Question

A light of wavelength of 550 nm is incident at interface of a medium of refractive index of n_1=1.5 and another medium of refractive index of n_2=1 at an angle of 60(in degree).Calculate the penetration depth of evanescent field in 2^(nd ) medium if the wavelength of incident light is reduced to half of its initial value?

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Answer 1

Given:

λ = 550 nm

μ1  = 1.5

μ2 = 1

θ = 60 °

To find:

d =?

Explanation:

Let $\displaystyle \alpha _2$ be the attenuation coefficient penetration in medium 2.

$\displaystyle \alpha _2 = \frac{2\pi n_2}{\lambda} \sqrt{(\frac{n_1}{n_2} )^{2} sin^2\theta -1 }$

$\displaystyle \alpha _2 = \frac{2\times 3.14 \times1}{500\times 10^{-9}} \sqrt{(\frac{1.5}{1} )^{2} sin^2(60) -1 }$

$\displaystyle \mathbf{\alpha _2}$ =  8.636 $\mathbf{\times 10^6}$

But the penetration depth is the inverse of the attenuation coefficient.

∴$\displaystyle \delta = \frac{1}{\alpha _2}$

     = $\displaystyle \frac{1}{8.636 \times 10^6}$

 δ = 0.1157 $\times 10^{-6$

δ = 0.12 μm