Table of contents

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

33. Geometric Optics

Refraction of Light & Snell's Law

Physics

33. Geometric Optics

Refraction of Light & Snell's Law

Previous problemNext problem

5:01 minutes

Problem 51

Textbook Question

A light beam strikes a piece of glass at a 55.00° incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively. What is the angle between the two refracted beams?

Verified step by step guidance

Identify the relevant physics principle: This problem involves Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. Snell's Law is expressed as:

n

₁⁢sin⁢θ₁=n₂⁢sin⁢θ₂, where

n

₁ and

n

₂ are the indices of refraction of the two media, and

θ

₁ and

θ

₂ are the angles of incidence and refraction, respectively.

Apply Snell's Law for the first wavelength (450.0 nm): Substitute the given values into Snell's Law. Here,

n

₁=1 (air),

θ

₁=55.00°, and

n

₂=1.4831. Solve for the refracted angle

θ

₂ using:

\sin θ

₂=n₁⁢sin⁢θ₁n₂.

Apply Snell's Law for the second wavelength (700.0 nm): Repeat the same process as in Step 2, but use the index of refraction for 700.0 nm, which is

n

₂=1.4754. Solve for the refracted angle

θ

₂ for this wavelength.

Determine the angle between the two refracted beams: The angle between the two refracted beams is the difference between the two refracted angles calculated in Steps 2 and 3. Subtract the smaller angle from the larger angle to find the result.

Summarize the process: The key steps involve applying Snell's Law for each wavelength, calculating the refracted angles, and then finding the difference between these angles to determine the angle between the two refracted beams.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Snell's Law

Snell's Law describes how light refracts when it passes from one medium to another. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant and is equal to the ratio of the indices of refraction of the two media. This principle is essential for calculating the angles of refraction for different wavelengths of light as they enter the glass.

Index of Refraction

The index of refraction is a dimensionless number that describes how much light slows down and bends when it enters a material. It varies with the wavelength of light, which is why different wavelengths can have different indices of refraction in the same medium. Understanding this concept is crucial for determining how much each wavelength of light will refract when passing through the glass.

Wavelength Dependence of Refraction

Wavelength dependence of refraction refers to the phenomenon where different wavelengths of light are refracted by different amounts when passing through a medium. This occurs because the index of refraction varies with wavelength, leading to phenomena such as dispersion. In this problem, the differing indices of refraction for the two wavelengths will result in a measurable angle between the two refracted beams.

Watch next

Master Snell's Law with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Textbook Question

(I) What is the speed of light in (a) ethyl alcohol, (b) lucite, (c) crown glass?

388

views

Textbook Question

(I) The speed of light in ice is 2.29 x 10⁸ m/s. What is the index of refraction of ice?

841

views

Textbook Question

A 50-year-old man uses +2.5 D lenses to read a newspaper 25 cm away. Ten years later, he must hold the paper 32 cm away to see clearly with the same lenses. What power lenses does he need now in order to hold the paper 25 cm away? (Distances are measured from the lens.)

views

Textbook Question

The critical angle for a certain liquid–air surface is 52.6°. What is the index of refraction of the liquid?

334

views

Textbook Question

(II) A beam of light is emitted in a pool of water from a depth of 65.0 cm. Where must it strike the air–water interface, relative to the spot directly above it, in order that the light does not exit the water?

437

views

Textbook Question

A beam of light is emitted 7.7 cm beneath the surface of a liquid and strikes the surface 7.2 cm from the point directly above the source. If total internal reflection occurs, what can you say about the index of refraction of the liquid?

545

views

Textbook Question

A light beam strikes a 2.5-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.8-cm-thick piece of glass for which n = 1.47. What is the distance D in Fig. 32–51?

221

views

Textbook Question

(II) A 13.0-cm-thick plane piece of glass (n = 1.54) lies on the surface of a 12.0-cm-deep pool of water. How far below the top of the glass does the bottom of the pool seem, as viewed from directly above?

386

views

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap