Reflection and Refraction

Master laws of reflection, refraction, mirror formula, and lens formula for JEE Physics

12 min read

Prerequisites

Before studying this topic, make sure you understand:

The Hook: Spider-Man and Mirror Tricks

Connect: Real Life → Physics

Remember in Spider-Man: No Way Home when Doctor Strange uses mirror dimension magic and Peter sees infinite reflections of himself? Or when he tricks enemies using reflections in glass buildings? That’s the power of understanding reflection and refraction!

Ever wondered why a straw in water looks bent? Why mirrors in cars say “Objects are closer than they appear”? Why eyeglasses work? All answered by two simple laws!

JEE Weightage: HIGH — Expect 2-3 questions in JEE Main, 1-2 in Advanced

The Core Concept

Light behaves predictably when it hits boundaries between different media. Two fundamental phenomena occur:

  1. Reflection - Light bounces back into the same medium
  2. Refraction - Light bends as it enters a different medium

Law of Reflection

The Two Laws

Law 1: The incident ray, reflected ray, and normal all lie in the same plane.

Law 2: Angle of incidence = Angle of reflection

$$\boxed{i = r}$$

In simple terms: Light bounces off a mirror like a ball bounces off a wall — the angle going in equals the angle coming out!

Laws of Refraction (Snell’s Law)

When light travels from medium 1 (refractive index $n_1$) to medium 2 (refractive index $n_2$):

$$\boxed{n_1 \sin i = n_2 \sin r}$$

Alternative forms:

  • $\frac{\sin i}{\sin r} = \frac{n_2}{n_1} = \frac{v_1}{v_2}$ (ratio of speeds)
  • For air to medium: $n = \frac{\sin i}{\sin r}$ (absolute refractive index)

In simple terms: Light bends toward the normal when entering a denser medium (like going from air to water), and away from the normal when entering a rarer medium.

Interstellar Connection
In Interstellar, light bends around the black hole (gravitational lensing). Similarly, light bends when changing medium — though here it’s due to speed change, not gravity!

Interactive Visualization

See how light behaves at different angles:

Reflection by Spherical Mirrors

Types of Mirrors

Mirror TypeShapeRay Convergence
ConcaveCurves inward (like a cave)Converging
ConvexCurves outward (like a dome)Diverging

Mirror Terminology

Key Terms
  • Pole (P): Center of mirror surface
  • Center of Curvature (C): Center of the sphere
  • Radius of Curvature (R): Distance PC
  • Principal Axis: Line through P and C
  • Focus (F): Point where parallel rays converge (or appear to diverge from)
  • Focal Length (f): Distance PF, where $f = \frac{R}{2}$

Sign Convention (New Cartesian)

Most Important — Learn This!

Measure ALL distances from the POLE:

  1. Object side (left) is negative, image side (right) is positive
  2. Above principal axis is positive, below is negative
  3. For mirrors:
    • Concave mirror: $f$ is negative
    • Convex mirror: $f$ is positive

Mirror Formula

$$\boxed{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}$$

Where:

  • $f$ = focal length
  • $v$ = image distance from pole
  • $u$ = object distance from pole

Memory Trick: “Finally Vishal Understood” → $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Magnification

$$\boxed{m = \frac{h_i}{h_o} = -\frac{v}{u}}$$
  • $m > 0$: Erect image (virtual)
  • $m < 0$: Inverted image (real)
  • $|m| > 1$: Magnified
  • $|m| < 1$: Diminished

Refraction by Spherical Surfaces

For refraction at a single spherical surface (radius R) separating media with refractive indices $n_1$ and $n_2$:

$$\boxed{\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}}$$

Sign Convention: Same as mirrors (measure from pole)

Special Cases:

  • Plane surface: $R = \infty$, so $\frac{v}{u} = \frac{n_1}{n_2}$
  • Object in denser medium: $v$ appears closer
Swimming Pool Effect

This is why swimming pools look shallower than they really are! The apparent depth is:

$$d_{apparent} = \frac{d_{real}}{n}$$

For water ($n = 1.33$), a 4-meter pool appears only 3 meters deep!

