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Physics Assignment

Physics Assignment - Phys 122 TUTORIAL 10 Refraction Optics...

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Phys 122 Introductory Physics II TUTORIAL 10 Optics – Reflection and refraction KEY. Refraction. 11. The refractive indices of materials A and B have a ratio of n A / n B = 1.33. The speed of light in material A is 1.25 × 10 8 m/s. What is the speed of light in material B ? SOLUTION Applying Equation 26.1 to both materials, we have A B A B and c c n n v v = = Dividing the equation for material A by that for material B gives A A B B B A / / n c v v n c v v = = Solving for v B , we find that ( 29 ( 29 8 8 A B A B 1.25 10 m/s 1.33 1.66 10 m/s n v v n = = × = × 12. The frequency of a light wave is the same when the light travels in ethyl alcohol as it is when it travels in carbon disulfide. Find the ratio of the wavelength of the light in ethyl alcohol to that in carbon disulfide. SOLUTION Using Equations 16.1 and 26.1, we find λ = v f = c / n f = c f n Using this result and recognizing that the frequency f and the speed c of light in a vacuum do not depend on the material, we obtain the ratio of the wavelengths as follows: λ alcohol λ disulfide = c f n alcohol c f n disulfide = c f 1 n alcohol c f 1 n disulfide = n disulfide n alcohol = 1.632 1.362 = 1.198 13. A ray of light traveling in material A strikes the interface between materials A and B at an angle of incidence of 72°. The angle of refraction is 56°. Find the ratio n A / n B of the refractive indices of the two materials. SOLUTION Using Snell’s law, we have 1

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A A B B sin56 sin 72 sin56 or 0.87 sin 72 n n n n ° ° = ° = = ° 14. A person working on the transmission of a car accidentally drops a bolt into a tray of oil. The oil is 5.00 cm deep. The bolt appears to be 3.40 cm beneath the surface of the oil, when viewed from directly above. What is the index of refraction of the oil? SOLUTION According to Equation 26.3, the apparent depth d of the bolt is 2 1 n d d n ′ = where d is the actual depth, n 1 is the refractive index of the medium (oil) in which the object is located, and n 2 is the medium (air) in which the observer is located directly above the object. Solving for n 1 and recognizing that the refractive index of air is n 2 = 1.00, we obtain ( 29 1 2 5.00 cm 1.00 1.47 3.40 cm d n n d = = = 15. The drawing shows a coin resting on the bottom of a beaker filled with an unknown liquid. A ray of light from the coin travels to the surface of the liquid and is refracted as it enters into the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light traveling in the liquid? From the drawing in the text, we see that the angle of incidence at the liquid-air interface is –1 1 5.00 cm tan 39.8 6.00 cm θ = = ° The drawing also shows that the angle of refraction is 90.0°. Thus, according to Snell's law (Equation 26.2: 1 1 2 2 sin sin n n θ θ = ), the index of refraction of the unknown liquid is 2 2 1 1 sin (1.000) (sin 90.0 ) 1.56 sin sin 39.8 n n θ θ ° = = = ° From Equation 26.1 ( n = c / v
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