★★ Intermediate
Optical fibre and diamond
An optical fibre has core n=1.55, cladding n=1.45. Find θc. Also find θc for diamond (n=2.42) in air.
Show solution
1
Optical fibre critical angle
sin θc = 1.45 / 1.55 = 0.9355 θc = arcsin(0.9355) ≈ 69.3°

Light stays trapped in the fibre core as long as it strikes the core-cladding wall at an angle greater than 69.3° from the normal.

2
Diamond critical angle
sin θc = 1.00 / 2.42 = 0.4132 θc = arcsin(0.4132) ≈ 24.4°

Very small critical angle — almost any ray inside a diamond undergoes TIR. This is why diamonds sparkle: light bounces around inside many times before escaping.

Fibre θc: 69.3° · Diamond θc: 24.4°
✅ Checkpoint 3
a) Find the critical angle for water (n=1.33) to air.
b) Can TIR occur when light goes from air into glass? Why or why not?
c) Why must an optical fibre's cladding have a lower index of refraction than the core?

a) sinθc = 1.00/1.33 = 0.752 → θc = 48.8°

b) No. TIR only occurs going from a denser medium to a less dense one. Air (n≈1.00) is less dense than glass (n≈1.50), so light going air→glass bends toward the normal and always passes through — it never hits the critical angle condition.

c) TIR requires going from high n to low n. If cladding had equal or higher n than the core, light would refract out of the core instead of reflecting, and the fibre would lose signal.

4
Mirrors
Concave vs convex — and reading the mirror equation

Curved mirrors focus or spread reflected rays. A concave (converging) mirror curves inward and can form real images. A convex (diverging) mirror curves outward and always forms virtual, upright, smaller images. Both obey the same mirror equation.

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Mirror equation and magnification:
1/f = 1/dₒ + 1/dᵢ    m = −dᵢ/dₒ
Sign conventions: concave f > 0, convex f < 0 · real image dᵢ > 0 (in front of mirror) · virtual image dᵢ < 0 (behind mirror) · m > 0 means upright, m < 0 means inverted.
Mirror equation
1/f = 1/dₒ + 1/dᵢ
Magnification
m = −dᵢ / dₒ = hᵢ / hₒ
Radius of curvature
R = 2f
C F Object real, inverted, reduced axis
Concave Mirror — Ray Diagram (object beyond C)
★ Easy
Concave mirror — real image
A concave mirror has f=15 cm. An object is placed 45 cm in front. Find dᵢ and m.
Show solution
1
Apply mirror equation
1/dᵢ = 1/f − 1/dₒ = 1/15 − 1/45 = 3/45 − 1/45 = 2/45 dᵢ = 45/2 = 22.5 cm

Positive dᵢ → real image, in front of the mirror.

2
Calculate magnification
m = −dᵢ/dₒ = −22.5/45 = −0.5

Negative m → inverted. |m| = 0.5 → image is half the size of the object.

dᵢ = 22.5 cm (real, in front) · m = −0.5 (inverted, half-size)
★★ Intermediate
Convex mirror — car side mirror
A convex mirror has f=−20 cm. Object is 30 cm in front. Find dᵢ and m, and describe the image.
Show solution
1
Apply mirror equation (f is negative for convex)
1/dᵢ = 1/f − 1/dₒ = 1/(−20) − 1/30 = −3/60 − 2/60 = −5/60 dᵢ = −12 cm

Negative dᵢ → virtual image, behind the mirror.

