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#angle of incidence

17 public questions tagged with this topic.

Wavefront and Huygens Principle

Why does the reflected wavefront from a plane surface maintain the same angle as the incident wavefront?

**Wavefront** is locus of points in same phase, spherical from point source, plane at large distance because radius large, Huygens principle every point on wavefront acts as secondary source of wavelets, new wavefront envelope of secondary wavelets, allows prediction of new wavefront shape from known wavefront, explains reflection and refraction. The wave model shows that secondary wavelets from the surface form a reflected wavefront where the angle of incidence equals the angle of reflection due to geometric symmetry. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC =

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Wavefront and Huygens Principle

What ensures that the laws of reflection hold true when derived using the wave model?

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. The wave model uses secondary wavelets, where the reflected wavefront’s envelope forms when the incident and reflected angles are equal, satisfying geometric congruence. 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), calcula

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Telescope, Human Eye and Defects of Vision

A ray of light passes from air (\( n = 1 \)) to glass (\( n = 1.5 \)) at an angle of incidence of \( 45^\circ \). What i

**Astronomical telescope** in normal adjustment M = -f_o/f_e, f_o objective focal length (m), f_e eyepiece focal length, length L = f_o+f_e, final image at infinity, angular magnification ratio of angles subtended by image and object. For f_o=100 cm, f_e=5 cm, M=-20, inverted, used for celestial observation. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), glass ( n₂ = 1.5 ), i = 45° . 1 × sin 45° = 1.5 × sin r . sin 45° = 0.707 ⇒ 0.707 = 1.5 sin r ⇒ sin r = (0.707/1.5) ≈ 0.471 . r = sin⁻¹(0.471) ≈ 28.1°

Ref: NCERT > Physics Book > Ray Optics > Telescope, Human Eye and Defects of Vision

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Telescope, Human Eye and Defects of Vision

A ray of light is incident at \( 60^\circ \) on a glass-air interface (refractive index of glass = 1.5). What is the ang

**Human eye** least distance D=25 cm, near point, far point infinity for normal eye, accommodation by ciliary muscles changing lens curvature, power ≈60 D total. Defects: myopia far point 25 cm corrected by converging lens, astigmatism cylindrical lens. Using Snell’s law: n₁ sin i = n₂ sin r . Glass ( n₁ = 1.5 ), air ( n₂ = 1 ), i = 60° . 1.5 sin 60° = 1 sin r . sin 60° = (√(3)/2) ≈ 0.866 ⇒ 1.5 × 0.866 = 1.299 . sin r = 1.299 > 1 , which is impossible, so

Ref: NCERT > Physics Book > Ray Optics > Telescope, Human Eye and Defects of Vision

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Telescope, Human Eye and Defects of Vision

A ray of light passes from air into water (\( n = 1.33 \)) at an angle of incidence of \( 45^\circ \). What is the angle

**Human eye** least distance D=25 cm, near point, far point infinity for normal eye, accommodation by ciliary muscles changing lens curvature, power ≈60 D total. Defects: myopia far point 25 cm corrected by converging lens, astigmatism cylindrical lens. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), water ( n₂ = 1.33 ), i = 45° . 1 × sin 45° = 1.33 × sin r . sin 45° = (1/√(2)) ≈ 0.707 ⇒ 0.707 = 1.33 sin r . sin r = (0.707/1.33) ≈ 0.532 ⇒ r = sin⁻¹(0.532)

Ref: NCERT > Physics Book > Ray Optics > Telescope, Human Eye and Defects of Vision

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Refraction Through Prism and Minimum Deviation

A ray of light passes from air (\( n = 1 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 60^\circ \). What

**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 . Air ( n₁ = 1 ), water ( n₂ = 1.33 ), i = 60° . 1 × sin 60° = 1.33 × sin r . sin 60° = 0.866 ⇒ 0.866 = 1.33 sin r ⇒ sin r = (0.866/1.33) ≈ 0.651 . r = sin⁻¹(0.651) ≈ 40.6° . Substituting values gives 41°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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Refraction Through Prism and Minimum Deviation

A ray of light passes from air (\( n = 1 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 30^\circ \). What

**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 . Air ( n₁ = 1 ), glass ( n₂ = 1.62 ), i = 30° . 1 × sin 30° = 1.62 × sin r . sin 30° = 0.5 ⇒ 0.5 = 1.62 sin r ⇒ sin r = (0.5/1.62) ≈ 0.309 . r = sin⁻¹(0.309) ≈ 18° . Substituting values gives 18°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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Total Internal Reflection and Critical Angle

A ray of light passes from water (\( n = 1.33 \)) to air at an angle of incidence of \( 50^\circ \). What happens?

**Total internal reflection** occurs when light travels from denser to rarer medium and incidence > C, condition sinC = 1/n for air interface. For glass n=1.52 C≈41°, water n=1.33 C≈48.8°, so at 49° water-air TIR occurs, explaining why ray does not emerge. Critical angle: sin i_c = (n₂/n₁) = (1/1.33) ≈ 0.752 ⇒ i_c ≈ 48.75° . Since i = 50° > i_c , total internal reflection occurs. Substituting values gives Total internal reflection, 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

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Total Internal Reflection and Critical Angle

A ray of light passes from glass (\( n = 1.52 \)) to air at an angle of incidence of \( 45^\circ \). What happens?

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Critical angle: sin i_c = (n₂/n₁) = (1/1.52) ≈ 0.658 ⇒ i_c ≈ 41.1° . Since i = 45° > i_c , total internal reflection occurs. Substituting values gives Total internal reflection, 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 conventio

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Total Internal Reflection and Critical Angle

A ray of light passes from air (\( n = 1 \)) to glass (\( n = 1.5 \)) at an angle of incidence of \( 40^\circ \). What i

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), glass ( n₂ = 1.5 ), i = 40° . 1 × sin 40° = 1.5 × sin r . sin 40° ≈ 0.643 ⇒ 0.643 = 1.5 sin r ⇒ sin r = (0.643/1.5) ≈ 0.429 . r = sin⁻¹(0.429) ≈ 25.4° . Substituting values gives 25°, which matches

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 60^\circ \) and refractive index \( 1.5 \) produces a minimum deviation of \( 30^\circ \). What is t

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. At minimum deviation: i = (A + D_m/2) . A = 60° , D_m = 30° . i = (60 + 30/2) = (90/2) = 45° . Substituting values gives 45°, 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 Plane Surfaces and Snell's Law

A ray of light passes from glass (\( n = 1.5 \)) to air at an angle of incidence of \( 40^\circ \). What is the angle of

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Snell’s law: n₁ sin i = n₂ sin r . Glass ( n₁ = 1.5 ), air ( n₂ = 1 ), i = 40° . 1.5 × sin 40° = 1 × sin r . sin 40° ≈ 0.643 ⇒ 1.5 × 0.643 ≈ 0.964 ⇒ sin r = 0.964 . r = sin⁻¹(0.964) ≈ 74.6° . Critical angle: sin i_c

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