Skip to content
Login Register

#light in medium

2 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

What is the wavelength of light in a medium with refractive index 1.25 if its wavelength in air is \( 625 \, \text{nm} \

**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. Wavelength in a medium λ_m = (λₐir/n) . Given λₐir = 625 nm , n = 1.25 , λ_m = (625/1.25) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a

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

View Details
Refraction, Refractive Index and Critical Angle

What is the wavelength of light in a medium with refractive index 1.4 if its wavelength in air is \( 560 \, \text{nm} \)

**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. Wavelength in a medium λ_m = (λₐir/n) . Given λₐir = 560 nm , n = 1.4 , λ_m = (560/1.4) = 400 nm . 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