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#incident angle

3 public questions tagged with this topic.

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

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

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Refraction Through Prism and Minimum Deviation

A ray of light passes from glass (\( n = 1.5 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 30^\circ \). W

**Prism formula** for small A, δ_m≈(n-1)A, for 30°, n=1.6, δ_m≈0.6×30°=18°, approximate, exact using sin formula. For 45°, n=1.6, sin[(45+δ_m)/2]=1.6×sin22.5°=1.6×0.3827=0.6123, (45+δ_m)/2=37.8°, δ_m=30.6°, showing deviation increases with A and n. Snell’s law: n₁ sin i = n₂ sin r . Glass ( n₁ = 1.5 ), water ( n₂ = 1.33 ), i = 30° . 1.5 × sin 30° = 1.33 × sin r . sin 30° = 0.5 ⇒ 1.5 × 0.5 = 1.33 sin r ⇒ 0.75 = 1.33 sin r . sin r = (0.75/1.33) ≈ 0.564 ⇒ r = sin⁻¹(0.564) ≈ 34.3° . Substituting values gives 34°, which matches expected image position

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation

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Refraction at Plane Surfaces and Snell's Law

A ray of light passes from air to crown glass (\( n = 1.52 \)) at an angle of incidence of \( 60^\circ \). What is the a

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), glass ( n₂ = 1.52 ), i = 60° . 1 × sin 60° = 1.52 × sin r . sin 60° = 0.866 ⇒ 0.866 = 1.52 sin r ⇒ sin r = (0.866/1.52) ≈ 0.57 . r = sin⁻¹(0.57) ≈ 34.8° . Substituting values gives 35°, which matches expected image position and magnification from

Ref: NCERT > Physics Book > Ray Optics > Refraction at Plane Surfaces and Snell's Law

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