2
Calculate magnification
m = −dᵢ/dₒ = −(−12)/30 = +0.4

Positive m → upright. |m| = 0.4 → image is 40% the size of the object. This is the typical car side-mirror result: smaller, upright, wider field of view.

dᵢ = −12 cm (virtual, behind mirror) · m = +0.4 (upright, reduced)
✅ Checkpoint 4
a) Concave mirror f=10 cm, object at 40 cm. Find dᵢ and m.
b) Concave mirror f=10 cm, object at 5 cm (inside focal point). Find dᵢ — what type of image?
c) Why does a convex mirror always produce a smaller, upright, virtual image?

a) 1/dᵢ = 1/10 − 1/40 = 4/40 − 1/40 = 3/40 → dᵢ = 13.3 cm (real). m = −13.3/40 = −0.33 (inverted, reduced).

b) 1/dᵢ = 1/10 − 1/5 = 1/10 − 2/10 = −1/10 → dᵢ = −10 cm (virtual, behind mirror). Image is upright and magnified — this is how a shaving/makeup mirror works.

c) For a convex mirror, f is negative. The mirror equation always gives dᵢ between −f and 0 (virtual, behind), and m between 0 and +1 (upright, smaller). The diverging nature of the mirror means reflected rays never converge to form a real image.

5
Lenses
Same equation as mirrors — different sign conventions for real images

Lenses refract light to form images. A converging (convex) lens brings parallel rays to a focal point on the far side; a diverging (concave) lens spreads them out. The thin lens equation is identical in form to the mirror equation — but for lenses, real images form on the opposite side from the object (dᵢ > 0).

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Mirror vs Lens — same equation, different geometry:
Mirror: real image in front (same side as object), dᵢ > 0.
Lens: real image behind (opposite side from object), dᵢ > 0.
Virtual images: behind mirror / same side as object for lens — both give dᵢ < 0.
Thin lens equation
1/f = 1/dₒ + 1/dᵢ
Magnification
m = −dᵢ / dₒ
Sign conventions
Converging: f > 0 · Diverging: f < 0
F F′ Object real, inverted Converging Lens axis
Converging Lens — Ray Diagram
★ Easy
Converging lens — real image
A converging lens has f=20 cm. Object at dₒ=60 cm. Find dᵢ and m.
Show solution
1
Apply thin lens equation
1/dᵢ = 1/f − 1/dₒ = 1/20 − 1/60 = 3/60 − 1/60 = 2/60 dᵢ = 30 cm

Positive → real image on the far side of the lens.

2
Magnification
m = −dᵢ/dₒ = −30/60 = −0.5

Inverted, half-size real image.

dᵢ = 30 cm (real) · m = −0.5 (inverted, half-size)
★★ Intermediate
Diverging lens — eyeglasses
A diverging lens has f=−25 cm. Object at dₒ=40 cm. Find dᵢ and describe the image.
Show solution
1
Apply thin lens equation (f negative)
1/dᵢ = 1/(−25) − 1/40 = −8/200 − 5/200 = −13/200 dᵢ = −200/13 ≈ −15.4 cm

Negative → virtual image on the same side as the object.

2
Magnification
m = −dᵢ/dₒ = −(−15.4)/40 = +0.385

Upright, reduced. This is how diverging lenses correct nearsightedness — they form a virtual image closer than the actual object, within the eye's focus range.

dᵢ = −15.4 cm (virtual, same side as object) · m = +0.385 (upright, reduced)
✅ Checkpoint 5
a) Converging lens f=12 cm, object at 36 cm. Find dᵢ and m.
b) Converging lens, object placed exactly at f. Where does the image form, and why?
c) What is the practical difference between a real image and a virtual image — which one can be projected onto a screen?

a) 1/dᵢ = 1/12 − 1/36 = 3/36 − 1/36 = 2/36 → dᵢ = 18 cm. m = −18/36 = −0.5 (inverted, half-size).

b) When dₒ = f: 1/dᵢ = 1/f − 1/f = 0 → dᵢ = . The refracted rays are parallel — the image forms at infinity. This is how a projector lens works (object just inside f → image at a finite but very large distance).

c) A real image is formed by actual converging rays — it can be projected onto a screen (cinema projector, camera sensor). A virtual image is formed by diverging rays that only appear to come from a point — it cannot be projected, only seen by looking through the lens or mirror (magnifying glass, plane mirror).