#critical angle

Optical Phenomena and Applications

What is the critical angle for light passing from a medium with refractive index 1.85 to air (refractive index 1.0)?

**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. sin i_c = (n₂/n₁) , where n₁ = 1.85 , n₂ = 1.0 . sin i_c = (1.0/1.85) ≈ 0.541 , i_c = sin⁻¹(0.541) ≈ 32.7° . 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 32.7°, illustrating int

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Optical Phenomena and Applications

What is the refractive index of a medium if the critical angle for light passing into air is \( 39^\circ \)?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 39° . sin 39° ≈ 0.629 , n₁ = (1.0/0.629) ≈ 1.59 . 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 1.59,

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Optical Phenomena and Applications

What is the angle of refraction when light is incident at the critical angle from a denser medium to a rarer medium?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. At the critical angle, the angle of refraction is 90° , as the refracted ray travels along the boundary between the two media. 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 90°, illustrating i

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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 g

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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 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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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the critical angle for light passing into air is \( 48.6^\circ \)?

**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. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 48.6° . sin 48.6° ≈ 0.75 , n₁ = (1.0/0.75) ≈ 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' = λ/n and A =

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Refraction, Refractive Index and Critical Angle

What is the condition for total internal reflection to occur?

**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. Total internal reflection occurs when the angle of incidence exceeds the critical angle, and light travels from a denser to a rarer medium. 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 happens to the refracted ray when the angle of incidence exceeds the critical angle?

**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. When the angle of incidence exceeds the critical angle, no refraction occurs, and the ray undergoes 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 Undergoes total

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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 \( 36.9^\circ \)?

**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 = 36.9° . sin 36.9° ≈ 0.6 , n₁ = (1.0/0.6) ≈ 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 at Spherical Surfaces and Lens Maker's Formula

What is the critical angle for a diamond (\( n = 2.42 \)) to water (\( n = 1.33 \)) interface?

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. Critical angle: sin i_c = (n₂/n₁) . Diamond ( n₁ = 2.42 ), water ( n₂ = 1.33 ). sin i_c = (1.33/2.42) ≈ 0.55 . i_c = sin⁻¹(0.55) ≈ 33.4° . Substituting values gives 33°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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Refraction at Spherical Surfaces and Lens Maker's Formula

What is the critical angle for a crown glass (\( n = 1.52 \)) to air interface?

**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Critical angle: sin i_c = (n₂/n₁) . Crown glass ( n₁ = 1.52 ), air ( n₂ = 1 ). sin i_c = (1/1.52) ≈ 0.658 . i_c = sin⁻¹(0.658) ≈ 41.1° . Substituting values gives 41°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refr

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