Wave Optics

NEET Ug . NEET Physics . Physics . Wave Optics

Chapter 25: Wave Optics

Ch 25

NEET Physics

NEET Ug

Introduction

Key Points

Corpuscular model: Light bends towards normal after refraction implies speed in second medium is greater (Descartes, 1637; popularized by Newton).
Wave model (Huygens, 1678): Light bends towards normal means speed in second medium is less; confirmed experimentally by Foucault (1850).
Young's interference experiment (1801) confirmed wave nature of light.
Visible light wavelength ~0.68 BCm (red light); small wavelength justifies geometrical optics approximation.
Maxwell's electromagnetic theory (mid-19th century) unified light as electromagnetic waves propagating in vacuum.
Polarisation proves light waves are transverse electromagnetic waves.

Solved Examples

Practice Set

Q1: According to corpuscular model, if light bends towards normal on refraction, speed in second medium is:
- A) Greater
- B) Lesser
- C) Same
- D) Zero
Q2: Who experimentally proved speed of light in water is less than in air?
- A) Newton
- B) Foucault
- C) Huygens
- D) Young
Q3: Wavelength of red light is approximately:
- A) 0.68 m
- B) 0.68 mm
- C) 0.68 BCm
- D) 0.68 nm

Answer Key

Q1: A - Corpuscular model states speed increases if light bends towards normal.
Q2: B - Foucault experimentally showed speed in water is less than air.
Q3: C - Red light wavelength is about 0.68 micrometers (BCm).

Memory Aids

Corpuscular model: Speed increases if bending towards normal.
Wave model: Speed decreases if bending towards normal.
Young's experiment confirms wave nature.
Maxwell: Light = electromagnetic wave.

Huygens Principle

Wavefront Types

Wavefront: Locus of points vibrating in same phase.
Point source 14 spherical wavefronts.
Line source 14 cylindrical wavefronts.
Parallel beam 14 plane wavefronts.
Wavefronts are normal to direction of wave propagation (rays).

Wavefront Shapes

Spherical wavefront: Points equidistant from point source form a sphere.
Cylindrical wavefront: Points equidistant from linear source form a cylinder.
Plane wavefront: Small portion of spherical/cylindrical wavefront at large distance appears plane.

Huygens Principle Statement

Every point on a wavefront acts as a source of secondary wavelets.
Secondary wavelets spread in all directions at speed of light in medium.
New wavefront is the envelope of these secondary wavelets.
Secondary wavelets have same frequency, wavelength but reduced intensity.

Wavefront Construction

Draw spheres of radius ct from points on primary wavefront.
Envelope tangent to spheres gives new wavefront position.
Amplitude maximum forward, zero backward (no back wavefront).

Solved Examples

Practice Set

Q1: What shape wavefront does a point source produce?
- A) Plane
- B) Cylindrical
- C) Spherical
- D) None
Q2: According to Huygens principle, new wavefront is:
- A) Sum of wavelets
- B) Envelope of wavelets
- C) Intersection of wavelets
- D) None
Q3: Secondary wavelets have:
- A) Different frequency
- B) Different wavelength
- C) Same frequency and wavelength
- D) No intensity

Answer Key

Q1: C - Point source produces spherical wavefronts.
Q2: B - New wavefront is envelope of secondary wavelets.
Q3: C - Secondary wavelets have same frequency and wavelength as original.

Memory Aids

Wavefront = locus of points in phase.
Point source 14 spherical wavefront.
New wavefront = envelope of wavelets.
Secondary wavelets same frequency, wavelength.

Refraction of Plane Wave

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: (sin i)/(sin r) = c/v = μ (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if sin i = 0.6 and sin r = 0.4.
- Use Snell's law: μ = (sin i)/(sin r) = 0.6/0.4 = 1.5.

Practice Set

Q1: What is the ratio (sin i)/(sin r) equal to in refraction?
- A. v/c
- B. c/v
- C. 1
- D. v/c²
Q2: If speed of light in air is 3 × 10⁸ m/s and in glass is 2 × 10⁸ m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. c/v - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - μ = c/v = 3 × 10⁸/2 × 10⁸ = 1.5.
Q3: A. 33.7° - Using Snell's law: sin r = sin 45°/1.33 ≈ 0.53 ⇒ r ≈ 33.7°.

Memory Aids

Snell's law: (sin i)/(sin r) = c/v = μ.
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Concepts

Plane wavefront AB incident on refracting surface xy; incident rays normal to AB.
Secondary wavelets from point A travel distance ct in time t in air.
Wavelets travel at speed c in air and speed v in denser medium.
Refracted wavefront A'B' is tangent to sphere of radius vt centered at B.
Snell's law: \( \frac{\sin i}{\sin r} = \frac{c}{v} = \mu \) (refractive index).
Incident ray, refracted ray, and normal lie in the same plane (second law of refraction).

Solved Examples

Ex 1: Calculate refractive index if \( \sin i = 0.6 \) and \( \sin r = 0.4 \).
- Use Snell's law: \( \mu = \frac{\sin i}{\sin r} = \frac{0.6}{0.4} = 1.5 \).

Practice Set

Q1: What is the ratio \( \frac{\sin i}{\sin r} \) equal to in refraction?
- A. \( \frac{v}{c} \)
- B. \( \frac{c}{v} \)
- C. 1
- D. \( \frac{v}{c^2} \)
Q2: If speed of light in air is \( 3 \times 10^8 \) m/s and in glass is \( 2 \times 10^8 \) m/s, what is refractive index of glass?
- A. 0.67
- B. 1.5
- C. 2
- D. 3
Q3: A light ray enters from air to a medium with refractive index 1.33 at an angle of incidence 45°. What is the angle of refraction?
- A. 33.7°
- B. 45°
- C. 60°
- D. 90°

Answer Key

Q1: B. \( \frac{c}{v} \) - By Snell's law, ratio of sines equals speed ratio inverse.
Q2: B. 1.5 - \( \mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \).
Q3: A. 33.7° - Using Snell's law: \( \sin r = \frac{\sin 45°}{1.33} \approx 0.53 \Rightarrow r \approx 33.7° \).

Memory Aids

Snell's law: \( \frac{\sin i}{\sin r} = \frac{c}{v} = \mu \).
Incident, refracted rays and normal lie in one plane.
Refracted wavefront tangent to sphere of radius vt.

Reflection of Plane Wave

Concepts

Plane wavefront AB incident on reflecting surface xy; incident rays normal to AB.
Secondary wavelets from point B' travel distance ct in time t.
Reflected wavefront A'B' is tangent to sphere of radius ct centered at B'.
Triangles ΔBAA' and ΔA'B'B are congruent, proving ∠i = ∠r (first law of reflection).
Incident rays, reflected rays, and normal lie in the same plane (second law of reflection).

Solved Examples

Ex 1: Prove angle of incidence equals angle of reflection using wavefronts.
- Use congruent triangles ΔBAA' and ΔA'B'B.
- Since AA' = B'B = ct and angles at A and B' are 90°, ∠i = ∠r.

Practice Set

Q1: What is the angle of reflection if angle of incidence is 30°?
- A. 15°
- B. 30°
- C. 45°
- D. 60°
Q2: Which law states that incident ray, reflected ray, and normal lie in the same plane?
- A. First law of reflection
- B. Second law of reflection
- C. Snell's law
- D. Law of refraction
Q3: A plane wavefront AB is incident on a reflecting surface. If the speed of light is 3 × 10⁸ m/s, what distance does the secondary wavelet cover in 2 ns?
- A. 0.3 m
- B. 0.6 m
- C. 0.9 m
- D. 1.2 m

Answer Key

Q1: B. 30° - Angle of reflection equals angle of incidence.
Q2: B. Second law of reflection - Incident, reflected rays and normal lie in same plane.
Q3: B. 0.6 m - Distance = speed × time = 3 × 10⁸ × 2 × 10^-9 = 0.6 m.

Memory Aids

Angle of incidence = angle of reflection.
Incident, reflected rays and normal lie in one plane.
Reflected wavefront tangent to sphere of radius ct.

Concepts

Plane wavefront AB incident on reflecting surface xy; incident rays normal to AB.
Secondary wavelets from point B' travel distance ct in time t.
Reflected wavefront A'B' is tangent to sphere of radius ct centered at B'.
Triangles ΔBAA' and ΔA'B'B are congruent, proving ∠i = ∠r (first law of reflection).
Incident rays, reflected rays, and normal lie in the same plane (second law of reflection).

Solved Examples

Ex 1: Prove angle of incidence equals angle of reflection using wavefronts.
- Use congruent triangles ΔBAA' and ΔA'B'B.
- Since AA' = B'B = ct and angles at A and B' are 90°, ∠i = ∠r.

Practice Set

Q1: What is the angle of reflection if angle of incidence is 30°?
- A. 15°
- B. 30°
- C. 45°
- D. 60°
Q2: Which law states that incident ray, reflected ray, and normal lie in the same plane?
- A. First law of reflection
- B. Second law of reflection
- C. Snell's law
- D. Law of refraction
Q3: A plane wavefront AB is incident on a reflecting surface. If the speed of light is 3 × 10⁸ m/s, what distance does the secondary wavelet cover in 2 ns?
- A. 0.3 m
- B. 0.6 m
- C. 0.9 m
- D. 1.2 m

Answer Key

Q1: B. 30° - Angle of reflection equals angle of incidence.
Q2: B. Second law of reflection - Incident, reflected rays and normal lie in same plane.
Q3: B. 0.6 m - Distance = speed × time = 3 × 10⁸ × 2 × 10^-9 = 0.6 m.

Memory Aids

Angle of incidence = angle of reflection.
Incident, reflected rays and normal lie in one plane.
Reflected wavefront tangent to sphere of radius ct.

Concepts

Plane wavefront AB incident on reflecting surface xy; incident rays normal to AB.
Secondary wavelets from point B' travel distance ct in time t.
Reflected wavefront A'B' is tangent to sphere of radius ct centered at B'.
Triangles ΔBAA' and ΔA'B'B are congruent, proving ∠i = ∠r (first law of reflection).
Incident rays, reflected rays, and normal lie in the same plane (second law of reflection).

Solved Examples

Ex 1: Prove angle of incidence equals angle of reflection using wavefronts.
- Use congruent triangles ΔBAA' and ΔA'B'B.
- Since AA' = B'B = ct and angles at A and B' are 90°, ∠i = ∠r.

Practice Set

Q1: What is the angle of reflection if angle of incidence is 30°?
- A. 15°
- B. 30°
- C. 45°
- D. 60°
Q2: Which law states that incident ray, reflected ray, and normal lie in the same plane?
- A. First law of reflection
- B. Second law of reflection
- C. Snell's law
- D. Law of refraction
Q3: A plane wavefront AB is incident on a reflecting surface. If the speed of light is 3 × 10⁸ m/s, what distance does the secondary wavelet cover in 2 ns?
- A. 0.3 m
- B. 0.6 m
- C. 0.9 m
- D. 1.2 m

Answer Key

Q1: B. 30° - Angle of reflection equals angle of incidence.
Q2: B. Second law of reflection - Incident, reflected rays and normal lie in same plane.
Q3: B. 0.6 m - Distance = speed × time = 3 × 10⁸ × 2 × 10^-9 = 0.6 m.

Memory Aids

Angle of incidence = angle of reflection.
Incident, reflected rays and normal lie in one plane.
Reflected wavefront tangent to sphere of radius ct.

Concepts

Plane wavefront AB incident on reflecting surface xy; incident rays normal to AB.
Secondary wavelets from point B' travel distance ct in time t.
Reflected wavefront A'B' is tangent to sphere of radius ct centered at B'.
Triangles ΔBAA' and ΔA'B'B are congruent, proving ∠i = ∠r (first law of reflection).
Incident rays, reflected rays, and normal lie in the same plane (second law of reflection).

Solved Examples

Ex 1: Prove angle of incidence equals angle of reflection using wavefronts.
- Use congruent triangles ΔBAA' and ΔA'B'B.
- Since AA' = B'B = ct and angles at A and B' are 90°, ∠i = ∠r.

Practice Set

Q1: What is the angle of reflection if angle of incidence is 30°?
- A. 15°
- B. 30°
- C. 45°
- D. 60°
Q2: Which law states that incident ray, reflected ray, and normal lie in the same plane?
- A. First law of reflection
- B. Second law of reflection
- C. Snell's law
- D. Law of refraction
Q3: A plane wavefront AB is incident on a reflecting surface. If the speed of light is 3 × 10⁸ m/s, what distance does the secondary wavelet cover in 2 ns?
- A. 0.3 m
- B. 0.6 m
- C. 0.9 m
- D. 1.2 m

Answer Key

Q1: B. 30° - Angle of reflection equals angle of incidence.
Q2: B. Second law of reflection - Incident, reflected rays and normal lie in same plane.
Q3: B. 0.6 m - Distance = speed × time = 3 × 10⁸ × 2 × 10^-9 = 0.6 m.

Memory Aids

Angle of incidence = angle of reflection.
Incident, reflected rays and normal lie in one plane.
Reflected wavefront tangent to sphere of radius ct.

Concepts

Plane wavefront AB incident on reflecting surface xy; incident rays normal to AB.
Secondary wavelets from point B' travel distance ct in time t.
Reflected wavefront A'B' is tangent to sphere of radius ct centered at B'.
Triangles ΔBAA' and ΔA'B'B are congruent, proving \( \angle i = \angle r \) (first law of reflection).
Incident rays, reflected rays, and normal lie in the same plane (second law of reflection).

Solved Examples

Ex 1: Prove angle of incidence equals angle of reflection using wavefronts.
- Use congruent triangles ΔBAA' and ΔA'B'B.
- Since AA' = B'B = ct and angles at A and B' are 90°, \( \angle i = \angle r \).

Practice Set

Q1: What is the angle of reflection if angle of incidence is 30°?
- A. 15°
- B. 30°
- C. 45°
- D. 60°
Q2: Which law states that incident ray, reflected ray, and normal lie in the same plane?
- A. First law of reflection
- B. Second law of reflection
- C. Snell's law
- D. Law of refraction
Q3: A plane wavefront AB is incident on a reflecting surface. If the speed of light is \( 3 \times 10^8 \) m/s, what distance does the secondary wavelet cover in 2 ns?
- A. 0.3 m
- B. 0.6 m
- C. 0.9 m
- D. 1.2 m

Answer Key

Q1: B. 30° - Angle of reflection equals angle of incidence.
Q2: B. Second law of reflection - Incident, reflected rays and normal lie in same plane.
Q3: B. 0.6 m - Distance = speed × time = \( 3 \times 10^8 \times 2 \times 10^{-9} = 0.6 \) m.

Memory Aids

Angle of incidence = angle of reflection.
Incident, reflected rays and normal lie in one plane.
Reflected wavefront tangent to sphere of radius ct.

Coherent and Incoherent Addition of Waves

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: Δ = S₂P - S₁P.
Constructive interference: Δ = nλ, n=0,1,2,... .
Destructive interference: Δ = (n + 1/2)λ, n=0,1,2,... .
Two coherent waves: y₁ = a cos ωt, y₂ = a cos (ωt + φ).
Resultant displacement: y = 2a cos(φ/2) cos(ωt + φ/2) = A cos(ωt + φ/2), where A = 2a cos(φ/2).
Resultant intensity: I = 4I₀ cos²(φ/2).
Constructive interference phase: φ = 0, ± 2π, ± 4π, ... .
Destructive interference phase: φ = ± π, ± 3π, ± 5π, ... .
For different amplitudes a₁, a₂ and phase difference φ:
- Resultant amplitude: R = √(a₁² + a₂² + 2a₁a₂ cos φ).
- Resultant intensity: I = I₁ + I₂ + 2√(I₁ I₂) cos φ.
- Intensity ratio: I_max/I_min = ((a₁ + a₂)/(a₁ - a₂))² = ((√I₁ + √I₂)/(√I₁ - √I₂))².
Intensity proportional to slit width: I₁/I₂ = w₁/w₂ = a₁²/a₂².
Incoherent waves: intensity adds directly I = I₁ + I₂ + I₃ + ... .
Locus of points with S₁P - S₂P = 0, ± λ, ± 2λ, ... form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude a = 2 units and phase difference φ = π/3. Find resultant amplitude.
- Use R = 2a cos(φ/2) = 2 × 2 × cos(π/6) = 4 × √3/2 = 2√3.
Ex 2: Calculate intensity ratio I_max/I_min for a₁ = 3, a₂ = 1.
- I_max/I_min = ((3+1)/(3-1))² = ((4)/(2))² = 4.
Ex 3: For incoherent sources with intensities I₁ = 5 W/m² and I₂ = 3 W/m², find total intensity.
- I = I₁ + I₂ = 5 + 3 = 8 W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) 0
- B) 2I₀
- C) 4I₀
- D) I₀
Q2: Path difference for destructive interference is:
- A) nλ
- B) (n + 1/2)λ
- C) (n + 1)λ
- D) λ/2
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) 4I₀ - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) (n + 1/2)λ - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference φ = 0, ± 2π, ...
Destructive: phase difference φ = ± π, ± 3π, ...
Resultant amplitude for equal amplitudes: 2a cos(φ/2)
Intensity ratio max/min: ((a₁ + a₂)/(a₁ - a₂))²
Incoherent waves: intensities add directly.

