Assignment 2 Solutions - Assignment 2 Solutions Question 1(8 Marks 4Al 3O 2 2Al2 O 3 3H 2 N 2 2NH 3 C3 H 8 5O 2 3CO 2 4H 2 O I 2 2Na 2S2 O 3 2NaI Na 2S4

# Assignment 2 Solutions - Assignment 2 Solutions Question...

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32º Figure 1 θ r n 1 = 1 n 2 = 1.37 Assignment 2 Solutions Question 1 (8 Marks) 3 2 2 O 2Al 3O 4Al + O 4H + 3CO 5O + H C 2 2 2 8 3 2NH N + 3H 3 2 2 6 4 2 3 2 2 2 O S Na + 2NaI O S 2Na + I Question 2 (8 Marks) (i) Calculating the angle of refraction We can calculate the angle θ r using Snell’s law; sin sin i 1 2 θ θ n n r = Remember that the angles are always taken with respect to the normal to the surface, so the angle of incidence is; º 8 5 32 - 90 = = i θ We are given that n 1 = 1, and n 2 = 1.37, then using Snell’s law to calculate the refracted angle gives; sin58 1 sin 37 . 1 = r θ º 24 . 38 1.37 58 sin arcsin = = r θ Therefore, the angle of refraction is 38º, or (this answer cannot be accurate to more than 2 significant figures). (ii) Calculating the wavelength of the light in the liquid We know that for the light beam in vacuum; Wavelength nm 470 0 = λ Velocity m/s 10 3 8 × = c And we know that the speed of light is different in different media depending on the refractive index;
v c n = , or n c v = (1) where n is the absolute refractive index of the medium, v is the velocity of light in the medium. The wavelength and frequency of a wave are related to its velocity by f λ = v , which means that the frequency is given by f=v/ λ . The frequency will be the same in both media – this is because the frequency represents the number of wave cycles arriving

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• Spring '08
• Beavan
• Light, Sin, refractive index, Geometrical optics