11. Find the index of refraction of water using the diagram provided separately

Question

11. Find the index of refraction of water using the diagram provided separately from Giancoli, 4/e, Fig. 32-21(a) 12. A triangular prism shown is made of glass with index of refraction 1.5. The prism is sitting on a tabletop, and a wall is 1.8 meters to the right of the prism. A thin horizontal beam of sunlight (coming through the shades near sunset) reaches the prism as shown. (Assume the size of the prism is much less than the distance to the wall.) (a)Why doesn't the light beam change direction when it enters the glass through the left side of the prism? Explain first using an analogy or other conceptual reasoning, and then using Snell's Law Is the center of the bright spot on wall above or below the prism? How far is it above or below the incoming horizontal light beam? Draw a clear diagram. (Since the size of the prism is much less than the distance to the wall, so you may assume the wall is 1.8m from the point where the light comes out of the prism.) As you know, the prism actually breaks the sunlight into its component colors. The index of refraction of a given material is not a constant; it actually depends slightly on wavelength. Specifically, the index of refraction for lower wavelengths is slightly greater than the index of refraction for higher wavelengths. For instance, the index of refraction of glass, though approximately equal to 1.5 for visible light, might actually be 1.49 for red light and 1.51 for blue light. Given this... in the spectrum you see on the wall, is the red on the top or the bottom? Explain briefly, and show in your diagram. 60 (b) (c)

Explanation / Answer

Q11.

diagram a:

let index of refraction of air =n1
index of refraction of water=n2

incident angle=theta1
refraction angle=theta2

using snell's law:

n1*sin(theta1)=n2*sin(theta2)

==>index of refraction of water=n2=n1*sin(theta1)/sin(theta2)

diagram b:

let index of refraction of air =n2
index of refraction of water=n1

incident angle=theta1
refraction angle=theta2

using snell's law:

n1*sin(theta1)=n2*sin(theta2)

==>index of refraction of water=n1=n2*sin(theta2)/sin(theta1)

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