#air

Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 480 \, \text{nm} \) in air, given the speed of light in air is \(

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Frequency nu = (c/λ) . λ = 4.8 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/4.8 × 10⁻⁷) = 6.25 × 10¹⁴ Hz . Using Δ = d

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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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 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

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

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 water-air interface (\( n_{\text{water}} = 1.33 \))?

**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) . Water ( n₁ = 1.33 ), air ( n₂ = 1 ). sin i_c = (1/1.33) ≈ 0.752 . i_c = sin⁻¹(0.752) ≈ 48.75° . Substituting values gives 49°, 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 refracti

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

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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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Wave Speed, Energy and Power

A sound wave in air has a frequency of 170 Hz and wavelength of 2 m. What is the speed of sound in air?

**Transverse wave velocity** depends on medium not frequency alone. For string under tension, v ∝ √(T/μ), calculation requires μ from mass and length, then square root evaluation, giving v in m/s, then f = v/λ for given wavelength. Speed: v = v λ = 170 × 2 = 340 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 340 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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Degrees of Freedom and Molar Specific Heat

The mean free path of air molecules at STP is 2.9 × 10⁻⁷ m. If temperature doubles at constant pressure, what is the new

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. l = (1)/(√(2) n π d²), n = (N)/(V), and PV = Nk_B T.At constant P, V ∝ T, so n ∝ (1)/(T). If T doubles, n halves, l doubles.New l = 2 × 2.9 × 10⁻⁷ = 5.8 × 10⁻⁷ m . Substituting values gives 5.8 × 10⁻⁷ m, which matches expected

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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