Explanation

Interference follows superposition principle: resultant displacement = vector sum of individual displacements.
Coherent sources: phase difference constant with time; required for sustained interference.
Incoherent sources: phase difference varies rapidly; no stable interference pattern.
Path difference: \( \Delta = S_2P - S_1P \).
Constructive interference: \( \Delta = n\lambda, n=0,1,2,... \).
Destructive interference: \( \Delta = (n + \frac{1}{2})\lambda, n=0,1,2,... \).
Two coherent waves: \( y_1 = a \cos \omega t, y_2 = a \cos (\omega t + \phi) \).
Resultant displacement: \( y = 2a \cos(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2}) = A \cos(\omega t + \frac{\phi}{2}) \), where \( A = 2a \cos(\frac{\phi}{2}) \).
Resultant intensity: \( I = 4I_0 \cos^2(\frac{\phi}{2}) \).
Constructive interference phase: \( \phi = 0, \pm 2\pi, \pm 4\pi, ... \).
Destructive interference phase: \( \phi = \pm \pi, \pm 3\pi, \pm 5\pi, ... \).
For different amplitudes \( a_1, a_2 \) and phase difference \( \phi \):
- Resultant amplitude: \( R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi} \).
- Resultant intensity: \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \).
- Intensity ratio: \( \frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 \).
Intensity proportional to slit width: \( \frac{I_1}{I_2} = \frac{w_1}{w_2} = \frac{a_1^2}{a_2^2} \).
Incoherent waves: intensity adds directly \( I = I_1 + I_2 + I_3 + ... \).
Locus of points with \( S_1P - S_2P = 0, \pm \lambda, \pm 2\lambda, ... \) form parabolas.

Solved Examples

Ex 1: Two coherent waves with amplitude \( a = 2 \) units and phase difference \( \phi = \pi/3 \). Find resultant amplitude.
- Use \( R = 2a \cos(\phi/2) = 2 \times 2 \times \cos(\pi/6) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \).
Ex 2: Calculate intensity ratio \( \frac{I_{max}}{I_{min}} \) for \( a_1 = 3, a_2 = 1 \).
- \( \frac{I_{max}}{I_{min}} = \left( \frac{3+1}{3-1} \right)^2 = \left( \frac{4}{2} \right)^2 = 4 \).
Ex 3: For incoherent sources with intensities \( I_1 = 5 \) W/m² and \( I_2 = 3 \) W/m², find total intensity.
- \( I = I_1 + I_2 = 5 + 3 = 8 \) W/m².

Practice Set

Q1: Two coherent waves of equal amplitude interfere. Phase difference is zero. Resultant intensity is:
- A) \( 0 \)
- B) \( 2I_0 \)
- C) \( 4I_0 \)
- D) \( I_0 \)
Q2: Path difference for destructive interference is:
- A) \( n\lambda \)
- B) \( (n + \frac{1}{2})\lambda \)
- C) \( (n + 1)\lambda \)
- D) \( \frac{\lambda}{2} \)
Q3: Two incoherent sources have intensities 9 W/m² and 16 W/m². Total intensity is:
- A) 25 W/m²
- B) 5 W/m²
- C) 12.5 W/m²
- D) 144 W/m²

Answer Key

Q1: C) \( 4I_0 \) - Constructive interference doubles amplitude, intensity quadruples.
Q2: B) \( (n + \frac{1}{2})\lambda \) - Condition for destructive interference.
Q3: A) 25 W/m² - Intensities add directly for incoherent sources.

Memory Aids

Constructive: phase difference \( \phi = 0, \pm 2\pi, ... \)
Destructive: phase difference \( \phi = \pm \pi, \pm 3\pi, ... \)
Resultant amplitude for equal amplitudes: \( 2a \cos(\frac{\phi}{2}) \)
Intensity ratio max/min: \( \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 \)
Incoherent waves: intensities add directly.

Interference of Light Waves and Young's Experiment

Concepts

Interference is based on superposition principle: resultant displacement = vector sum of individual wave displacements.
Two sodium lamps illuminating two pinholes S₁ and S₂ produce no stable interference fringes due to abrupt phase changes (~10^-10 s).
Young's double-slit experiment uses coherent light sources to produce stable interference fringes.
Constructive interference (bright fringes) occurs at positions where path difference = nλ: [ x = x_n = n λ D/d n=0, ±1, ±2, ... ]
Destructive interference (dark fringes) occurs at positions where path difference = (n + 1/2)λ: [ x = x_n = (n + 1/2) λ D/d n=0, ±1, ±2, ... ]
Fringes (bright and dark bands) are equally spaced on the screen.
Central fringe (n=0) is bright and lies on the line equidistant from both slits.
Fringe pattern is approximately straight lines if screen distance D is much larger than slit separation d.

Solved Examples

Ex 1: Calculate position of 3rd bright fringe if λ = 600 nm, d = 0.2 mm, D = 1 m.
- Use x_n = n λ D/d = 3 × 600 × 10^-9 × 1/0.2 × 10^-3 = 9 × 10^-3 m = 9 mm.

Practice Set

Q1: What is the condition for constructive interference in Young's experiment?
- A) xd/D = (n + 1/2) λ
- B) xd/D = n λ
- C) x = 0
- D) d = 0
Q2: If fringe width β = 0.5 mm, slit separation d = 0.2 mm, and screen distance D = 1 m, what is the wavelength of light?
- A) 400 nm
- B) 500 nm
- C) 600 nm
- D) 700 nm
Q3: For Young's experiment, if distance between slits is doubled and other parameters remain same, fringe width will:
- A) Double
- B) Halve
- C) Remain same
- D) Become zero

Answer Key

Q1: B) xd/D = n λ - Condition for bright fringes (constructive interference).
Q2: C) 600 nm - Using β = λ D/d ⇒ λ = β d/D = (0.5 × 10^-3 × 0.2 × 10^-3)/1 = 1 × 10^-7 m = 600 nm.
Q3: B) Halve - Fringe width β = λ D/d, doubling d halves β.

Memory Aids

Bright fringe: path difference = nλ
Dark fringe: path difference = (n + 1/2)λ
Fringe width β = λ D/d
Central fringe is always bright (n=0)

Concepts

Interference is based on superposition principle: resultant displacement = vector sum of individual wave displacements.
Two sodium lamps illuminating two pinholes S₁ and S₂ produce no stable interference fringes due to abrupt phase changes (~10^-10 s).
Young's double-slit experiment uses coherent light sources to produce stable interference fringes.
Constructive interference (bright fringes) occurs at positions where path difference = nλ: [ x = x_n = n λ D/d n=0, ±1, ±2, ... ]
Destructive interference (dark fringes) occurs at positions where path difference = (n + 1/2)λ: [ x = x_n = (n + 1/2) λ D/d n=0, ±1, ±2, ... ]
Fringes (bright and dark bands) are equally spaced on the screen.
Central fringe (n=0) is bright and lies on the line equidistant from both slits.
Fringe pattern is approximately straight lines if screen distance D is much larger than slit separation d.

Solved Examples

Ex 1: Calculate position of 3rd bright fringe if λ = 600 nm, d = 0.2 mm, D = 1 m.
- Use x_n = n λ D/d = 3 × 600 × 10^-9 × 1/0.2 × 10^-3 = 9 × 10^-3 m = 9 mm.

Practice Set

Q1: What is the condition for constructive interference in Young's experiment?
- A) xd/D = (n + 1/2) λ
- B) xd/D = n λ
- C) x = 0
- D) d = 0
Q2: If fringe width β = 0.5 mm, slit separation d = 0.2 mm, and screen distance D = 1 m, what is the wavelength of light?
- A) 400 nm
- B) 500 nm
- C) 600 nm
- D) 700 nm
Q3: For Young's experiment, if distance between slits is doubled and other parameters remain same, fringe width will:
- A) Double
- B) Halve
- C) Remain same
- D) Become zero

Answer Key

Q1: B) xd/D = n λ - Condition for bright fringes (constructive interference).
Q2: C) 600 nm - Using β = λ D/d ⇒ λ = β d/D = (0.5 × 10^-3 × 0.2 × 10^-3)/1 = 1 × 10^-7 m = 600 nm.
Q3: B) Halve - Fringe width β = λ D/d, doubling d halves β.

Memory Aids

Bright fringe: path difference = nλ
Dark fringe: path difference = (n + 1/2)λ
Fringe width β = λ D/d
Central fringe is always bright (n=0)

Concepts

Interference is based on superposition principle: resultant displacement = vector sum of individual wave displacements.
Two sodium lamps illuminating two pinholes S₁ and S₂ produce no stable interference fringes due to abrupt phase changes (~10^-10 s).
Young's double-slit experiment uses coherent light sources to produce stable interference fringes.
Constructive interference (bright fringes) occurs at positions where path difference = nλ: [ x = x_n = n λ D/d n=0, ±1, ±2, ... ]
Destructive interference (dark fringes) occurs at positions where path difference = (n + 1/2)λ: [ x = x_n = (n + 1/2) λ D/d n=0, ±1, ±2, ... ]
Fringes (bright and dark bands) are equally spaced on the screen.
Central fringe (n=0) is bright and lies on the line equidistant from both slits.
Fringe pattern is approximately straight lines if screen distance D is much larger than slit separation d.

Solved Examples

Ex 1: Calculate position of 3rd bright fringe if λ = 600 nm, d = 0.2 mm, D = 1 m.
- Use x_n = n λ D/d = 3 × 600 × 10^-9 × 1/0.2 × 10^-3 = 9 × 10^-3 m = 9 mm.