Refraction by Lenses

Lens Formula

$$\boxed{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}$$

Notice the minus sign! Different from mirror formula.

Magnification

$$\boxed{m = \frac{h_i}{h_o} = \frac{v}{u}}$$

(No minus sign in $\frac{v}{u}$ for lenses)

Sign Convention for Lenses

  • Object always on left (incident light side)
  • For convex (converging) lens: $f$ is positive
  • For concave (diverging) lens: $f$ is negative

Ray Diagrams - Key Rays

For any lens, draw these three rays:

  1. Parallel ray → passes through focus (or appears to come from focus)
  2. Central ray → passes through optical center undeviated
  3. Focal ray → passing through focus emerges parallel

Where all three meet (or appear to meet) is where the image forms!

Memory Tricks & Patterns

Mnemonic for Mirror Formula

Memory Trick: “Friends Visit United” → $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Mnemonic for Lens Formula

Memory Trick: “Finally Virat Undermined” (notice the MINUS) → $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Pattern Recognition

Power Law Pattern

All formulas involve reciprocals (powers of -1):

  • $\frac{1}{f}$, $\frac{1}{v}$, $\frac{1}{u}$ → Think in terms of “optical power”
  • Closer objects → larger $\frac{1}{u}$ → larger $\frac{1}{v}$ → closer image

Quick Decision Tree

Light hitting a boundary?
├─ Bounces back? → Use Law of Reflection (i = r)
└─ Passes through? → Use Snell's Law (n₁sin i = n₂sin r)

Curved surface?
├─ Mirror? → Use mirror formula (1/f = 1/v + 1/u)
└─ Lens? → Use lens formula (1/f = 1/v - 1/u)

When to Use This

Decision Tree

Use Reflection when:

  • Light bounces off a surface
  • Mirrors are involved
  • Asked about virtual images behind mirrors

Use Refraction when:

  • Light enters a different medium
  • Lenses, prisms, or water/glass interfaces
  • Bent rays or apparent positions

Use Ray Diagrams when:

  • Asked about image characteristics
  • Quick qualitative answer needed
  • Checking your calculation

Common Mistakes to Avoid

Trap #1: Sign Convention Confusion

Wrong: Treating all distances as positive

Correct:

  • For mirrors: object always on left (negative $u$)
  • For lenses: object on left, but check focal length sign
  • Measure everything from the pole/optical center

Exam Tip: Draw a quick diagram and mark signs BEFORE substituting!

Trap #2: Mirror vs Lens Formula Mix-up

Wrong: Using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ for lenses

Correct:

  • Mirrors: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ (PLUS)
  • Lenses: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (MINUS)

Quick Check: If it bends light (lens) → use minus; if it bounces light (mirror) → use plus

Trap #3: Magnification Sign

Wrong: Ignoring the sign of magnification

Correct:

  • Negative m → inverted image
  • Positive m → erect image
  • This tells you if image is real or virtual!
Trap #4: Real vs Virtual Image

Common Error: Thinking virtual images can’t be seen

Truth:

  • Real image: Light actually converges (can be caught on screen)
  • Virtual image: Light appears to come from a point (seen in mirror/lens, can’t be projected)
  • Both are visible to the eye!

Practice Problems

Level 1: Foundation (NCERT-type)

Problem 1: Basic Reflection

A ray of light is incident on a plane mirror at an angle of 30° with the mirror surface. What is the angle of reflection?

Solution: Angle with mirror = 30° Angle with normal = 90° - 30° = 60°

By law of reflection: $i = r = 60°$

Answer: 60° with the normal

Problem 2: Snell's Law

Light travels from air (n = 1) into glass (n = 1.5) at an incident angle of 30°. Find the angle of refraction.

Solution: Using Snell’s law: $n_1 \sin i = n_2 \sin r$

$1 \times \sin 30° = 1.5 \times \sin r$ $0.5 = 1.5 \sin r$ $\sin r = \frac{0.5}{1.5} = \frac{1}{3} = 0.333$ $r = \sin^{-1}(0.333) \approx 19.5°$

Answer: 19.5°

Problem 3: Mirror Formula

An object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the image position and magnification.

Solution: Given: $u = -20$ cm (object on left), $f = -15$ cm (concave)

Using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:

$$\frac{1}{-15} = \frac{1}{v} + \frac{1}{-20}$$ $$\frac{1}{v} = \frac{1}{-15} + \frac{1}{20} = \frac{-4 + 3}{60} = \frac{-1}{60}$$ $$v = -60 \text{ cm}$$

Magnification: $m = -\frac{v}{u} = -\frac{(-60)}{(-20)} = -3$

Answer: Image at 60 cm in front of mirror (same side as object), magnified 3 times, inverted (real image)

Level 2: JEE Main

Problem 4: Apparent Depth

A coin lies at the bottom of a water tank 1.5 m deep (n = 4/3). An observer looks from above. What is the apparent depth?

Solution:

$$d_{apparent} = \frac{d_{real}}{n} = \frac{1.5}{4/3} = 1.5 \times \frac{3}{4} = 1.125 \text{ m}$$

Answer: 1.125 m (appears 37.5 cm shallower!)

Problem 5: Two Mirror System

Two plane mirrors are placed at 90° to each other. A ray strikes one mirror at 30° to the normal. After reflecting from both mirrors, what angle does the final ray make with the incident ray?

Solution: For two mirrors at angle θ, the ray deviates by 2θ regardless of incident angle.

Here θ = 90°, so deviation = 2 × 90° = 180°

Answer: 180° (ray reverses direction — principle used in retroreflectors)

Problem 6: Lens Formula

A convex lens of focal length 20 cm forms an image twice the size of the object. Find the object and image distances.

Solution: Given: $f = +20$ cm, $m = \pm 2$ (two cases)

Using $m = \frac{v}{u}$:

Case 1: $m = -2$ (inverted, real image) $v = -2u$

Using $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:

$$\frac{1}{20} = \frac{1}{-2u} - \frac{1}{u} = \frac{-1 - 2}{2u} = \frac{-3}{2u}$$ $$2u = -60 \implies u = -30 \text{ cm}$$ $$v = -2(-30) = 60 \text{ cm}$$

Case 2: $m = +2$ (erect, virtual image) $v = 2u$

$$\frac{1}{20} = \frac{1}{2u} - \frac{1}{u} = \frac{1 - 2}{2u} = \frac{-1}{2u}$$ $$2u = -20 \implies u = -10 \text{ cm}$$ $$v = 2(-10) = -20 \text{ cm}$$

Answer:

  • Real image: $u = -30$ cm, $v = +60$ cm
  • Virtual image: $u = -10$ cm, $v = -20$ cm

Level 3: JEE Advanced

Problem 7: Spherical Surface Refraction

A point object is placed in air 20 cm from the curved surface of a glass sphere (n = 1.5, R = 10 cm). Find the image position.

Solution: For refraction at spherical surface:

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

Here: $n_1 = 1$ (air), $n_2 = 1.5$ (glass), $u = -20$ cm, $R = -10$ cm (center on object side)

$$\frac{1.5}{v} - \frac{1}{-20} = \frac{1.5 - 1}{-10}$$ $$\frac{1.5}{v} + \frac{1}{20} = \frac{-0.5}{10} = -0.05$$ $$\frac{1.5}{v} = -0.05 - 0.05 = -0.1$$ $$v = \frac{1.5}{-0.1} = -15 \text{ cm}$$

Answer: Virtual image 15 cm from surface on same side as object

Problem 8: Critical Angle Application

A light ray traveling in glass (n = 1.5) strikes the glass-air boundary. At what minimum angle will total internal reflection occur?

Solution: At critical angle: $n_1 \sin \theta_c = n_2 \sin 90°$

$$1.5 \times \sin \theta_c = 1 \times 1$$ $$\sin \theta_c = \frac{1}{1.5} = \frac{2}{3}$$ $$\theta_c = \sin^{-1}(0.667) \approx 41.8°$$

Answer: 41.8° (any angle > this gives total internal reflection — principle of fiber optics!)

Problem 9: Compound System

An object is placed 15 cm from a concave mirror (f = 10 cm). A convex lens (f = 10 cm) is placed 30 cm from the mirror. Find the final image position.

Solution: Step 1: Image by mirror

$$\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}$$ $$\frac{1}{-10} = \frac{1}{v_1} + \frac{1}{-15}$$ $$\frac{1}{v_1} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = \frac{-1}{30}$$ $$v_1 = -30 \text{ cm}$$

Image 1 forms 30 cm in front of mirror (at lens position!)

Step 2: This becomes object for lens Object at lens, so $u_2 = 0$ (special case — object at optical center)

When object is AT the lens, it emerges undeviated at the same position.

Alternative: If we consider $u_2 = 0^-$ (infinitesimally on left), then the image is at the same point.

Answer: Final image at the lens position (30 cm from mirror)

JEE Trick: When an image from first element coincides with the second element, check if it’s at a special point (focus, center, etc.)

Quick Revision Box

SituationFormula/Approach
Plane mirror$v = -u$, $m = 1$, virtual erect image
Concave mirror (object beyond C)Real, inverted, diminished
Concave mirror (object between F and C)Real, inverted, magnified
Concave mirror (object at F)Image at infinity
Concave mirror (object within F)Virtual, erect, magnified
Convex mirrorAlways virtual, erect, diminished
Convex lens (object beyond 2F)Real, inverted, diminished
Convex lens (object at 2F)Real, inverted, same size
Convex lens (object between F and 2F)Real, inverted, magnified
Convex lens (object within F)Virtual, erect, magnified
Concave lensAlways virtual, erect, diminished
Light entering denser mediumBends toward normal
Light entering rarer mediumBends away from normal
Apparent depth in water$d_{app} = \frac{d_{real}}{n}$

JEE Exam Strategy

Time-Saving Tips
  1. Draw diagrams: Even rough sketches prevent sign errors (saves 2 minutes)
  2. Check special cases: Object at focus? At center? Lens/mirror at zero position?
  3. Use symmetry: For two lenses/mirrors, check if you can use symmetry instead of calculating twice
  4. Approximations: For $\sin \theta$ when $\theta$ is small: $\sin \theta \approx \theta$ (in radians)
  5. Ratio method: Sometimes $\frac{v_1}{v_2}$ is asked, not individual values — saves calculation!

Common JEE Patterns:

  • Combination of lens + mirror (always 2 steps)
  • Apparent depth with multiple layers
  • Finding where to place object for desired magnification
  • Critical angle and total internal reflection applications

Within Optics

Connected Chapters

Math Connections

Teacher’s Summary

Key Takeaways
  1. Master sign conventions FIRST — 80% of errors come from wrong signs
  2. Mirror formula has +, lens formula has − between $\frac{1}{v}$ and $\frac{1}{u}$
  3. Always draw a diagram — it’s faster than you think and prevents mistakes
  4. Ray diagrams give qualitative answers instantly — use for checking calculations
  5. Real images can be projected; virtual images cannot (but both are visible!)
  6. Snell’s law works both ways — light bends toward normal in denser medium

“In optics, the right sign makes all the difference. Draw first, substitute second, and you’ll never go wrong!”

JEE Success Mantra: Practice 10 problems each on mirrors and lenses with ray diagrams — after that, these become the easiest marks in Physics!