Skip to content
Login Register

#light source

5 public questions tagged with this topic.

Superposition, Resultant Amplitude and Intensity

Why can’t two ordinary light sources, like sodium lamps, produce a stable interference pattern when illuminating two sli

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Ordinary sources emit light with rapidly changing phase differences, making them incoherent and unable to maintain a stable interference pattern. 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 They are incoherent due to random

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

View Details
Photon Energy, Momentum, Power and X-rays

A light source emits photons of energy \( 3.0 \times 10^{-19} \, \text{J} \) at a rate of \( 4.0 \times 10^{15} \) photo

**Photon power** light source emits 1.0×10¹⁶ photons/s power 4.0 mW, energy per photon E= P/N =4×10⁻³/10¹⁶=4×10⁻¹⁹ J, wavelength λ= h c/E=6.63×10⁻³⁴×3×10⁸/4×10⁻¹⁹=497 nm, for 6×10¹⁵ photons/s 3.0 mW E=5×10⁻¹⁹ J, for beam 2.5×10¹⁵ photons/s 4×10⁻¹⁹ J each power=10⁻³ W=1 mW, illustrating P = N h f. Power P = N × E . P = 4.0 × 10¹⁵ × 3.0 × 10⁻¹⁹ = 1.2 × 10⁻³ W = 1.2 mW . 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) and f_max = e V/h,

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

View Details
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

View Details
Microscopy

In phase contrast microscopy, what is used to generate a hollow cone of light?

Phase contrast requires annular illumination to separate background light from light scattered by the specimen. A phase annulus, a circular slit located in the condenser front focal plane, produces a hollow cone of rays focused on the sample. The direct hollow cone and the diffracted rays from cellular structures travel different paths. At the objective, a complementary phase plate with a ring retards the direct light by quarter wavelength. Their interference converts small phase differences caused by refractive index variations into amplitude contrast, enabling unstained cells to be visualize

Ref: NCERT Biology Class XII Principles on Klenow fill-in labeling, Lehninger Chapter 9 DNA cloning techniques, and Molecular Cloning by Sambrook Chapter 10 documenting end-labeling of cohesive termini.

View Details