Solve because the angle 2 θ is very small 3 20 10 m

Info icon This preview shows pages 44–50. Sign up to view the full content.

View Full Document Right Arrow Icon
Solve: Because the angle 2 θ is very small, 3 2.0 10 m tan2 2 5.0 m θ θ × = ( ) 1 180 rad degrees 0.011 5000 5000 θ π = = = °
Image of page 44

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

View Full Document Right Arrow Icon
23.45. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize: For angles θ water that are less than the critical angle, light will be refracted into the air. Solve: Snell’s law at the water-air boundary is air air water water sin sin . n n θ θ = Because the maximum angle of θ air is 90 , ° we have ( ) water 1.0 sin90 1.33sin θ ° = 1 water 1 sin 48.75 1.33 θ = = ° Applying Snell’s law again to the glass-water boundary, glass glass water water sin sin n n θ θ = ( ) 1 1 water glass water glass 1.33 sin48.75 sin sin sin 42 1.50 n n θ θ ° = = = ° Thus 42 ° is the maximum angle of incidence onto the glass for which the ray emerges into the air.
Image of page 45
23.46. Model: Use the ray model of light. Visualize: Solve: When the plastic is in place, the microscope focuses on the virtual image of the dot. From the figure, we note that s = 1.0 cm and s = 1.0 cm 0.4 cm = 0.6 cm. The rays are paraxial, and the object and image distances are measured relative to the plastic-air boundary. Using Equation 23.13, air plastic n s s n ′ = ( ) plastic 1.0 0.6 cm 1.0 cm n = plastic 1.0 cm 1.67 0.6 cm n = =
Image of page 46

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

View Full Document Right Arrow Icon
23.47. Model: Use the ray model of light and the law of refraction. Solve: Snell’s law at the air-glass boundary is air air glass glass sin sin . n n θ θ = We require 1 glass air 2 , θ θ = so ( ) air glass glass glass sin 2 sin n n θ θ = ( ) air glass glass glass glass 2sin cos sin n n θ θ θ = glass 1 1 glass air air 1.50 cos cos 41.4 82.8 2 2 1.0 n n θ θ = = = ° ⇒ = ° ×
Image of page 47
23.48. Model: Use the ray model of light and the law of refraction. Visualize: Solve: (a) The ray of light strikes the meter stick at empty P , which is a distance L from the zero mark of the meter stick. So, tan60 50 cm L ° = ( ) 50 cm tan60 86.6 cm L = ° = (b) The ray of light refracts at P half and strikes the meter stick a distance 1 2 x x + from the zero of the meter stick. We can find x 1 from the triangle P full P half O : 1 tan60 25 cm x ° = ( ) 1 25 cm tan 60 43.30 cm x = ° = We also have ( ) 2 half 25 cm tan . x φ = Using Snell’s law, air water half sin60 sin n n φ ° = 1 half sin60 sin 40.63 1.33 φ ° = = ° ( ) 2 25 cm tan40.63 21.45 cm x = ° = 1 2 43.30 cm 21.45 cm 64.8 cm x x + = + = (c) The ray of light experiences refraction at P full and the angle of refraction is the same as in part (b). We get ( ) 3 full 3 tan 50 cm tan40.63 42.9 cm 50 cm x x φ = = ° =
Image of page 48

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

View Full Document Right Arrow Icon
23.49. Model: Use the ray model of light. Light undergoes total internal reflection if it is incident on a boundary at an angle greater than the critical angle. Visualize: Solve: (a) To reach your eye, a light ray must refract through the top surface of the water and into the air. You can see in the figure that rays coming from the bottom of the tank are incident on the top surface at fairly small
Image of page 49
Image of page 50
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