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Tamil Nadu Class 12 Physics — Optics & Wave Optics Notes, Formulae & Important Questions

Complete Samacheer Kalvi Class 12 Physics Optics guide with refraction, TIR, lenses, Young's double slit, diffraction, polarisation, worked numericals, and board Q&A.

Updated 22 May 202614 min read

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1. Refraction of Light

Snell's Law

When light travels from medium 1 (refractive index n₁) to medium 2 (refractive index n₂):

n₁ sinθ₁ = n₂ sinθ₂

where θ₁ is the angle of incidence and θ₂ is the angle of refraction, both measured from the normal.

Key points:

  • When light goes from rarer to denser (n₁ < n₂): it bends towards the normal (θ₂ < θ₁).
  • When light goes from denser to rarer (n₁ > n₂): it bends away from the normal (θ₂ > θ₁).
  • The refractive index n = c/v where c is the speed of light in vacuum and v is the speed in the medium.

Absolute and Relative Refractive Index

  • Absolute refractive index of a medium: n = c/v
  • Relative refractive index: ₁n₂ = n₂/n₁ = v₁/v₂ = sinθ₁/sinθ₂

Cauchy's relation (dispersion): n = A + B/λ² — refractive index decreases as wavelength increases, which is why violet light bends more than red in a prism.


2. Total Internal Reflection (TIR)

Critical Angle

When light travels from a denser medium (n₁) to a rarer medium (n₂), and the angle of incidence reaches the critical angle C, the refracted ray grazes the interface (θ₂ = 90°).

From Snell's law at the critical angle:

n₁ sinC = n₂ sin90° = n₂ ∴ sinC = n₂/n₁

If n₂ = 1 (air): sinC = 1/n₁

For glass (n = 1.5): sinC = 1/1.5 → C ≈ 41.8° For water (n = 1.33): sinC = 1/1.33 → C ≈ 48.8° For diamond (n = 2.42): sinC = 1/2.42 → C ≈ 24.4° (sparkle effect!)

Applications of TIR

ApplicationExplanation
Optical fibreLight travels through core (denser) by repeated TIR at core-cladding boundary
Prism binocularsRight-angle prisms reflect light by TIR — no silvering needed
Mirage (hot road)Layers of hot air near road are rarer; light from sky undergoes TIR
EndoscopeFlexible optical fibre transmits images by TIR
Sparkling of diamondVery small critical angle (24°) — most light undergoes TIR inside

3. Refraction Through Lenses

Lens Maker's Equation

For a thin lens made of material with refractive index n in air:

1/f = (n − 1)(1/R₁ − 1/R₂)

Sign convention (Cartesian):

  • Distances measured from optical centre O.
  • Distances in the direction of incident light: positive.
  • Distances opposite to incident light: negative.
  • For a convex lens: R₁ > 0, R₂ < 0 → f > 0 (converging)
  • For a concave lens: R₁ < 0, R₂ > 0 → f < 0 (diverging)

Thin Lens Formula

1/v − 1/u = 1/f

Magnification: m = v/u = height of image / height of object

Power of a Lens

P = 1/f (f in metres, P in dioptres D)

Equivalently: 1/f_eq = 1/f₁ + 1/f₂

  • Convex lens: P positive; concave lens: P negative.
  • Combination of lenses in contact: P_total = P₁ + P₂ + P₃ ...

Why use lens combination? To reduce aberrations and increase magnification in instruments like microscopes and cameras.


4. Huygens' Wave Theory

Wavefronts

A wavefront is a surface connecting all points that are in the same phase of vibration at a given instant.

Source typeWavefront shape
Point sourceSpherical
Line sourceCylindrical
Very distant sourcePlane (effectively flat)

Huygens' Principle

1. Every point on a wavefront acts as a secondary source and emits secondary wavelets in all forward directions. 2. The new wavefront at any later time is the common tangential envelope (forward tangent surface) of all secondary wavelets.

Refraction and Reflection from Huygens' Principle

Refraction: When a plane wavefront hits a refracting surface at an angle, parts of the wavefront in the denser medium travel slower. The new wavefront tilts, explaining Snell's law geometrically.

Reflection: Secondary wavelets from the reflecting surface reconstruct a reflected wavefront — the angle of incidence equals the angle of reflection.


5. Interference of Light — Young's Double Slit Experiment (YDSE)

Setup

Two narrow parallel slits S₁ and S₂, separated by distance d, are illuminated by coherent monochromatic light of wavelength λ. A screen is placed at distance D from the slits.

Path Difference

For a point P at height y on the screen:

Path difference Δ = S₂P − S₁P = yd/D (for small angles)

Conditions

Position: y_n = nλD/d

Position: y_n = (2n−1)λD/2d

  • Constructive interference (bright fringe): Δ = nλ where n = 0, ±1, ±2, ...
  • Destructive interference (dark fringe): Δ = (2n−1)λ/2

Fringe Width

Distance between two consecutive bright (or dark) fringes:

β = λD/d

To increase fringe width: increase D (move screen farther), increase λ (use red light), or decrease d (bring slits closer).

Intensity Distribution

I = 4I₀ cos²(δ/2)

where δ = 2πΔ/λ is the phase difference. Maximum intensity = 4I₀ (at bright fringes), minimum = 0 (at dark fringes).

Conditions for Sustained Interference

1. Sources must be coherent — constant phase difference. 2. Light must be monochromatic — single wavelength. 3. The two sources must have the same amplitude for complete destructive interference (I_min = 0). 4. Slits must be very narrow and closely spaced.


6. Diffraction of Light

Meaning

Diffraction is the bending of light around the edges of an obstacle or through a narrow slit. It is evidence of the wave nature of light.

Single Slit Diffraction

A slit of width a is illuminated by monochromatic light of wavelength λ.

Condition for minima:

a sinθ = nλ (n = ±1, ±2, ±3, ...)
  • The central maximum is the brightest and widest band.
  • Width of central maximum = 2λD/a (from first minima on both sides)
  • Secondary maxima occur between minima, each half as wide as the central maximum.

Effect of slit width:

  • Narrower slit (smaller a) → wider diffraction pattern.
  • Wider slit (larger a) → narrower, more focused pattern.

Difference: Interference vs Diffraction

FeatureInterferenceDiffraction
OriginTwo or more coherent sources (slits)Single slit or obstacle
Fringe widthAll fringes equal widthCentral maximum is widest
Fringe intensityAll bright fringes equalIntensity decreases from centre
PatternAlternating bright and darkCentral bright, then decreasing fringes

7. Polarisation

Nature of Light

Light is a transverse electromagnetic wave — the electric and magnetic field vectors oscillate perpendicular to the direction of propagation. Sound waves are longitudinal and cannot be polarised; this proves light is transverse.

Unpolarised light: Electric field vibrates in all planes perpendicular to propagation. Plane-polarised light: Electric field vibrates in only one plane.

Methods of Polarisation

1. Polaroids (selective absorption / dichroism): A polaroid transmits only one direction of vibration. 2. Reflection (Brewster's law): At a specific angle θ_B (Brewster's angle), the reflected ray is completely plane-polarised. tanθ_B = n 3. Scattering: Sunlight scattered by molecules (blue sky) is partially polarised. 4. Double refraction (birefringence): Certain crystals like calcite split incident light into two polarised rays (ordinary and extraordinary).

Malus's Law

When plane-polarised light of intensity I₀ passes through an analyser whose axis makes angle θ with the polarisation direction:

I = I₀ cos²θ
  • θ = 0°: I = I₀ (maximum transmission)
  • θ = 90°: I = 0 (complete extinction)

Brewster's Law

At Brewster's angle θ_B (also called the polarising angle):

tan θ_B = n (where n is the refractive index of the reflecting medium)

At this angle, the reflected and refracted rays are perpendicular to each other (θ_r + θ_refracted = 90°).

For glass (n = 1.5): tan θ_B = 1.5 → θ_B ≈ 56.3°


Worked Numerical Examples

Numerical 1 — Critical Angle and Total Internal Reflection

Problem: A ray of light travels from glass (n = 1.5) to water (n = 1.33). Find the critical angle for this interface.

Solution
Step 1: Identify the denser and rarer media. n_glass = 1.5 (denser), n_water = 1.33 (rarer). Light travels from glass → water, so TIR is possible. Step 2: Apply the critical angle formula. sinC = n_rarer / n_denser = n_water / n_glass sinC = 1.33 / 1.5 = 0.8867 Step 3: Find C. C = sin⁻¹(0.8867) ≈ 62.5° Answer: The critical angle is approximately 62.5°. Any ray in glass hitting this interface at more than 62.5° will undergo TIR.

Numerical 2 — Thin Lens: Image Position and Magnification

Problem: An object 5 cm tall is placed 30 cm in front of a convex lens of focal length 20 cm. Find (a) the image distance, (b) the magnification, and (c) the image height.

Solution
Given: u = −30 cm (object on left, so negative), f = +20 cm (convex lens) Step 1: Apply thin lens formula. 1/v − 1/u = 1/f 1/v − 1/(−30) = 1/20 1/v + 1/30 = 1/20 1/v = 1/20 − 1/30 = 3/60 − 2/60 = 1/60 v = +60 cm (a) Image is 60 cm on the other side of the lens (real image). Step 2: Calculate magnification. m = v/u = 60/(−30) = −2 (b) Magnification = −2 (negative sign → inverted image, |m| = 2 means magnified 2×) Step 3: Image height. h_image = m × h_object = −2 × 5 = −10 cm (c) Image height = 10 cm (inverted).

Numerical 3 — Young's Double Slit: Fringe Width

Problem: In a YDSE, the slits are separated by 0.5 mm and the screen is 1.0 m away. Monochromatic light of wavelength 589 nm is used. Find (a) the fringe width and (b) the distance of the 3rd bright fringe from the central maximum.

Solution
Given: d = 0.5 mm = 0.5 × 10⁻³ m D = 1.0 m λ = 589 nm = 589 × 10⁻⁹ m Step 1: Fringe width β. β = λD/d β = (589 × 10⁻⁹ × 1.0) / (0.5 × 10⁻³) β = 589 × 10⁻⁹ / 5 × 10⁻⁴ β = 1.178 × 10⁻³ m = 1.178 mm (a) Fringe width ≈ 1.18 mm Step 2: Position of 3rd bright fringe (n = 3). y₃ = nλD/d = 3β = 3 × 1.178 mm = 3.53 mm (b) The 3rd bright fringe is 3.53 mm from the centre.

Numerical 4 — Malus's Law: Intensity After Analyser

Problem: Unpolarised light of intensity I₀ falls on a polariser. The transmitted polarised light then falls on an analyser. When the analyser is rotated so that its axis makes 60° with the polariser axis, find the final intensity.

Solution
Step 1: After the polariser. Unpolarised light through a polariser: only half the intensity passes. I₁ = I₀/2 Step 2: Apply Malus's law at the analyser. I₂ = I₁ cos²θ = (I₀/2) × cos²60° cos 60° = 0.5, so cos²60° = 0.25 I₂ = (I₀/2) × 0.25 = I₀/8 Answer: Final intensity = I₀/8

Board Examination Questions and Answers

One-Mark Questions

Q1. State the condition for total internal reflection. A: (1) Light must travel from optically denser to rarer medium. (2) The angle of incidence must exceed the critical angle.

Q2. What is the fringe width in Young's double slit experiment? A: β = λD/d, where λ is wavelength, D is screen distance, and d is slit separation.

Q3. What does Malus's law state? A: The intensity of polarised light transmitted through an analyser is I = I₀cos²θ, where θ is the angle between the analyser axis and the plane of polarisation.


Two-Mark Questions

Q4. Distinguish between interference and diffraction. A:
Answer
  • Interference is produced by superposition of waves from two or more coherent sources; diffraction is produced by a single source through a narrow aperture or around an obstacle.
  • In interference, all bright fringes have equal intensity; in diffraction, the central maximum is most intense and secondary maxima decrease in intensity.
Q5. A glass slab has refractive index 1.5. Find its critical angle when surrounded by air. A: sinC = 1/n = 1/1.5 = 0.667 C = sin⁻¹(0.667) ≈ 41.8°
Q6. Two convex lenses of focal lengths 20 cm and 30 cm are placed in contact. Find the effective focal length and power of the combination. A: 1/f = 1/f₁ + 1/f₂ = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 f = 12 cm = 0.12 m P = 1/f = 1/0.12 ≈ +8.33 D

Five-Mark Questions

Q7. Explain Young's double slit experiment. Derive the expression for fringe width and state the conditions for constructive and destructive interference.
Answer

Setup: Two narrow parallel slits S₁ and S₂ (separation d) are made on an opaque screen and illuminated by a monochromatic source of wavelength λ. A viewing screen is placed at distance D (D >> d).

Theory: At any point P on the screen at height y from the centre, the path difference between the two waves is:

Δ = S₂P − S₁P = yd/D

Constructive interference (bright fringe): Path difference Δ = nλ (n = 0, ±1, ±2, ...) Position: y_n = nλD/d

Destructive interference (dark fringe): Path difference Δ = (2n−1)λ/2 Position: y_n = (2n−1)λD/(2d)

Fringe width derivation: Distance between consecutive bright fringes: β = y_(n+1) − y_n = (n+1)λD/d − nλD/d = λD/d

This is the fringe width formula. It is the same for dark fringes, showing that fringes are equally spaced.

Conditions for sustained interference: 1. Sources must be coherent (constant phase difference). 2. Light must be monochromatic. 3. Slit separation d must be much smaller than D.


Q8. What is polarisation of light? Explain (a) Malus's law, (b) Brewster's law, and (c) two applications of polaroids.
Answer

Polarisation: In ordinary unpolarised light, the electric field vibrates in all directions perpendicular to propagation. When light is restricted to vibrate in only one plane, it is called plane-polarised light. Polarisation is possible only for transverse waves, confirming the transverse nature of light.

(a) Malus's Law: When plane-polarised light of intensity I₀ is incident on an analyser whose transmission axis makes angle θ with the polarisation direction, the transmitted intensity is: I = I₀cos²θ At θ = 0°, I = I₀ (maximum). At θ = 90°, I = 0 (extinction). This law is used to verify the degree of polarisation of light.

(b) Brewster's Law: When unpolarised light is reflected from a surface at the Brewster angle θ_B, the reflected beam is completely plane-polarised parallel to the surface. Brewster's relation: tanθ_B = n where n is the refractive index of the reflecting medium. At this angle, the reflected and refracted rays are mutually perpendicular. For glass (n = 1.5), θ_B = 56.3°.

(c) Applications of polaroids: 1. Polaroid sunglasses: Horizontally polarised glare (from road and water surfaces) is blocked by vertically oriented polaroid lenses, reducing eye strain. 2. 3D movies: Two projectors emit light polarised at 90° to each other. Viewers wear polaroid glasses with orthogonal lenses — each eye sees only one image, creating depth perception.


NEET and Board Exam Weightage

TopicNEET (expected questions)Board marks
Refraction and TIR1–23–5
Lenses and lens formula1–23–5
YDSE and fringe width1–25
Diffraction12–3
Polarisation and Malus's law13–5

Focus areas for Tamil Nadu board exam:

  • Derive fringe width formula (5-mark question, appears in almost every year).
  • Numerical on thin lens formula — practice sign convention carefully.
  • Critical angle calculation — straightforward 2-mark question.
  • Distinguish interference vs diffraction — standard 2-mark question.
  • Malus's law derivation and numerical — frequently asked.

Frequently Asked Questions

Total internal reflection (TIR) occurs when light travels from an optically denser medium to a rarer medium and the angle of incidence exceeds the critical angle. The two conditions are: (1) light must travel from a denser to a rarer medium (n₁ > n₂), and (2) the angle of incidence must be greater than the critical angle C, where sinC = n₂/n₁.