Refraction Of Light Video Lessons

Concept

# Problem: The refractive index of silica versus wavelength can be described using: Where A1 = 0.696749, A2 = 0.408218, A3 = 0.890815, λ1 = 0.0690660 µm, λ2 = 0.115662 µm, λ3 = 9.900559 µm(a) Consider the case of oblique incidence from air to silica. If the incident light is in air and its incidence angle is θi = 89°, calculate the transmission angle θt if the optical wavelength in air is 400 nm, 600 nm, and 1.5 µm(b) An optical beam encounters a right triangle made of silica as shown in (a) in normal incidence configuration. The input power is Pi = 10 mW. Determine the output beam power P0. Perform the calculation at three different wavelengths as above.

###### FREE Expert Solution

From Snell's law:

$\overline{){{\mathbf{\eta }}}_{{\mathbf{i}}}{{\mathbf{sin\theta }}}_{{\mathbf{i}}}{\mathbf{=}}{{\mathbf{\eta }}}_{{\mathbf{t}}}{{\mathbf{sin\theta }}}_{{\mathbf{t}}}}$

(a)

Refractive index at λ = 400 nm = 0.4 μm

$\begin{array}{rcl}\mathbf{\eta }& \mathbf{=}& \sqrt{\mathbf{1}\mathbf{+}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{696749}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{0690660}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{408218}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{115662}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{890815}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}{\mathbf{4}}^{\mathbf{2}}\mathbf{-}\mathbf{9}\mathbf{.}{\mathbf{900559}}^{\mathbf{2}}}\mathbf{\right)}}\end{array}$

η = 1.47

From Snell's law:

$\begin{array}{rcl}{\mathbf{\theta }}_{\mathbf{t}}& \mathbf{=}& \mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{{\mathbf{n}}_{\mathbf{i}}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{\theta }}_{\mathbf{i}}}{{\mathbf{\eta }}_{\mathbf{t}}}\mathbf{\right]}\\ & \mathbf{=}& \mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{89}\mathbf{°}\mathbf{\right)}}{\mathbf{1}\mathbf{.}\mathbf{47}}\mathbf{\right]}\end{array}$

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###### Problem Details

The refractive index of silica versus wavelength can be described using:

Where A1 = 0.696749, A2 = 0.408218, A3 = 0.890815, λ1 = 0.0690660 µm, λ2 = 0.115662 µm, λ3 = 9.900559 µm

(a) Consider the case of oblique incidence from air to silica. If the incident light is in air and its incidence angle is θi = 89°, calculate the transmission angle θt if the optical wavelength in air is 400 nm, 600 nm, and 1.5 µm

(b) An optical beam encounters a right triangle made of silica as shown in (a) in normal incidence configuration. The input power is Pi = 10 mW. Determine the output beam power P0. Perform the calculation at three different wavelengths as above.