Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
  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

01

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.

02

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: \(470{\rm{ }}nm\).

03

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: \({\rm{19}}{\rm{.1}}{{\rm{5}}^{\rm{o}}}\).

04

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: \(0.03\;mm\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free