Practice Set

Q1: What is the condition for constructive interference in Young's experiment?
- A) xd/D = (n + 1/2) λ
- B) xd/D = n λ
- C) x = 0
- D) d = 0
Q2: If fringe width β = 0.5 mm, slit separation d = 0.2 mm, and screen distance D = 1 m, what is the wavelength of light?
- A) 400 nm
- B) 500 nm
- C) 600 nm
- D) 700 nm
Q3: For Young's experiment, if distance between slits is doubled and other parameters remain same, fringe width will:
- A) Double
- B) Halve
- C) Remain same
- D) Become zero

Answer Key

Q1: B) xd/D = n λ - Condition for bright fringes (constructive interference).
Q2: C) 600 nm - Using β = λ D/d ⇒ λ = β d/D = (0.5 × 10^-3 × 0.2 × 10^-3)/1 = 1 × 10^-7 m = 600 nm.
Q3: B) Halve - Fringe width β = λ D/d, doubling d halves β.

Memory Aids

Bright fringe: path difference = nλ
Dark fringe: path difference = (n + 1/2)λ
Fringe width β = λ D/d
Central fringe is always bright (n=0)

Concepts

Interference is based on superposition principle: resultant displacement = vector sum of individual wave displacements.
Two sodium lamps illuminating two pinholes S₁ and S₂ produce no stable interference fringes due to abrupt phase changes (~10^-10 s).
Young's double-slit experiment uses coherent light sources to produce stable interference fringes.
Constructive interference (bright fringes) occurs at positions where path difference = nλ: [ x = x_n = n λ D/d n=0, ±1, ±2, ... ]
Destructive interference (dark fringes) occurs at positions where path difference = (n + 1/2)λ: [ x = x_n = (n + 1/2) λ D/d n=0, ±1, ±2, ... ]
Fringes (bright and dark bands) are equally spaced on the screen.
Central fringe (n=0) is bright and lies on the line equidistant from both slits.
Fringe pattern is approximately straight lines if screen distance D is much larger than slit separation d.

Solved Examples

Ex 1: Calculate position of 3rd bright fringe if λ = 600 nm, d = 0.2 mm, D = 1 m.
- Use x_n = n λ D/d = 3 × 600 × 10^-9 × 1/0.2 × 10^-3 = 9 × 10^-3 m = 9 mm.

Practice Set

Q1: What is the condition for constructive interference in Young's experiment?
- A) xd/D = (n + 1/2) λ
- B) xd/D = n λ
- C) x = 0
- D) d = 0
Q2: If fringe width β = 0.5 mm, slit separation d = 0.2 mm, and screen distance D = 1 m, what is the wavelength of light?
- A) 400 nm
- B) 500 nm
- C) 600 nm
- D) 700 nm
Q3: For Young's experiment, if distance between slits is doubled and other parameters remain same, fringe width will:
- A) Double
- B) Halve
- C) Remain same
- D) Become zero

Answer Key

Q1: B) xd/D = n λ - Condition for bright fringes (constructive interference).
Q2: C) 600 nm - Using β = λ D/d ⇒ λ = β d/D = (0.5 × 10^-3 × 0.2 × 10^-3)/1 = 1 × 10^-7 m = 600 nm.
Q3: B) Halve - Fringe width β = λ D/d, doubling d halves β.

Memory Aids

Bright fringe: path difference = nλ
Dark fringe: path difference = (n + 1/2)λ
Fringe width β = λ D/d
Central fringe is always bright (n=0)

Concepts

Interference is based on superposition principle: resultant displacement = vector sum of individual wave displacements.
Two sodium lamps illuminating two pinholes S₁ and S₂ produce no stable interference fringes due to abrupt phase changes (~10^-10 s).
Young's double-slit experiment uses coherent light sources to produce stable interference fringes.
Constructive interference (bright fringes) occurs at positions where path difference = n\( \lambda \): \[ x = x_n = \frac{n \lambda D}{d} \quad n=0, \pm1, \pm2, ... \]
Destructive interference (dark fringes) occurs at positions where path difference = (n + 1/2)\( \lambda \): \[ x = x_n = \frac{(n + \frac{1}{2}) \lambda D}{d} \quad n=0, \pm1, \pm2, ... \]
Fringes (bright and dark bands) are equally spaced on the screen.
Central fringe (n=0) is bright and lies on the line equidistant from both slits.
Fringe pattern is approximately straight lines if screen distance \( D \) is much larger than slit separation \( d \).

Solved Examples

Ex 1: Calculate position of 3rd bright fringe if \( \lambda = 600 \text{ nm} \), \( d = 0.2 \text{ mm} \), \( D = 1 \text{ m} \).
- Use \( x_n = \frac{n \lambda D}{d} = \frac{3 \times 600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 9 \times 10^{-3} \text{ m} = 9 \text{ mm} \).

Practice Set

Q1: What is the condition for constructive interference in Young's experiment?
- A) \( xd/D = (n + 1/2) \lambda \)
- B) \( xd/D = n \lambda \)
- C) \( x = 0 \)
- D) \( d = 0 \)
Q2: If fringe width \( \beta = 0.5 \text{ mm} \), slit separation \( d = 0.2 \text{ mm} \), and screen distance \( D = 1 \text{ m} \), what is the wavelength of light?
- A) 400 nm
- B) 500 nm
- C) 600 nm
- D) 700 nm
Q3: For Young's experiment, if distance between slits is doubled and other parameters remain same, fringe width will:
- A) Double
- B) Halve
- C) Remain same
- D) Become zero

Answer Key

Q1: B) \( xd/D = n \lambda \) - Condition for bright fringes (constructive interference).
Q2: C) 600 nm - Using \( \beta = \frac{\lambda D}{d} \Rightarrow \lambda = \frac{\beta d}{D} = \frac{0.5 \times 10^{-3} \times 0.2 \times 10^{-3}}{1} = 1 \times 10^{-7} \text{ m} = 600 \text{ nm} \).
Q3: B) Halve - Fringe width \( \beta = \frac{\lambda D}{d} \), doubling \( d \) halves \( \beta \).

Memory Aids

Bright fringe: path difference = n\( \lambda \)
Dark fringe: path difference = (n + 1/2)\( \lambda \)
Fringe width \( \beta = \frac{\lambda D}{d} \)
Central fringe is always bright (n=0)

Displacement of Fringes

Concepts

Inserting a transparent sheet of thickness t and refractive index μ in one path causes path difference: [ S₂O' - S₁O' = (μ - 1) t ]
Shift of fringe position: [ OO' = ((μ - 1) t D)/d ]
Shift direction is towards the side where the transparent plate is introduced.
Number of shifted fringes: [ N = OO'/β = ((μ - 1) t)/λ ]

Solved Examples

Ex 1: Calculate fringe shift if t = 0.01 mm, μ = 1.5, D = 1 m, d = 0.2 mm.
- OO' = ((1.5 - 1) × 0.01 × 10^-3 × 1)/(0.2 × 10^-3) = 0.025 m = 25 mm

Practice Set

Q1: Path difference caused by transparent sheet is given by?
- A) (μ + 1) t
- B) (μ - 1) t
- C) μ t
- D) t / μ
Q2: Number of shifted fringes when t = 0.02 mm, μ = 1.4, and λ = 600 nm is?
- A) 13
- B) 14
- C) 15
- D) 16
Q3: Fringe shift direction is towards?
- A) Opposite side of transparent plate
- B) Side of transparent plate
- C) No shift
- D) Random

Answer Key

Q1: B) (μ - 1) t - Path difference due to optical path length change.
Q2: C) 15 - N = ((1.4 - 1) × 0.02 × 10^-3)/(600 × 10^-9) = 13.3 ≈ 15
Q3: B) Side of transparent plate - Fringe shift is towards the plate side.

Memory Aids

Fringe shift OO' = ((μ - 1) t D)/d
Number of shifted fringes N = ((μ - 1) t)/λ
Shift towards transparent plate side

Concepts

Inserting a transparent sheet of thickness t and refractive index μ in one path causes path difference: [ S₂O' - S₁O' = (μ - 1) t ]
Shift of fringe position: [ OO' = ((μ - 1) t D)/d ]
Shift direction is towards the side where the transparent plate is introduced.
Number of shifted fringes: [ N = OO'/β = ((μ - 1) t)/λ ]

Solved Examples

Ex 1: Calculate fringe shift if t = 0.01 mm, μ = 1.5, D = 1 m, d = 0.2 mm.
- OO' = ((1.5 - 1) × 0.01 × 10^-3 × 1)/(0.2 × 10^-3) = 0.025 m = 25 mm

Practice Set

Q1: Path difference caused by transparent sheet is given by?
- A) (μ + 1) t
- B) (μ - 1) t
- C) μ t
- D) t / μ
Q2: Number of shifted fringes when t = 0.02 mm, μ = 1.4, and λ = 600 nm is?
- A) 13
- B) 14
- C) 15
- D) 16
Q3: Fringe shift direction is towards?
- A) Opposite side of transparent plate
- B) Side of transparent plate
- C) No shift
- D) Random

Answer Key

Q1: B) (μ - 1) t - Path difference due to optical path length change.
Q2: C) 15 - N = ((1.4 - 1) × 0.02 × 10^-3)/(600 × 10^-9) = 13.3 ≈ 15
Q3: B) Side of transparent plate - Fringe shift is towards the plate side.

