#medium

Optical Phenomena and Applications

What is the wavelength of light in a medium with refractive index 1.5 if its wavelength in vacuum is \( 750 \, \text{nm}

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 750 nm , n = 1.5 , λ_m = (750/1.5) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 500 nm, illustrating interf

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Optical Phenomena and Applications

What is the refractive index of a medium if the speed of light in it is \( 1.8 \times 10^8 \, \text{m/s} \) and in vacuu

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 1.8 × 10⁸ m/s . n = (3.0 × 10⁸/1.8 × 10⁸) ≈ 1.67 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.8, given the speed in vacuum is \( 3.0 \times 10^8 \, \te

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Speed in a medium v = (c/n) . Given n = 1.8 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.8) ≈ 1.67 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.5 \times 10^8 \, \text{m/s} \) and in vacuu

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.5 × 10⁸ m/s . n = (3.0 × 10⁸/2.5 × 10⁸) = 1.2 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.4, given the speed in vacuum is \( 3.0 \times 10^8 \, \te

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Speed in a medium v = (c/n) . Given n = 1.4 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.4) ≈ 2.14 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.4 \times 10^8 \, \text{m/s} \) and in vacuu

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.4 × 10⁸ m/s , n = (3.0 × 10⁸/2.4 × 10⁸) = 1.25 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A =

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Refraction, Refractive Index and Critical Angle

What is the wavelength of light in a medium with refractive index 1.5 if its wavelength in air is \( 720 \, \text{nm} \)

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Wavelength in a medium λ_m = (λₐir/n) . Given λₐir = 720 nm , n = 1.5 , λ_m = (720/1.5) = 480 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.25 \times 10^8 \, \text{m/s} \) and in vacu

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.25 × 10⁸ m/s . n = (3.0 × 10⁸/2.25 × 10⁸) = 1.33 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Energy, Intensity and Momentum of EM Waves

What property of electromagnetic waves explains why their speed in a medium is less than in vacuum?

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. The speed in a medium is reduced due to the medium’s permittivity ( ε ) and permeability ( μ ), giving v = (1/√(μ ε)) , which is less than c because ε > ε₀ and μ > μ₀ . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Medium’s electric properties, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Stationary Waves and Standing Waves in Strings

Which condition is essential for the formation of a standing wave in a medium?

**Quantization due to boundaries** leads to discrete harmonic series. Frequency difference between harmonics is f₁, so f₃ - f₁ = 2f₁ = v/L. Understanding node-antinode pattern explains resonance and overtones in strings. Standing waves require waves of equal frequency to interfere, typically from reflection, creating a stationary pattern of nodes and antinodes. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Equal frequency of interfering waves, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Stationary Waves and Standing Waves in Strings

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