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monochromatic light of wavelength λ=800nm illuminates normally a thin liquid film with nf =4/3 and t=500nm lying on a glass of refractive index n=2.5 The liquids refractive index can be continuously increased up to some limiting value by applying a variable voltage across it.

a) If the applied voltage increases nf from (4/3) to(3.0),how many times and for what values of nf does the surface of the film become very bright during this process?

b) If the refractive index is kept constant at nf =2.0 ,calculate the intensity of reflected light in terms of intensity I0 of the incident light.

User Icon for Explain Bot Explain Bot answered
1 year ago

a) To determine when the surface of the film becomes very bright, we need to consider the conditions for constructive interference. Constructive interference occurs when the change in phase between the two reflected rays is an integer multiple of 2π.

The phase change for a reflection at a thin film depends on the refractive indices and the thickness of the film. In this case, the film has a refractive index of nf and a thickness of t.

The phase change can be calculated using the formula:
Δφ = (2π/λ) * 2nt

Initially, nf = 4/3, so the initial phase change is:
Δφ_init = (2π/λ) * 2(4/3)t

When nf increases to 3.0, the phase change becomes:
Δφ_final = (2π/λ) * 2(3.0)t

For the surface to become very bright, the phase change should be an integer multiple of 2π:
Δφ_final - Δφ_init = 2π * n

Substituting the values, we have:
(2π/λ) * 2(3.0)t - (2π/λ) * 2(4/3)t = 2π * n

Simplifying the equation, we get:
(2π/λ) * 2(3/3)t - (2π/λ) * 2(4/3)t = 2π * n
(2π/λ) * 2t - (2π/λ) * (8/3)t = 2π * n
(2π/λ) * (6/3)t - (2π/λ) * (8/3)t = 2π * n
(2π/λ) * (-2/3)t = 2π * n
(-4/3) * (2/λ)t = n

Simplifying further, we get:
(8/9) * (1/λ)t = n

Since nf is increasing from 4/3 to 3.0, the surface becomes bright when the limiting value nf is reached. Therefore, we need to find the values of nf that satisfy:
nf >= (8/9) * (1/λ)t

b) When the refractive index is kept constant at nf = 2.0, we can calculate the intensity of the reflected light in terms of the intensity of the incident light, I0.

The intensity of the reflected light can be determined using the equation:
Intensity_reflected = (|(n - nf)/(n + nf)|)^2 * I0

Substituting the values, we have:
Intensity_reflected = (|(2.5 - 2.0)/(2.5 + 2.0)|)^2 * I0
= (|0.5/4.5|)^2 * I0
= (1/9)^2 * I0
= 1/81 * I0

Therefore, the intensity of the reflected light is (1/81) times the intensity of the incident light.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

a) To determine the values of nf for which the surface of the film becomes very bright, we need to consider the conditions for constructive interference.

When monochromatic light passes through a thin film, it undergoes both reflection and refraction at the interfaces. Constructive interference occurs when the path difference between the reflected and transmitted waves is a whole number of wavelengths.

The path difference (Δp) is given by:
Δp = 2nt - 2nf t cosθ

In this case:
nt = 2.5 (refractive index of the glass)
t = 500 nm (thickness of the liquid film)
θ = 0° (illumination is normal)

For constructive interference, Δp should be an integer multiple of the wavelength λ.

Let's calculate the values for nf:

For nf = 4/3:
Δp = 2(2.5)(500 nm) - 2(4/3)(500 nm) cos(0°)
= 2500 nm - (4/3)(500 nm)
= 2500 nm - 666.67 nm
= 1833.33 nm

Since 1833.33 nm is not a whole number multiple of 800 nm, the surface does not appear very bright for nf = 4/3.

For nf = 3.0:
Δp = 2(2.5)(500 nm) - 2(3.0)(500 nm) cos(0°)
= 2500 nm - (3.0)(500 nm)
= 2500 nm - 1500 nm
= 1000 nm

Since 1000 nm is a whole number multiple of 800 nm (1000 nm = 1.25λ), the surface appears very bright for nf = 3.0.

Therefore, the surface becomes very bright once for nf = 3.0 during this process.

b) When the refractive index is kept constant at nf = 2.0, we can calculate the intensity of reflected light (I) in terms of the intensity of incident light (I0) using the Fresnel equations.

The reflectance (R) can be calculated as R = [(n1 - n2)/(n1 + n2)]^2, where n1 is the refractive index of the first medium (air) and n2 is the refractive index of the second medium (glass).

In this case:
n1 = 1 (refractive index of air)
n2 = 2.5 (refractive index of glass)

R = [(1 - 2.5)/(1 + 2.5)]^2
= [-1.5/3.5]^2
= 0.51^2
= 0.2601

The intensity of reflected light (I) is given by I = R * I0.

Therefore, the intensity of reflected light in terms of the intensity of incident light is 0.2601 * I0.