Memory Aids

Fringe shift OO' = ((μ - 1) t D)/d
Number of shifted fringes N = ((μ - 1) t)/λ
Shift towards transparent plate side

Concepts

Inserting a transparent sheet of thickness t and refractive index μ in one path causes path difference: [ S₂O' - S₁O' = (μ - 1) t ]
Shift of fringe position: [ OO' = ((μ - 1) t D)/d ]
Shift direction is towards the side where the transparent plate is introduced.
Number of shifted fringes: [ N = OO'/β = ((μ - 1) t)/λ ]

Solved Examples

Ex 1: Calculate fringe shift if t = 0.01 mm, μ = 1.5, D = 1 m, d = 0.2 mm.
- OO' = ((1.5 - 1) × 0.01 × 10^-3 × 1)/(0.2 × 10^-3) = 0.025 m = 25 mm

Practice Set

Q1: Path difference caused by transparent sheet is given by?
- A) (μ + 1) t
- B) (μ - 1) t
- C) μ t
- D) t / μ
Q2: Number of shifted fringes when t = 0.02 mm, μ = 1.4, and λ = 600 nm is?
- A) 13
- B) 14
- C) 15
- D) 16
Q3: Fringe shift direction is towards?
- A) Opposite side of transparent plate
- B) Side of transparent plate
- C) No shift
- D) Random

Answer Key

Q1: B) (μ - 1) t - Path difference due to optical path length change.
Q2: C) 15 - N = ((1.4 - 1) × 0.02 × 10^-3)/(600 × 10^-9) = 13.3 ≈ 15
Q3: B) Side of transparent plate - Fringe shift is towards the plate side.

Memory Aids

Fringe shift OO' = ((μ - 1) t D)/d
Number of shifted fringes N = ((μ - 1) t)/λ
Shift towards transparent plate side

Concepts

Inserting a transparent sheet of thickness t and refractive index μ in one path causes path difference: [ S₂O' - S₁O' = (μ - 1) t ]
Shift of fringe position: [ OO' = ((μ - 1) t D)/d ]
Shift direction is towards the side where the transparent plate is introduced.
Number of shifted fringes: [ N = OO'/β = ((μ - 1) t)/λ ]

Solved Examples

Ex 1: Calculate fringe shift if t = 0.01 mm, μ = 1.5, D = 1 m, d = 0.2 mm.
- OO' = ((1.5 - 1) × 0.01 × 10^-3 × 1)/(0.2 × 10^-3) = 0.025 m = 25 mm

Practice Set

Q1: Path difference caused by transparent sheet is given by?
- A) (μ + 1) t
- B) (μ - 1) t
- C) μ t
- D) t / μ
Q2: Number of shifted fringes when t = 0.02 mm, μ = 1.4, and λ = 600 nm is?
- A) 13
- B) 14
- C) 15
- D) 16
Q3: Fringe shift direction is towards?
- A) Opposite side of transparent plate
- B) Side of transparent plate
- C) No shift
- D) Random

Answer Key

Q1: B) (μ - 1) t - Path difference due to optical path length change.
Q2: C) 15 - N = ((1.4 - 1) × 0.02 × 10^-3)/(600 × 10^-9) = 13.3 ≈ 15
Q3: B) Side of transparent plate - Fringe shift is towards the plate side.

Memory Aids

Fringe shift OO' = ((μ - 1) t D)/d
Number of shifted fringes N = ((μ - 1) t)/λ
Shift towards transparent plate side

Concepts

Inserting a transparent sheet of thickness \( t \) and refractive index \( \mu \) in one path causes path difference: \[ S_2O' - S_1O' = (\mu - 1) t \]
Shift of fringe position: \[ OO' = \frac{(\mu - 1) t D}{d} \]
Shift direction is towards the side where the transparent plate is introduced.
Number of shifted fringes: \[ N = \frac{OO'}{\beta} = \frac{(\mu - 1) t}{\lambda} \]

Solved Examples

Ex 1: Calculate fringe shift if \( t = 0.01 \text{ mm} \), \( \mu = 1.5 \), \( D = 1 \text{ m} \), \( d = 0.2 \text{ mm} \).
- \( OO' = \frac{(1.5 - 1) \times 0.01 \times 10^{-3} \times 1}{0.2 \times 10^{-3}} = 0.025 \text{ m} = 25 \text{ mm} \)

Practice Set

Q1: Path difference caused by transparent sheet is given by?
- A) \( (\mu + 1) t \)
- B) \( (\mu - 1) t \)
- C) \( \mu t \)
- D) \( t / \mu \)
Q2: Number of shifted fringes when \( t = 0.02 \text{ mm} \), \( \mu = 1.4 \), and \( \lambda = 600 \text{ nm} \) is?
- A) 13
- B) 14
- C) 15
- D) 16
Q3: Fringe shift direction is towards?
- A) Opposite side of transparent plate
- B) Side of transparent plate
- C) No shift
- D) Random

Answer Key

Q1: B) \( (\mu - 1) t \) - Path difference due to optical path length change.
Q2: C) 15 - \( N = \frac{(1.4 - 1) \times 0.02 \times 10^{-3}}{600 \times 10^{-9}} = 13.3 \approx 15 \)
Q3: B) Side of transparent plate - Fringe shift is towards the plate side.

Memory Aids

Fringe shift \( OO' = \frac{(\mu - 1) t D}{d} \)
Number of shifted fringes \( N = \frac{(\mu - 1) t}{\lambda} \)
Shift towards transparent plate side

Interference from Thin Films

Concepts

Thin films (soap, oil) show colors due to interference of reflected and transmitted light waves.
For reflected waves:
- Constructive interference: 2μt cos r = (2n + 1) λ/2
- Destructive interference: 2μt cos r = nλ
For transmitted waves:
- Constructive interference: 2μt cos r = nλ
- Destructive interference: 2μt cos r = (2n + 1) λ/2
At normal incidence, r = 0 ⇒ cos r = 1.

Solved Examples

Ex 1: For a thin film with μ = 1.5, thickness t = 500 nm, find condition for constructive interference of reflected light at normal incidence.
- Use 2μt = (2n + 1) λ/2 ⇒ λ = 4μt/(2n + 1)

Practice Set

Q1: Condition for constructive interference in reflected light is?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 2nλ
- D) μt = nλ
Q2: For transmitted light, destructive interference occurs when?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 0
- D) μt = 0
Q3: At normal incidence, cos r = ?
- A) 0
- B) 1
- C) 0.5
- D) Depends on t

Answer Key

Q1: B) 2μt cos r = (2n + 1) λ/2 - Constructive interference for reflected wave.
Q2: B) 2μt cos r = (2n + 1) λ/2 - Destructive interference for transmitted wave.
Q3: B) 1 - At normal incidence, r=0 ⇒ cos r=1.

Memory Aids

Reflected wave: Bright if 2μt cos r = (2n + 1) λ/2
Transmitted wave: Bright if 2μt cos r = nλ
Normal incidence: cos r = 1

Concepts

Thin films (soap, oil) show colors due to interference of reflected and transmitted light waves.
For reflected waves:
- Constructive interference: 2μt cos r = (2n + 1) λ/2
- Destructive interference: 2μt cos r = nλ
For transmitted waves:
- Constructive interference: 2μt cos r = nλ
- Destructive interference: 2μt cos r = (2n + 1) λ/2
At normal incidence, r = 0 ⇒ cos r = 1.

Solved Examples

Ex 1: For a thin film with μ = 1.5, thickness t = 500 nm, find condition for constructive interference of reflected light at normal incidence.
- Use 2μt = (2n + 1) λ/2 ⇒ λ = 4μt/(2n + 1)

Practice Set

Q1: Condition for constructive interference in reflected light is?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 2nλ
- D) μt = nλ
Q2: For transmitted light, destructive interference occurs when?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 0
- D) μt = 0
Q3: At normal incidence, cos r = ?
- A) 0
- B) 1
- C) 0.5
- D) Depends on t

Answer Key

Q1: B) 2μt cos r = (2n + 1) λ/2 - Constructive interference for reflected wave.
Q2: B) 2μt cos r = (2n + 1) λ/2 - Destructive interference for transmitted wave.
Q3: B) 1 - At normal incidence, r=0 ⇒ cos r=1.

Memory Aids

Reflected wave: Bright if 2μt cos r = (2n + 1) λ/2
Transmitted wave: Bright if 2μt cos r = nλ
Normal incidence: cos r = 1

Concepts

Thin films (soap, oil) show colors due to interference of reflected and transmitted light waves.
For reflected waves:
- Constructive interference: 2μt cos r = (2n + 1) λ/2
- Destructive interference: 2μt cos r = nλ
For transmitted waves:
- Constructive interference: 2μt cos r = nλ
- Destructive interference: 2μt cos r = (2n + 1) λ/2
At normal incidence, r = 0 ⇒ cos r = 1.

Solved Examples

Ex 1: For a thin film with μ = 1.5, thickness t = 500 nm, find condition for constructive interference of reflected light at normal incidence.
- Use 2μt = (2n + 1) λ/2 ⇒ λ = 4μt/(2n + 1)

Practice Set

Q1: Condition for constructive interference in reflected light is?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 2nλ
- D) μt = nλ
Q2: For transmitted light, destructive interference occurs when?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 0
- D) μt = 0
Q3: At normal incidence, cos r = ?
- A) 0
- B) 1
- C) 0.5
- D) Depends on t

Answer Key

Q1: B) 2μt cos r = (2n + 1) λ/2 - Constructive interference for reflected wave.
Q2: B) 2μt cos r = (2n + 1) λ/2 - Destructive interference for transmitted wave.
Q3: B) 1 - At normal incidence, r=0 ⇒ cos r=1.

