Refraction, Refractive Index and Critical Angle

Refraction, Refractive Index and Critical Angle

What determines the intensity of light in the wave picture?

**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. In the wave picture, intensity is proportional to the square of the amplitude of the wave. 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 Square of the amplitude, illustrating interference, diffraction and polarization principles.

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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

Why does the angle of refraction increase beyond 90° become impossible when light moves from a denser to a rarer medium?

**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. Beyond the critical angle, the sine of the refraction angle exceeds 1, which is mathematically impossible, leading to total internal reflection. 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 Sine exceeds unity, illustrating interference, diffraction and polarization principles.

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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

Why does the wave theory of light predict that the speed of light decreases when it bends towards the normal during refr

**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. The wave theory suggests that light slows down in a denser medium, causing the wavefront to bend towards the normal as it takes less time to travel a shorter path in the slower medium. 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 glass if its refractive index is 1.6 and the speed of light in vacuum is \( 3.0 \times 10^

**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 = 1.6 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.6) = 1.875 × 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 explains the absence of a refracted ray when the angle of incidence exceeds the critical angle?

**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. Beyond the critical angle, the refracted ray would require a sine greater than 1, which is impossible, leading to total internal reflection. 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 Sine of refraction angle

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Refraction, Refractive Index and Critical Angle

What is the critical angle for light passing from a medium with refractive index 1.9 to air (refractive index 1.0)?

**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. sin i_c = (n₂/n₁) , where n₁ = 1.9 , n₂ = 1.0 . sin i_c = (1.0/1.9) ≈ 0.526 , i_c = sin⁻¹(0.526) ≈ 31.8° . 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

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with a refractive index of 1.5, given the speed of light in vacuum is \( 3.0 \tim

**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) , where c is the speed in vacuum and v is the speed in the medium. Given n = 1.5 , v = (c/n) = (3.0 × 10⁸/1.5) = 2.0 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.45, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**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 = 1.45 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.45) ≈ 2.07 × 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 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 critical angle for light passing into air is \( 42^\circ \)?

**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. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 42° . sin 42° ≈ 0.669 , n₁ = (1.0/0.669) ≈ 1.49 . 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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