Come on Ray, just give me the equation!

A reminder: Please try the simulations below. They will go a long way to giving you an intuitive understanding of how the light rays move in both the cases of refraction and reflection. If you want a preview of tomorrow's lesson, try the lenses in the bottom simulation.

Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html

Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html


Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.

Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (d_i = -2.09 cm).

13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.

14. concave mirror, R = 5.66 cm

27. n = c/v = 1.310

34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)


We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.

35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.