Chapter 26 ---------- 5). The index of the refraction, n, is the ratio of the speed of light in vacuum c, to the speed of light in a medium, v. The speed of light in glass is v = c/n = 3*10^8 m/s / 1.5 = 2*10^(8) m/s The time of passage through a glass pane of thickness d is t = d/v = 4*10^(-3) m * 2*10^(8) m/s = = 2*10^(-11) sec 6). The index of refraction, n, is the ratio of speed of light in the vacuum, c, to the speed of light in the material, v. n(material A)/n(material B) = = c/v(material A)/(c/v(material B)) = v(material B/v(material A) = = 1/1.25 = 0.8 9). The angle of the reflection is equal to the angle of incidence and is, therefore, equalt to 43 degrees. In order to find the angle of refraction, we need to use Snell's law (see equation 26.2), n(air) * sin(incidence angle) = n(water) * sin(refraction angle) From Table 26.1 n(air) = 1.000293 n(water) = 1.333 1.000293 * sin(43) = 1.333 * sin(X) sin(X) = sin(43)/1.333 = 0.511 => X = 30.7 degrees 14). Again apply the Snell's Law to extract real from apparent depth. d'=d * n_2/n_1 d'=2.5 cm * 1.546/1.000 = 3.86 cm 29). The shark cannot be seen by an observer beyond certain distance because the incident angle of a light ray bounced from the shark and coming towards the observer is larger than the angle of total internal refraction. The maximum distance when shark is visible is given by the condition for critical angle: tan(critical_angle) = distance / depth and the Snell's Law n(water) * sin(critical_angle) = n(air) * sin(90 degrees) = 1; sin(critical_angle) = 1/n(water) = 1/1.333 = 0.7502 critical angle = 48.6 degrees distance = depth * tan (critical angle) = 4.5 m * tan(48.6 degrees) = 5.1 m 51). The image distance d_i can be found from the lens equation (26.6) 1/f = 1/d_i + 1/d_o => 1/d_i = 1/12 - 1/8 => d_i = -24 cm magnification m = -d_i/d_o8 = -(-24 cm)/8 cm = 3