Two different glasses - glass A and glass B - are separated by a planar boundary as shown in the figure. y (cm) 9 8 7 6 5 4 3 2 1 1 2 3 glass A 4 glass B 5 6 x (cm) 7 8 9 A light ray shown in the lower left part of the figure - approaches and then hits the boundary. Part of the ray reflects back to glass A, part of the ray enters glass B. The index of refraction of glass A is nA = 1.100. Determine the index of refraction of glass B. (Notice that the lines representing the light rays travel through grid intersection points.)

Two different glasses - glass A and glass B - are separated by a planar boundary as shown in the figure. y (cm) 9 8 7 6 5 4 3 2 1 1 2 3 glass A 4 glass B 5 6 x (cm) 7 8 9 A light ray shown in the lower left part of the figure - approaches and then hits the boundary. Part of the ray reflects back to glass A, part of the ray enters glass B. The index of refraction of glass A is nA = 1.100. Determine the index of refraction of glass B. (Notice that the lines representing the light rays travel through grid intersection points.)

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter38: Refraction And Images Formed By Refraction

Section: Chapter Questions

Problem 20PQ

See similar textbooks

Similar questions

Curved glassair interfaces like those observed in an empty shot glass make it possible for total internal reflection to occur at the shot glasss internal surface. Consider a glass cylinder (n = 1.54) with an outer radius of 2.50 cm and an inner radius of 2.00 cm as shown in Figure P38.105. Find the minimum angle i such that there is total internal reflection at the inner surface of the shot glass. FIGURE P38.105 Problems 105 and 106.
An object is placed a distance of 10.0 cm to the left of a thin converging lens of focal length f = 8.00 cm, and a concave spherical mirror with radius of curvature +18.0 cm is placed a distance of 45.0 cm to the right of the lens (Fig. P38.129). a. What is the location of the final image formed by the lensmirror combination as seen by an observer positioned to the left of the object? b. What is the magnification of the final image as seen by an observer positioned to the left of the object? c. Is the final image formed by the lensmirror combination upright or inverted? FIGURE P38.129
Figure P34.50 shows a top view of a square enclosure. The inner surfaces are plane mirrors. A ray of light enters a small hole in the center of one mirror. (a) At what angle must the ray enter if it exits through the hole after being reflected once by each of the other three mirrors? (b) What If? Are there other values of for which the ray can exit after multiple reflections? If so, sketch one of the rays paths. Figure P34.50
A person looking into an empty container is able to see the far edge of the containers bottom, as shown in Figure P22.23a. The height of the container is h, and its width is d. When the container is completely filled with a fluid of index of refraction n and viewed from the same angle, the person can see the center of a coin at the middle of the containers bottom, as shown in Figure P22.23b. (a) Show that the ratio h/d is given by hd=n214n2 (b) Assuming the container has a width of 8.00 cm and is filled with water, use the expression above to find the height of the container.
A light ray is incident on an interface between water (n = 1.333) and air (n = 1.0002926) from within the air. If the angle of incidence in the air is 30.0, what is the angle of the refracted ray in the water?
Figure P35.20 (page 958) shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to Snells law, n1 sin 1 = n2 sin 2. For paraxial rays, we assume 1, and 2 are small, so we may write n1 tan 1 = n2 tan 2. The magnification is defined as M = h/h. Prove that the magnification is given by M = n1q/n2p. Figure P35.20