#Snell's law

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

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

View Details

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,

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

View Details

Refraction, Refractive Index and Critical Angle

Why does the angle of refraction increase beyond 90° become impossible when light moves from a denser to a rarer medium?

**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 sine of the refraction angle exceeds 1, which is mathematically 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), calculati

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

View Details

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 =

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

View Details

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

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

View Details

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

View Details

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

View Details

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

View Details

Refraction Through Prism and Minimum Deviation

A ray of light passes from water (\( n = 1.33 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 45^\circ \).

**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. Snell’s law: n₁ sin i = n₂ sin r . Water ( n₁ = 1.33 ), glass ( n₂ = 1.62 ), i = 45° . 1.33 × sin 45° = 1.62 × sin r . sin 45° = 0.707 ⇒ 1.33 × 0.707 = 1.62 sin r ⇒ 0.941 = 1.62 sin r . sin r = (0.941/1.62) ≈ 0.581 ⇒ r = sin⁻¹(0.581) ≈ 35.5° . Substituting values gives 36°, which matches expected

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

View Details

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

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

View Details

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

View Details

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

View Details