Visualize solve the critical angle of incidence is

Info icon This preview shows pages 17–24. Sign up to view the full content.

View Full Document Right Arrow Icon
Visualize: Solve: The critical angle of incidence is given by Equation 23.9: 1 1 oil c glass 1.46 sin sin 76.7 1.50 n n θ = = = ° Assess: The critical angle exists because n oil < n glass .
Image of page 17

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.18. Model: Represent the can as a point source and use the ray model of light. Visualize: Paraxial rays from the can refract into the water and enter into the fish’s eye. Solve: The object distance from the edge of the aquarium is s . From the water side, the can appears to be at an image distance s = 30 cm. Using Equation 23.13, 2 water 1 air 1.33 1.0 n n s s s s n n ′ = = = 30 cm 23 cm 1.33 s = =
Image of page 18
23.19. Model: Represent the beetle as a point source and use the ray model of light. Visualize: Paraxial rays from the beetle refract into the air and then enter into the observer’s eye. The rays in the air when extended into the plastic appear to be coming from the beetle at a shallower location, a distance s from the plastic-air boundary. Solve: The actual object distance is s and the image distance is s = 2.0 cm. Using Equation 23.13, 2 air 1 plastic n n s s s n n ′ = = 1.0 2.0 cm 1.59 s = s = 3.2 cm Assess: The beetle is much deeper in the plastic than it appears to be.
Image of page 19

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.20. Model: Represent the diver’s head and toes as point sources. Use the ray model of light. Visualize: Paraxial rays from the head and the toes of the diver refract into the air and then enter into your eyes. When these refracted rays are extended into the water, the head and the toes appear elevated toward you. Solve: Using Equation 23.13, 2 air T T T 1 water n n s s s n n ′ = = air H H water n s s n ′ = Subtracting the two equations, her apparent height is ( ) ( ) air H T H T water 1.0 150 cm 113 cm 1.33 n s s s s n = = =
Image of page 20
23.21. Model: Represent the aquarium’s wall as a point source, and use the ray model of light. Visualize: Paraxial rays from the outer edge (O) are refracted into the water and then enter into the fish’s eye. When extended into the wall, these rays will appear to be coming from O rather from O. The point on the inside edge (I) of the wall will not change its apparent location. Solve: We are given that s O s I = 4.00 mm and O I 3.50 mm. s s = Using Equation 23.13, water O O wall n s s n ′ = water I I wall n s s n ′ = ( ) water O I O I wall n s s s s n = ( ) wall 1.33 3.50 mm 4.00 mm n = ( ) wall 4.00 mm 1.33 1.52 3.50 mm n = =
Image of page 21

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.22. Model: Use the ray model of light. Visualize: Solve: Using Snell’s law, air red red sin30 sin n n θ ° = 1 red sin30 sin 19.2 1.52 θ ° = = ° air violet violet sin30 sin n n θ ° = 1 violet sin30 sin 18.8 1.55 θ ° = = ° Thus the angular spread is red violet 19.2 18.8 0.4 θ θ θ Δ = = °− ° = °
Image of page 22
23.23. Model: Use the ray model of light and the phenomenon of dispersion.
Image of page 23

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 24
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern