#physics calculation

Optical Phenomena and Applications

What is the speed of light in a medium if its refractive index is 1.2 and the speed in vacuum is \( 3.0 \times 10^8 \, \

**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. Speed in a medium v = (c/n) . Given n = 1.2 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.2) = 2.5 × 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' = λ/n and A = 2a cos(φ/2), calculation gives

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

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Wave Properties, Frequency and Energy Conservation

What is the distance of the fifth bright fringe from the central maximum in a double-slit experiment if \( \lambda = 490

**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. Bright fringe position x_n = (n λ D/d) . For the fifth bright fringe, n = 5 . λ = 4.9 × 10⁻⁷ m , d = 5.0 × 10⁻⁴ m , D = 1.0 m

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

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Wave Properties, Frequency and Energy Conservation

What is the speed of light in a medium with refractive index 1.25, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**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. Speed in a medium v = (c/n) . Given n = 1.25 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.25) = 2.4 × 10⁸ m/s . Using Δ = d sinθ,

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

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.45, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**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. Speed in a medium v = (c/n) . Given n = 1.45 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.45) ≈ 2.07 × 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,

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

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.1 \times 10^8 \, \text{m/s} \) and in vacuu

**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. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.1 × 10⁸ m/s . n = (3.0 × 10⁸/2.1 × 10⁸) ≈ 1.43 . 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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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of a particle of mass \( 2.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.5 \times 10^6 \, \

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. p = m v = 2.0 × 10⁻³⁰ × 1.5 × 10⁶ = 3.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.0 × 10⁻²⁴) = 2.21 × 10⁻¹⁰ m = 0.221 nm . Applying E = h f = h c/λ, p =

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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De Broglie Wavelength and Matter Waves

Light of frequency \( 8.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with threshold frequency \( 4.0 \times 1

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. E = h v = 6.63 × 10⁻³⁴ × 8.5 × 10¹⁴ = 5.6355 × 10⁻¹⁹ J . Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 4.0 × 10¹⁴ = 2.652 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 5.6355 × 10⁻¹⁹ - 2.652

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Photon Energy, Momentum, Power and X-rays

The threshold frequency of a metal is \( 4.8 \times 10^{14} \, \text{Hz} \). What is the maximum kinetic energy for ligh

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 4.8 × 10¹⁴ = 3.1824 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 6.8 × 10¹⁴ = 4.5084 × 10⁻¹⁹ J . Kₘₐₓ = E

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Photon Energy, Momentum, Power and X-rays

A light source emits \( 3.0 \times 10^{16} \) photons per second with a power of \( 9.0 \, \text{mW} \). What is the wav

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. E = (P/N) = (9.0 × 10⁻³/3.0 × 10¹⁶) = 3.0 × 10⁻¹⁹ J . λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/3.0 × 10⁻¹⁹) = 6.63 × 10⁻⁷ m = 663 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Photon Energy, Momentum, Power and X-rays

A particle of mass \( 1.0 \times 10^{-28} \, \text{kg} \) has the same momentum as a photon of wavelength \( 600 \, \tex

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Photon momentum p = (h/λ) = (6.63 × 10⁻³⁴/600 × 10⁻⁹) = 1.105 × 10⁻²⁷ kg m/s . For particle, p = m v ⇒ v = (p/m) = (1.105 × 10⁻²⁷/1.0 × 10⁻²⁸) = 1.105 × 10¹ =

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Photon Energy, Momentum, Power and X-rays

A light source emits photons of energy \( 4.5 \times 10^{-19} \, \text{J} \) at a rate of \( 2.0 \times 10^{16} \) photo

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.5 × 10⁻¹⁹) = 4.42 × 10⁻⁷ m = 442 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V)

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Photoelectric Effect - Work Function and Threshold

The threshold frequency of a metal is \( 5.0 \times 10^{14} \, \text{Hz} \). What is the stopping potential for light of

**Work function** Φ minimum energy to escape, for Φ=2.2 eV λ₀=1240/2.2=563.6 nm, Φ=3.0 eV threshold frequency f₀=Φ/h=3×1.6×10⁻¹⁹/6.63×10⁻³⁴=7.24×10¹⁴ Hz, stopping potential V₀ = (h f - Φ)/e, for 0.7 V K_max=0.7 eV, for 1.5 V K_max=1.5 eV, intensity increase increases photocurrent not K_max, frequency below threshold no emission regardless of intensity. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.315 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 4.641 × 10⁻¹⁹ - 3.315 × 10⁻¹⁹ = 1.326 × 10⁻¹⁹ J .

Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold

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