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#angle of incidence

15 public questions tagged with this topic.

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°

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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

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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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Refraction at Plane Surfaces and Snell's Law

A ray of light is incident from air (\( n = 1 \)) into glass (\( n = 1.5 \)) at \( 30^\circ \). What is the angle of ref

**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 = 1.5 ), i = 30° . 1 × sin 30° = 1.5 × sin r . sin 30° = 0.5 ⇒ 0.5 = 1.5 sin r ⇒ sin r = (0.5/1.5) = 0.333 . r = sin⁻¹(0.333) ≈ 19.5° . Substituting values gives 19°, which matches expected image position and magnification from

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Refraction at Plane Surfaces and Snell's Law

A ray of light passes from air (\( n = 1 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 35^\circ \). What

**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 . Air ( n₁ = 1 ), water ( n₂ = 1.33 ), i = 35° . 1 × sin 35° = 1.33 × sin r . sin 35° ≈ 0.574 ⇒ 0.574 = 1.33 sin r ⇒ sin r = (0.574/1.33) ≈ 0.432 . r = sin⁻¹(0.432) ≈ 25.6° . Substituting values

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