Chapter 35. Refraction
The Index of Refraction (Refer to Table 35-1 for values of n.)
35-1. The speed of light through a certain medium is 1.6 x 108 m/s in a transparent medium, what is the index of refraction in that medium?
n = 1.88
35-2. If the speed of light is to be reduced by one-third, what must be the index of refraction for the medium through which the light travels?The speed c is reduced by a third, so that:
and n = 1.50
35-3. Compute the speed of light in (a) crown glass, (b) diamond, (c) water, and (d) ethyl alcohol. (Since n = c/v, we find that v = c/n for each of these media.)
(a) vg = 1.97 x 108 m/s (b) vd = 1.24 x 108 m/s
(a) v = 2.26x 108 m/s (b) va = 2.21 x 108 m/s
35-4 If light travels at 2.1 x 108 m/s in atransparent medium, what is the index of refraction?
n = 1.43
The Laws of Refraction
35-5. Light is incident at an angle of 370 from air to flint glass (n = 1.6). What is the angle of refraction into the glass?
θg = 22.10
35-6. A beam of light makes an angle of 600 with the surface of water. What is the angle of refraction into the water?θw = 40.60
35-7. Light passes from water (n = 1.33) to air. The beam emerges into air at an angle of 320 with the horizontal water surface? What is the angle of incidence inside the water?
θ = 900 – 320 = 580;
θw = 39.60
35-8. Light in air is incident at 600 and is refracted into an unknown medium at an angle of400. What is the index of refraction for the unknown medium?
n = 1.35
35-9. Light strikes from medium A into medium B at an angle of 350 with the horizontal boundary. If the angle of refraction is also 350, what is the relative index of refraction between the two media? [ θA = 900 – 350 = 550. ]
nr = 1.43
35-10. Light incident from air at450 is refracted into a transparent medium at an angle of 340. What is the index of refraction for the material?
nm = 1.23
*35-11. A ray of light originating in air (Fig. 35-20) is incident on water (n = 1.33) at an angle of 600. It then passes through the water entering glass (n = 1.50) and finally emerging back into air again. Compute the angle of emergence.
The angle of refractioninto one medium becomes the
angle of incidence for the next, and so on . . .
nair sin θair = nw sin θw = ng sin θg = nair sin θair
Thus it is seen that a ray emerging into the same medium
as that from which it originally entered has the same angle: θe = θi = 600
*35-12. Prove that, no matter how many parallel layers of different media are traversed by light, the entranceangle and the final emergent angle will be equal as long as the initial and final media are the same. The prove is the same as shown for Problem 35-11:
nair sin θair = nw sin θw = ng sin θg = nair sin θair; θe = θi = 600
Wavelength and Refraction
35-13. The wavelength of sodium light is 589 nm in air. Find its wavelength in glycerine.
From Table 28-1, the index for glycerin is:n = 1.47.
λg = 401 nm
35-14. The wavelength decreases by 25 percent as it goes from air to an unknown medium. What is the index of refraction for that medium?
A decrease of 25% means λx is equal to ¾ of its air value: nx = 1.33
35-15. A beam of light has a wavelength of 600 nm in air. What is the wavelength of this light as it passes into glass (n = 1.50)?
λg = 400nm
35-16. Red light (620 nm) changes to blue light (478 nm) when it passes into a liquid. What is the index of refraction for the liquid? What is the velocity of the light in the liquid?
nL = 1.30
*35-17. A ray of monochromatic light of wavelength 400 nm in medium A is incident at 300 at the boundary of another medium B. If the ray is refracted at an angle of 500, what is its...
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