Fermat’s principle slates that “light travels between two points along the path that requires the least time, as compared to other nearby paths.” From Fermat’s principle derive (a) the law of reflection (θi = θr) and (b) the law of refraction (Snell’s law). [Hint: Choose two appropriate points so that a ray between them can undergo reflection or refraction. Draw a rough path for a ray between these points, and write down an expression of the time required for light to travel the arbitrary path chosen. Then take the derivative to find the minimum.]
Want to see the full answer?
Check out a sample textbook solutionChapter 32 Solutions
PHYSICS F./SCI... W/MOD V.II W/KIT
Additional Science Textbook Solutions
Life in the Universe (4th Edition)
Essential University Physics: Volume 2 (3rd Edition)
University Physics with Modern Physics (14th Edition)
Conceptual Integrated Science
University Physics Volume 2
An Introduction to Thermal Physics
- Light traveling in a medium of index of refraction n1 is incident on another medium having an index of refraction n2. Under which of the following conditions can total internal reflection occur at the interface of the two media? (a) The indices of refraction have the relation n2 n1. (b) The indices of refraction have the relation n1 n2. (c) Light travels slower in the second medium than in the first. (d) The angle of incidence is less than the critical angle. (e) The angle of incidence must equal the angle of refraction.arrow_forward(a) In the figure, light from ray A refracts from material 1 into a thin layer of material 2, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (i) What is the value of incident angle θA? Draw a sketch of the situation. (ii) If θA is decreased, does part of the light refract into material 3? (b) Light from ray B refracts from material 1 into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (iii) What is the value of incident angle θB? Draw a sketch of the situation. (iv) If θB is decreased, does part of the light refract into material 3? Answer: 54.3°, yes, 51.1°, noarrow_forwardIn the figure, light from ray A refracts from material 1 (n1 = 1.60) into a thin layer of material 2 (n2 = 1.80), crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3 (n3 = 1.3). (a) What is the value of incident angle θA? (b) If θA is decreased, does part of the light refract into material 3? Light from ray B refracts from material 1 into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (c) What is the value of incident angle θB? (d) If θB is decreased, does part of the light refract into material 3?arrow_forward
- A ray of light is incident upon a plane with an angle of (3.60x10^1)° relative to a line perpendicular to its surface (called the normal). What is its angle of reflection (in deg) relative to the surface? Note: Your answer is assumed to be reduced to the highest power possible.arrow_forwardA ray of light crosses the boundary between some substance with n = 1.46 and air, going from the substance into air. If the angle of incidence is 26◦ what is the angle of refraction? Calculate to 1decimal.arrow_forwardWhen a ray of light strikes a nonmetallic surface, some of the light is reflected and some is refracted. Brewster’s angle is the angle of incidence for which the reflected and refracted beams are 90° apart. The significance of this angle is that it leads to a reflected ray which is completely polarized.(a) Sketch the incident, reflected, and refracted rays for a situation involving Brewster’s angle.(b) Compute Brewster’s angle for water. (Hint: the index of refraction of water is 1.33 and you might find the trigonometric identity sin(90° − θ ) = cosθ to be useful.)arrow_forward
- In the figure, light from ray A refracts from material 1 (n₁ = 1.73) into a thin layer of material 2 (n2 = 1.80), crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3 (n3 = 1.40). (a) What is the value of incident angle BA? (b) If 8A is decreased, does part of the light refract into material 3? Light from ray B refracts from material 1 into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (c) What is the value of incident angle Og? (d) If Og is decreased, does part of the light refract into material 3? OB I ng no 121arrow_forwardDerive the optical law of reflection. Hint: Let light go from the point A = (x1, Y1) to B = (x2, Y2) via an arbitrary point P = (x, 0) on a mirror along the x axis. Set dt/dx = (n/c)dD/dx = 0, where D = distance APB, and show that then 0 = 4. 1. Barrow_forwardA ray of light is incident on an air/water interface. The ray makes an angle of θ1 = 34 degrees with respect to the normal of the surface. The index of the air is n1 = 1 while water is n2 = 1.33. Part (a) Numerically, what is the angle in degrees? θ2 =? Part (b) Write an expression for the reflection angle ψ, with respect to the surface. ψ =? Part (c) Numerically, what is this angle in degrees? ψ =?arrow_forward
- A light ray falls on the left face of a prism (see below) at the angle of incidence 0 for which the emerging beam also has an angle of refraction 0 at the right face. Use geometry and trigonometry to show in detail that the index of refraction n of the prism is given by sin /(a+b) where is the vertex angle of the prism and a is the angle through which the n = sin beam has been deviated due to the prism (a is known to be the angle of deviation). If a = 36.0° and the two base angles of the prism are each 50.0°, what is n?arrow_forward(Snell's Law) Let v be the velocity of light in air and w be the velocity of light in water. According to Fermat's principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken, Find the ratio sin(01) sin(02) in terms of v and w. A C 02 Barrow_forwardA narrow beam of light is incident from air onto a polystyrene surface with index of refraction 1.49. Find the angle of incidence for which the corresponding angle of refraction is one-half the angle of incidence. (Hint: You might want to use the trigonometric identity sin (20) = 2 sin(0) cos(0).)arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning