#glass

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'

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

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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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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 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

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

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Total Internal Reflection and Critical Angle

What is the critical angle for a diamond (\( n = 2.42 \)) to glass (\( n = 1.5 \)) interface?

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. Critical angle: sin i_c = (n₂/n₁) . Diamond ( n₁ = 2.42 ), glass ( n₂ = 1.5 ). sin i_c = (1.5/2.42) ≈ 0.620 . i_c = sin⁻¹(0.620) ≈ 38.4° . Substituting values gives 38°, 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.

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

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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

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

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Refraction at Plane Surfaces and Snell's Law

What is the critical angle for a glass (\( n = 1.5 \)) to water (\( n = 1.33 \)) interface?

**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. Critical angle: sin i_c = (n₂/n₁) . Glass ( n₁ = 1.5 ), water ( n₂ = 1.33 ). sin i_c = (1.33/1.5) ≈ 0.887 . i_c = sin⁻¹(0.887) ≈ 62.5° . Substituting values gives 62°, 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.

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

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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

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

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Total Internal Reflection and Critical Angle

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

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. Critical angle: sin i_c = (n₂/n₁) . Glass ( n₁ = 1.52 ), air (コンパクト 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 refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

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