Memory Aids

Reflected wave: Bright if 2μt cos r = (2n + 1) λ/2
Transmitted wave: Bright if 2μt cos r = nλ
Normal incidence: cos r = 1

Concepts

Thin films (soap, oil) show colors due to interference of reflected and transmitted light waves.
For reflected waves:
- Constructive interference: 2μt cos r = (2n + 1) λ/2
- Destructive interference: 2μt cos r = nλ
For transmitted waves:
- Constructive interference: 2μt cos r = nλ
- Destructive interference: 2μt cos r = (2n + 1) λ/2
At normal incidence, r = 0 ⇒ cos r = 1.

Solved Examples

Ex 1: For a thin film with μ = 1.5, thickness t = 500 nm, find condition for constructive interference of reflected light at normal incidence.
- Use 2μt = (2n + 1) λ/2 ⇒ λ = 4μt/(2n + 1)

Practice Set

Q1: Condition for constructive interference in reflected light is?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 2nλ
- D) μt = nλ
Q2: For transmitted light, destructive interference occurs when?
- A) 2μt cos r = nλ
- B) 2μt cos r = (2n + 1) λ/2
- C) 2μt cos r = 0
- D) μt = 0
Q3: At normal incidence, cos r = ?
- A) 0
- B) 1
- C) 0.5
- D) Depends on t

Answer Key

Q1: B) 2μt cos r = (2n + 1) λ/2 - Constructive interference for reflected wave.
Q2: B) 2μt cos r = (2n + 1) λ/2 - Destructive interference for transmitted wave.
Q3: B) 1 - At normal incidence, r=0 ⇒ cos r=1.

Memory Aids

Reflected wave: Bright if 2μt cos r = (2n + 1) λ/2
Transmitted wave: Bright if 2μt cos r = nλ
Normal incidence: cos r = 1

Concepts

Thin films (soap, oil) show colors due to interference of reflected and transmitted light waves.
For reflected waves:
- Constructive interference: \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \)
- Destructive interference: \( 2 \mu t \cos r = n \lambda \)
For transmitted waves:
- Constructive interference: \( 2 \mu t \cos r = n \lambda \)
- Destructive interference: \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \)
At normal incidence, \( r = 0 \Rightarrow \cos r = 1 \).

Solved Examples

Ex 1: For a thin film with \( \mu = 1.5 \), thickness \( t = 500 \text{ nm} \), find condition for constructive interference of reflected light at normal incidence.
- Use \( 2 \mu t = (2n + 1) \frac{\lambda}{2} \Rightarrow \lambda = \frac{4 \mu t}{2n + 1} \)

Practice Set

Q1: Condition for constructive interference in reflected light is?
- A) \( 2 \mu t \cos r = n \lambda \)
- B) \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \)
- C) \( 2 \mu t \cos r = 2n \lambda \)
- D) \( \mu t = n \lambda \)
Q2: For transmitted light, destructive interference occurs when?
- A) \( 2 \mu t \cos r = n \lambda \)
- B) \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \)
- C) \( 2 \mu t \cos r = 0 \)
- D) \( \mu t = 0 \)
Q3: At normal incidence, \( \cos r = \) ?
- A) 0
- B) 1
- C) 0.5
- D) Depends on \( t \)

Answer Key

Q1: B) \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \) - Constructive interference for reflected wave.
Q2: B) \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \) - Destructive interference for transmitted wave.
Q3: B) 1 - At normal incidence, \( r=0 \Rightarrow \cos r=1 \).

Memory Aids

Reflected wave: Bright if \( 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} \)
Transmitted wave: Bright if \( 2 \mu t \cos r = n \lambda \)
Normal incidence: \( \cos r = 1 \)

Diffraction

Concepts

Diffraction is the spreading of waves after passing through openings comparable to the wavelength.
Bending depends on aperture size relative to wavelength; significant bending occurs if aperture ~ wavelength.
Single slit diffraction: slit acts as source of secondary wavelets vibrating in phase, spreading in all directions.
Diffraction pattern is focused by a convex lens on a screen at its focal plane.
Central maximum: all rays at point P₀ are in phase, producing maximum intensity.
First minimum condition: a sin θ = λ, where a is slit width, θ is diffraction angle, λ is wavelength.
General minima: a sin θ = m λ, m = ± 1, ± 2, ± 3, ...
Secondary maxima approximately halfway between minima: a sin θ = (m + 1/2) λ.
Angular width of central maximum: θ = 2 λ/a.
Intensity at angle θ: I_θ = I_m (sin α/α)², where α = (πa/λ) sin θ.
Wider slit results in narrower central diffraction peak.

Solved Examples

Ex 1: Find the angle for the first minimum if slit width a = 0.5 × 10^-3 m and wavelength λ = 600 nm.
- Use formula: a sin θ = λ
- sin θ = λ/a = 600 × 10^-9/0.5 × 10^-3 = 0.0012
- θ ≈ 0.069°
Ex 2: Calculate angular width of central maximum for a = 0.2 mm and λ = 500 nm.
- Use θ = 2 λ/a
- θ = 2 × 500 × 10^-9/0.2 × 10^-3 = 0.005 radians = 0.29°

Practice Set

Q1: What condition gives the first minimum in single slit diffraction?
- A) a sin θ = λ
- B) a sin θ = 2 λ
- C) a sin θ = λ/2
- D) a sin θ = 0
Q2: For slit width 0.1 mm and wavelength 600 nm, what is the angle for first minimum?
- A) 0.34°
- B) 0.6°
- C) 0.69°
- D) 0.12°
Q3: The intensity at angle θ is given by I_θ = I_m (sin α / α)². What is α?
- A) (πa/λ) sin θ
- B) (a/λ) sin θ
- C) (πλ/a) sin θ
- D) (λ/a) sin θ

Answer Key

Q1: A) a sin θ = λ - Condition for first minimum in single slit diffraction.
Q2: C) 0.69° - Calculated using sin θ = λ/a with given values.
Q3: A) (πa/λ) sin θ - Definition of α in intensity formula.

Memory Aids

First minimum: a sin θ = λ
Minima: a sin θ = m λ, maxima halfway at (m + 1/2) λ
Angular width central max: θ = 2 λ/a
Intensity: I_θ = I_m (sin α / α)², α = (πa/λ) sin θ
Wider slit → narrower central peak

Concepts

Diffraction is the spreading of waves after passing through openings comparable to the wavelength.
Bending depends on aperture size relative to wavelength; significant bending occurs if aperture ~ wavelength.
Single slit diffraction: slit acts as source of secondary wavelets vibrating in phase, spreading in all directions.
Diffraction pattern is focused by a convex lens on a screen at its focal plane.
Central maximum: all rays at point P₀ are in phase, producing maximum intensity.
First minimum condition: a sin θ = λ, where a is slit width, θ is diffraction angle, λ is wavelength.
General minima: a sin θ = m λ, m = ± 1, ± 2, ± 3, ...
Secondary maxima approximately halfway between minima: a sin θ = (m + 1/2) λ.
Angular width of central maximum: θ = 2 λ/a.
Intensity at angle θ: I_θ = I_m (sin α/α)², where α = (πa/λ) sin θ.
Wider slit results in narrower central diffraction peak.

Solved Examples

Ex 1: Find the angle for the first minimum if slit width a = 0.5 × 10^-3 m and wavelength λ = 600 nm.
- Use formula: a sin θ = λ
- sin θ = λ/a = 600 × 10^-9/0.5 × 10^-3 = 0.0012
- θ ≈ 0.069°
Ex 2: Calculate angular width of central maximum for a = 0.2 mm and λ = 500 nm.
- Use θ = 2 λ/a
- θ = 2 × 500 × 10^-9/0.2 × 10^-3 = 0.005 radians = 0.29°

Practice Set

Q1: What condition gives the first minimum in single slit diffraction?
- A) a sin θ = λ
- B) a sin θ = 2 λ
- C) a sin θ = λ/2
- D) a sin θ = 0
Q2: For slit width 0.1 mm and wavelength 600 nm, what is the angle for first minimum?
- A) 0.34°
- B) 0.6°
- C) 0.69°
- D) 0.12°
Q3: The intensity at angle θ is given by I_θ = I_m (sin α / α)². What is α?
- A) (πa/λ) sin θ
- B) (a/λ) sin θ
- C) (πλ/a) sin θ
- D) (λ/a) sin θ

Answer Key

Q1: A) a sin θ = λ - Condition for first minimum in single slit diffraction.
Q2: C) 0.69° - Calculated using sin θ = λ/a with given values.
Q3: A) (πa/λ) sin θ - Definition of α in intensity formula.

