Question

A parallel beam of light containing orange (\(610{\rm{ }}nm\)) and violet (\(410{\rm{ }}nm\)) wavelengths goes from fused quartz to water, striking the surface between them at a\({60.0^ \circ }\)incident angle. What is the angle between the two colours in water?

Short answer

The angle between the two colors is obtained as: \({0.12^o}\).

Step-by-step solution

Define Geometric Optics

Geometrical optics, often known as ray optics, is an optics model that describes light propagation using rays. In geometric optics, a ray is an abstraction that can be used to approximate the routes along which light propagates under particular conditions.

Given Data

The angle striking the surface between both is: \({60.0^ \circ }\)

The value of orange light is:\(610{\rm{ }}nm\).

The value of violet light is: \(410{\rm{ }}nm\).

Evaluating the angles

To evaluate the angle of refraction for each light color using Snell's law:\({n_i}\sin {\theta _i}{\rm{ }} = {\rm{ }}{n_f}\sin {\theta _f}\)

The angle evaluated for orange light is:

\(\begin{array}{c}{\theta _f} = {\sin ^{ - 1}}\left( {\frac{{{n_i}\sin {\theta _i}}}{{{n_f}}}} \right)\\{\theta _f} = {\sin ^{ - 1}}\left( {\frac{{1.456{\rm{ }} \times {\rm{ }}\sin {{60}^ \circ }}}{{1.332}}} \right)\\ = {71.20^ \circ }\end{array}\)

The angle evaluated for violet light is:

\(\begin{array}{c}{\theta _f} = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{n_i}{\rm{sin}}{\theta _i}}}{{{n_f}}}} \right)\\{\theta _f} = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{1}}{\rm{.468 }} \times {\rm{ sin6}}{{\rm{0}}^ \circ }}}{{{\rm{1}}{\rm{.332}}}}} \right)\\ = {\rm{71}}{\rm{.3}}{{\rm{2}}^ \circ }\end{array}\)

Evaluating the difference between both the colors

The difference between both the colours is evaluated as:

\(\begin{array}{c}\theta = {\rm{71}}{\rm{.3}}{{\rm{2}}^ \circ } - {\rm{71}}{\rm{.}}{{\rm{2}}^ \circ }\\ = {\rm{ 0}}{\rm{.1}}{{\rm{2}}^ \circ }\end{array}\)

Therefore, the difference is: \({0.12^o}\).