Question

A narrow beam of light containing red (\(660{\rm{ }}nm\)) and blue (\(470{\rm{ }}nm\)) wavelengths travels from air through a\(1.00{\rm{ }}cm\)thick flat piece of crown glass and back to air again. The beam strikes at a\({30.0^ \circ }\)incident angle. (a) At what angles do the two colors emerge? (b) By what distance are the red and blue separated when they emerge?

Short answer

1. The red angle is obtained as:\({\rm{19}}{\rm{.3}}{{\rm{1}}^{\rm{o}}}\)and blue angle is obtained as:\({\rm{19}}{\rm{.1}}{{\rm{5}}^{\rm{o}}}\).
2. The distance at which both the colors are separated is obtained as: \({\rm{0}}{\rm{.03}}\;{\rm{mm}}\).

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 incident angle is: \({30.0^ \circ }\)

The thickness of piece of crown glass is:\(1.00{\rm{ }}cm\).

The value of red light is:\(660{\rm{ }}nm\).

The value of blue light is: \(470{\rm{ }}nm\).

Evaluating the angles

The angle of refraction of each color can be obtained using the Snell's law:

\({n_i}\sin {\theta _i}{\rm{ }} = {\rm{ }}{n_f}\sin {\theta _f}\)

The angle of red light is:

\(\begin{array}{c}{\theta _{\rm{2}}}{\rm{ = sin(}}\frac{{{n_{\rm{1}}}{\rm{sin(}}{\theta _{\rm{1}}}{\rm{)}}}}{{{n_2}}}{\rm{)}}\\{\rm{ = sin}}\left( {\frac{{{\rm{1}}{\rm{.000}} \times {\rm{sin(30}}{\rm{.0}}{{\rm{0}}^{\rm{o}}}{\rm{)}}}}{{{\rm{1}}{\rm{.512}}}}} \right)\\{\rm{ = 19}}{\rm{.3}}{{\rm{1}}^{\rm{o}}}\end{array}\)

Therefore, the angle of red light is:\({\rm{19}}{\rm{.3}}{{\rm{1}}^{\rm{o}}}\).

The angle of BLUE light is:

\(\begin{array}{c}{\theta _{\rm{2}}}{\rm{ = sin(}}\frac{{{n_1}{\rm{sin(}}{\theta _{\rm{1}}}{\rm{)}}}}{{{n_1}}}{\rm{)}}\\{\rm{ = sin}}\left( {\frac{{{\rm{1}}{\rm{.000}} \times {\rm{sin(30}}{\rm{.0}}{{\rm{0}}^{\rm{o}}}{\rm{)}}}}{{{\rm{1}}{\rm{.524}}}}} \right)\\{\rm{ = 19}}{\rm{.1}}{{\rm{5}}^{\rm{o}}}\end{array}\)

Therefore, the angle of blue light is: \({\rm{19}}{\rm{.1}}{{\rm{5}}^{\rm{o}}}\).

Evaluating the distance

In the trigonometry, we know that the tangent of an angle is equal to the opposite over the adjacent, which means that:

\({\rm{tan}}\theta {\rm{ = }}\frac{{\Delta y}}{{\Delta x}}\).

Here, the adjacent depicted as\(\Delta x\)is said to be the thickness of the glass piece, and the opposite is depicted as\(\Delta y\)is said to be the horizontal displacement for each color from the normal to the point at which each color has entered (incident) the crown glass.

So, by knowing the refraction angle for each light color\({\theta _{\rm{f}}}\)from the part (a). Then we solve for\(\Delta y\);

Then, for the red color, we get:

\(\begin{array}{c}\Delta {\rm{y}}{}_{\rm{r}}{\rm{ = }}\Delta {\rm{x}} \times {\rm{tan(}}{\theta _{\rm{f}}}{\rm{)}}\\{\rm{ = 0}}{\rm{.01}}\;{\rm{m}} \times {\rm{tan(19}}{\rm{.3}}{{\rm{1}}^{\rm{o}}}{\rm{)}}\\{\rm{ = 3}}{\rm{.50}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{m}}\\{\rm{ = 3}}{\rm{.50}}\;{\rm{mm}}\end{array}\)

For the blue color, we obtain:

\(\begin{array}{c}\Delta y{}_b = \Delta x \times \tan ({\theta _f})\\ = 0.01\;m \times \tan ({19.15^o})\\ = 3.47 \times {10^{ - 3}}\;m\\ = 3.47\;mm\end{array}\)

The distance by which the red and blue lights are separated when they emerge is obtained as:

\(3.50\;mm{\rm{ }} - {\rm{ }}3.47\;mm{\rm{ }} = {\rm{ }}0.03\;mm\)

Therefore, the distance is: \(0.03\;mm\).