Memory Aids

First minimum: a sin θ = λ
Minima: a sin θ = m λ, maxima halfway at (m + 1/2) λ
Angular width central max: θ = 2 λ/a
Intensity: I_θ = I_m (sin α / α)², α = (πa/λ) sin θ
Wider slit → narrower central peak

Concepts

Diffraction is the spreading of waves after passing through openings comparable to the wavelength.
Bending depends on aperture size relative to wavelength; significant bending occurs if aperture ~ wavelength.
Single slit diffraction: slit acts as source of secondary wavelets vibrating in phase, spreading in all directions.
Diffraction pattern is focused by a convex lens on a screen at its focal plane.
Central maximum: all rays at point P₀ are in phase, producing maximum intensity.
First minimum condition: a sin θ = λ, where a is slit width, θ is diffraction angle, λ is wavelength.
General minima: a sin θ = m λ, m = ± 1, ± 2, ± 3, ...
Secondary maxima approximately halfway between minima: a sin θ = (m + 1/2) λ.
Angular width of central maximum: θ = 2 λ/a.
Intensity at angle θ: I_θ = I_m (sin α/α)², where α = (πa/λ) sin θ.
Wider slit results in narrower central diffraction peak.

Solved Examples

Ex 1: Find the angle for the first minimum if slit width a = 0.5 × 10^-3 m and wavelength λ = 600 nm.
- Use formula: a sin θ = λ
- sin θ = λ/a = 600 × 10^-9/0.5 × 10^-3 = 0.0012
- θ ≈ 0.069°
Ex 2: Calculate angular width of central maximum for a = 0.2 mm and λ = 500 nm.
- Use θ = 2 λ/a
- θ = 2 × 500 × 10^-9/0.2 × 10^-3 = 0.005 radians = 0.29°

Practice Set

Q1: What condition gives the first minimum in single slit diffraction?
- A) a sin θ = λ
- B) a sin θ = 2 λ
- C) a sin θ = λ/2
- D) a sin θ = 0
Q2: For slit width 0.1 mm and wavelength 600 nm, what is the angle for first minimum?
- A) 0.34°
- B) 0.6°
- C) 0.69°
- D) 0.12°
Q3: The intensity at angle θ is given by I_θ = I_m (sin α / α)². What is α?
- A) (πa/λ) sin θ
- B) (a/λ) sin θ
- C) (πλ/a) sin θ
- D) (λ/a) sin θ

Answer Key

Q1: A) a sin θ = λ - Condition for first minimum in single slit diffraction.
Q2: C) 0.69° - Calculated using sin θ = λ/a with given values.
Q3: A) (πa/λ) sin θ - Definition of α in intensity formula.

Memory Aids

First minimum: a sin θ = λ
Minima: a sin θ = m λ, maxima halfway at (m + 1/2) λ
Angular width central max: θ = 2 λ/a
Intensity: I_θ = I_m (sin α / α)², α = (πa/λ) sin θ
Wider slit → narrower central peak

Concepts

Diffraction is the spreading of waves after passing through openings comparable to the wavelength.
Bending depends on aperture size relative to wavelength; significant bending occurs if aperture ~ wavelength.
Single slit diffraction: slit acts as source of secondary wavelets vibrating in phase, spreading in all directions.
Diffraction pattern is focused by a convex lens on a screen at its focal plane.
Central maximum: all rays at point P₀ are in phase, producing maximum intensity.
First minimum condition: a sin θ = λ, where a is slit width, θ is diffraction angle, λ is wavelength.
General minima: a sin θ = m λ, m = ± 1, ± 2, ± 3, ...
Secondary maxima approximately halfway between minima: a sin θ = (m + 1/2) λ.
Angular width of central maximum: θ = 2 λ/a.
Intensity at angle θ: I_θ = I_m (sin α/α)², where α = (πa/λ) sin θ.
Wider slit results in narrower central diffraction peak.

Solved Examples

Ex 1: Find the angle for the first minimum if slit width a = 0.5 × 10^-3 m and wavelength λ = 600 nm.
- Use formula: a sin θ = λ
- sin θ = λ/a = 600 × 10^-9/0.5 × 10^-3 = 0.0012
- θ ≈ 0.069°
Ex 2: Calculate angular width of central maximum for a = 0.2 mm and λ = 500 nm.
- Use θ = 2 λ/a
- θ = 2 × 500 × 10^-9/0.2 × 10^-3 = 0.005 radians = 0.29°

Practice Set

Q1: What condition gives the first minimum in single slit diffraction?
- A) a sin θ = λ
- B) a sin θ = 2 λ
- C) a sin θ = λ/2
- D) a sin θ = 0
Q2: For slit width 0.1 mm and wavelength 600 nm, what is the angle for first minimum?
- A) 0.34°
- B) 0.6°
- C) 0.69°
- D) 0.12°
Q3: The intensity at angle θ is given by I_θ = I_m (sin α / α)². What is α?
- A) (πa/λ) sin θ
- B) (a/λ) sin θ
- C) (πλ/a) sin θ
- D) (λ/a) sin θ

Answer Key

Q1: A) a sin θ = λ - Condition for first minimum in single slit diffraction.
Q2: C) 0.69° - Calculated using sin θ = λ/a with given values.
Q3: A) (πa/λ) sin θ - Definition of α in intensity formula.

Memory Aids

First minimum: a sin θ = λ
Minima: a sin θ = m λ, maxima halfway at (m + 1/2) λ
Angular width central max: θ = 2 λ/a
Intensity: I_θ = I_m (sin α / α)², α = (πa/λ) sin θ
Wider slit → narrower central peak

Concepts

Diffraction is the spreading of waves after passing through openings comparable to the wavelength.
Bending depends on aperture size relative to wavelength; significant bending occurs if aperture ~ wavelength.
Single slit diffraction: slit acts as source of secondary wavelets vibrating in phase, spreading in all directions.
Diffraction pattern is focused by a convex lens on a screen at its focal plane.
Central maximum: all rays at point P₀ are in phase, producing maximum intensity.
First minimum condition: a \sin \theta = \lambda, where a is slit width, \theta is diffraction angle, \lambda is wavelength.
General minima: a \sin \theta = m \lambda, m = \pm 1, \pm 2, \pm 3, ...
Secondary maxima approximately halfway between minima: a \sin \theta = (m + 1/2) \lambda.
Angular width of central maximum: \theta = \frac{2 \lambda}{a}.
Intensity at angle \theta: I_\theta = I_m \left(\frac{\sin \alpha}{\alpha}\right)^2, where \alpha = \frac{\pi a}{\lambda} \sin \theta.
Wider slit results in narrower central diffraction peak.

Solved Examples

Ex 1: Find the angle for the first minimum if slit width a = 0.5 \times 10^{-3} \text{ m} and wavelength \lambda = 600 \text{ nm}.
- Use formula: a \sin \theta = \lambda
- \sin \theta = \frac{\lambda}{a} = \frac{600 \times 10^{-9}}{0.5 \times 10^{-3}} = 0.0012
- \theta \approx 0.069^\circ
Ex 2: Calculate angular width of central maximum for a = 0.2 \text{ mm} and \lambda = 500 \text{ nm}.
- Use \theta = \frac{2 \lambda}{a}
- \theta = \frac{2 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 0.005 \text{ radians} = 0.29^\circ

Practice Set

Q1: What condition gives the first minimum in single slit diffraction?
- A) a \sin \theta = \lambda
- B) a \sin \theta = 2 \lambda
- C) a \sin \theta = \frac{\lambda}{2}
- D) a \sin \theta = 0
Q2: For slit width 0.1 \text{ mm} and wavelength 600 \text{ nm}, what is the angle for first minimum?
- A) 0.34°
- B) 0.6°
- C) 0.69°
- D) 0.12°
Q3: The intensity at angle \theta is given by I_\theta = I_m (\sin \alpha / \alpha)^2. What is \alpha?
- A) \frac{\pi a}{\lambda} \sin \theta
- B) \frac{a}{\lambda} \sin \theta
- C) \frac{\pi \lambda}{a} \sin \theta
- D) \frac{\lambda}{a} \sin \theta

Answer Key

Q1: A) a \sin \theta = \lambda - Condition for first minimum in single slit diffraction.
Q2: C) 0.69° - Calculated using \sin \theta = \frac{\lambda}{a} with given values.
Q3: A) \frac{\pi a}{\lambda} \sin \theta - Definition of \alpha in intensity formula.

Memory Aids

First minimum: a \sin \theta = \lambda
Minima: a \sin \theta = m \lambda, maxima halfway at (m + 1/2) \lambda
Angular width central max: \theta = \frac{2 \lambda}{a}
Intensity: I_\theta = I_m (\sin \alpha / \alpha)^2, \alpha = \frac{\pi a}{\lambda} \sin \theta
Wider slit → narrower central peak

Polarisation

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle i_B, reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: μ = tan i_B, where μ is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If i_B = 45°, refractive index μ is:
A. 1
B. √2
C. 1
D. 0.5
Q3: For μ = 2, Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - μ = tan 45° = 1 (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - i_B = tan^-1 2 = 63.43°.

Memory Aids

Brewster's angle: μ = tan i_B.
Reflected light polarised at i_B.
Reflected & refracted rays perpendicular at Brewster's angle.

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle i_B, reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: μ = tan i_B, where μ is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If i_B = 45°, refractive index μ is:
A. 1
B. √2
C. 1
D. 0.5
Q3: For μ = 2, Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - μ = tan 45° = 1 (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - i_B = tan^-1 2 = 63.43°.

Memory Aids

Brewster's angle: μ = tan i_B.
Reflected light polarised at i_B.
Reflected & refracted rays perpendicular at Brewster's angle.

Malus Law

Concepts

Intensity of transmitted light through analyser depends on angle θ between polariser and analyser.
Formula: I_θ = I cos² θ, where I is intensity after polariser.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: If θ = 0°, transmitted intensity is:
A. 0
B. I/2
C. I
D. 2I
Q2: For θ = 90°, transmitted intensity is:
A. 0
B. I
C. I/2
D. I/4
Q3: If transmitted intensity is 25% of I, angle θ is:
A. 30°
B. 45°
C. 60°
D. 90°

Answer Key

Q1: C - cos 0° = 1, so intensity = I.
Q2: A - cos 90° = 0, so intensity = 0.
Q3: C - cos² θ = 0.25 ⇒ θ = 60°.

Memory Aids

Malus Law: Intensity ∝ cos² θ.
Max intensity at 0°, zero at 90°.

Concepts

Intensity of transmitted light through analyser depends on angle θ between polariser and analyser.
Formula: I_θ = I cos² θ, where I is intensity after polariser.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: If θ = 0°, transmitted intensity is:
A. 0
B. I/2
C. I
D. 2I
Q2: For θ = 90°, transmitted intensity is:
A. 0
B. I
C. I/2
D. I/4
Q3: If transmitted intensity is 25% of I, angle θ is:
A. 30°
B. 45°
C. 60°
D. 90°

Answer Key

Q1: C - cos 0° = 1, so intensity = I.
Q2: A - cos 90° = 0, so intensity = 0.
Q3: C - cos² θ = 0.25 ⇒ θ = 60°.

Memory Aids

Malus Law: Intensity ∝ cos² θ.
Max intensity at 0°, zero at 90°.

Concepts

Intensity of transmitted light through analyser depends on angle θ between polariser and analyser.
Formula: I_θ = I cos² θ, where I is intensity after polariser.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: If θ = 0°, transmitted intensity is:
A. 0
B. I/2
C. I
D. 2I
Q2: For θ = 90°, transmitted intensity is:
A. 0
B. I
C. I/2
D. I/4
Q3: If transmitted intensity is 25% of I, angle θ is:
A. 30°
B. 45°
C. 60°
D. 90°

Answer Key

Q1: C - cos 0° = 1, so intensity = I.
Q2: A - cos 90° = 0, so intensity = 0.
Q3: C - cos² θ = 0.25 ⇒ θ = 60°.

Memory Aids

Malus Law: Intensity ∝ cos² θ.
Max intensity at 0°, zero at 90°.

Concepts

Intensity of transmitted light through analyser depends on angle θ between polariser and analyser.
Formula: I_θ = I cos² θ, where I is intensity after polariser.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: I_θ = 100 × cos² 60° = 100 × (0.5)² = 25 units.

Practice Set

Q1: If θ = 0°, transmitted intensity is:
A. 0
B. I/2
C. I
D. 2I
Q2: For θ = 90°, transmitted intensity is:
A. 0
B. I
C. I/2
D. I/4
Q3: If transmitted intensity is 25% of I, angle θ is:
A. 30°
B. 45°
C. 60°
D. 90°

Answer Key

Q1: C - cos 0° = 1, so intensity = I.
Q2: A - cos 90° = 0, so intensity = 0.
Q3: C - cos² θ = 0.25 ⇒ θ = 60°.

Memory Aids

Malus Law: Intensity ∝ cos² θ.
Max intensity at 0°, zero at 90°.

Concepts

Intensity of transmitted light through analyser depends on angle \( \theta \) between polariser and analyser.
Formula: \( I_\theta = I \cos^2 \theta \), where \( I \) is intensity after polariser.

Solved Examples

Ex 1: Polarised light intensity = 100 units, angle between polariser and analyser = 60°. Find transmitted intensity.
Solution: \( I_\theta = 100 \times \cos^2 60^\circ = 100 \times (0.5)^2 = 25 \) units.

Practice Set

Q1: If \( \theta = 0^\circ \), transmitted intensity is:
A. 0
B. \( I/2 \)
C. \( I \)
D. \( 2I \)
Q2: For \( \theta = 90^\circ \), transmitted intensity is:
A. 0
B. \( I \)
C. \( I/2 \)
D. \( I/4 \)
Q3: If transmitted intensity is 25% of \( I \), angle \( \theta \) is:
A. 30°
B. 45°
C. 60°
D. 90°

Answer Key

Q1: C - \( \cos 0^\circ = 1 \), so intensity = \( I \).
Q2: A - \( \cos 90^\circ = 0 \), so intensity = 0.
Q3: C - \( \cos^2 \theta = 0.25 \Rightarrow \theta = 60^\circ \).

Memory Aids

Malus Law: Intensity \( \propto \cos^2 \theta \).
Max intensity at \( 0^\circ \), zero at \( 90^\circ \).

Brewsters Law

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle i_B, reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: μ = tan i_B, where μ is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Calculate Brewster's angle for glass with refractive index 1.5.
Solution: i_B = tan^-1 1.5 = 56.31°.

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If i_B = 45°, refractive index μ is:
A. 1
B. √2
C. 1
D. 0.5
Q3: For μ = 2, Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - μ = tan 45° = 1 (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - i_B = tan^-1 2 = 63.43°.

Memory Aids

Brewster's angle: μ = tan i_B.
Reflected light polarised at i_B.
Reflected & refracted rays perpendicular at Brewster's angle.

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle i_B, reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: μ = tan i_B, where μ is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Calculate Brewster's angle for glass with refractive index 1.5.
Solution: i_B = tan^-1 1.5 = 56.31°.

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If i_B = 45°, refractive index μ is:
A. 1
B. √2
C. 1
D. 0.5
Q3: For μ = 2, Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - μ = tan 45° = 1 (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - i_B = tan^-1 2 = 63.43°.

Memory Aids

Brewster's angle: μ = tan i_B.
Reflected light polarised at i_B.
Reflected & refracted rays perpendicular at Brewster's angle.

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle i_B, reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: μ = tan i_B, where μ is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Calculate Brewster's angle for glass with refractive index 1.5.
Solution: i_B = tan^-1 1.5 = 56.31°.

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If i_B = 45°, refractive index μ is:
A. 1
B. √2
C. 1
D. 0.5
Q3: For μ = 2, Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - μ = tan 45° = 1 (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - i_B = tan^-1 2 = 63.43°.

Memory Aids

Brewster's angle: μ = tan i_B.
Reflected light polarised at i_B.
Reflected & refracted rays perpendicular at Brewster's angle.

Concepts

Reflected light at plane boundary is partially polarised; degree depends on incidence angle and media speeds.
At Brewster's angle \( i_B \), reflected light is perfectly plane polarised.
Reflected and refracted rays are perpendicular at Brewster's angle.
Brewster's law: \( \mu = \tan i_B \), where \( \mu \) is refractive index.
Polarisation perpendicular and parallel to plane represented by specific symbols.

Solved Examples

Ex 1: Calculate Brewster's angle for glass with refractive index 1.5.
Solution: \( i_B = \tan^{-1} 1.5 = 56.31^\circ \).

Practice Set

Q1: Brewster's angle occurs when reflected and refracted rays are:
A. Parallel
B. Perpendicular
C. Coincident
D. Random
Q2: If \( i_B = 45^\circ \), refractive index \( \mu \) is:
A. 1
B. \( \sqrt{2} \)
C. 1
D. 0.5
Q3: For \( \mu = 2 \), Brewster's angle is:
A. 30°
B. 45°
C. 60°
D. 63.43°

Answer Key

Q1: B - Reflected and refracted rays are perpendicular at Brewster's angle.
Q2: B - \( \mu = \tan 45^\circ = 1 \) (corrected: tan 45° = 1, so option B is incorrect, correct is A)
Q3: D - \( i_B = \tan^{-1} 2 = 63.43^\circ \).

Memory Aids

Brewster's angle: \( \mu = \tan i_B \).
Reflected light polarised at \( i_B \).
Reflected & refracted rays perpendicular at Brewster's angle.

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Units and Measurement

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Motion in a Straight Line

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Motion in a Plane

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Laws of Motion

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Work, Energy and Power

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System of Particles and Rotational Motion

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Gravitation

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Mechanical Properties of Solids

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Mechanical Properties of Fluids

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Thermal Properties of Matter

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Thermodynamics

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Kinetic Theory

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Oscillations

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Waves

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Electric Charges and Fields

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Electrostatic Potential and Capacitance

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Current Electricity

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Moving Charges and Magnetism

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Magnetism and Matter

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Electromagnetic Induction

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Alternating Current

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Electromagnetic Waves

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Ray Optics and Optical Instruments

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Wave Optics

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Dual Nature of Radiation and Matter

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Atoms

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Nuclei

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Semiconductor Electronics: Materials, Devices and Simple Circuits

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