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1 In air, light travels at roughly c = 3.0 × 108 m/s Therefore, for t = 1.0 ns, we have a distance of d = ct = (3.0 × 108 m / s) (1.0 × 10−9 s) = 0.30 m (a) From Fig 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the larger wavelength to be approximately 610 nm (c) From Fig 33-2 the wavelength at which the eye is most sensitive is about 555 nm (d) Using the result in (c), we have f = c 3.00 ×108 m/s = = 5.41×1014 Hz λ 555 nm (e) The period is (5.41 × 1014 Hz)–1 = 1.85 × 10–15 s (a) The frequency of the radiation is 3.0 × 108 m / s c f = = = 4.7 × 10−3 Hz λ (10 × 10 )(6.4 × 10 m) (b) The period of the radiation is T= 1 = = 212 s = 32 s f 4.7 × 10−3 Hz Since ∆λ θc, there is no refracted light 110 (a) The diagram below shows a cross section, through the center of the cube and parallel to a face L is the length of a cube edge and S labels the spot A portion of a ray from the source to a cube face is also shown Light leaving the source at a small angle θ is refracted at the face and leaves the cube; light leaving at a sufficiently large angle is totally reflected The light that passes through the cube face forms a circle, the radius r being associated with the critical angle for total internal reflection If θc is that angle, then sin θ c = n where n is the index of refraction for the glass As the diagram shows, the radius of the circle is given by r = (L/2) tan θc Now, tan θ c = sin θ c sin θ c 1/ n = = cosθ c − sin θ c 1− 1/ n b g = n2 − and the radius of the circle is r= L n2 − = 10 mm b g 2 1.5 − = 4.47 mm If an opaque circular disk with this radius is pasted at the center of each cube face, the spot will not be seen (provided internally reflected light can be ignored) (b) There must be six opaque disks, one for each face The total area covered by disks is 6πr2 and the total surface area of the cube is 6L2 The fraction of the surface area that must be covered by disks is b b g 6πr πr π 4.47 mm f = = = L L2 10 mm g = 0.63 111 (a) Suppose that at time t1, the moon is starting a revolution (on the verge of going behind Jupiter, say) and that at this instant, the distance between Jupiter and Earth is A1 The time of the start of the revolution as seen on Earth is t1* = t1 + A1 / c Suppose the moon starts the next revolution at time t2 and at that instant, the Earth-Jupiter distance is A The start of the revolution as seen on Earth is t 2* = t + A / c Now, the actual period of the moon is given by T = t2 – t1 and the period as measured on Earth is T ∗ = t2∗ − t1∗ = t2 − t1 + A A1 A −A − =T + c c c The period as measured on Earth is longer than the actual period This is due to the fact that Earth moves during a revolution, and light takes a finite time to travel from Jupiter to Earth For the situation depicted in Fig 33-80, light emitted at the end of a revolution travels a longer distance to get to Earth than light emitted at the beginning Suppose the position of Earth is given by the angle θ, measured from x Let R be the radius of Earth’s orbit and d be the distance from the Sun to Jupiter The law of cosines, applied to the triangle with the Sun, Earth, and Jupiter at the vertices, yields A = d + R − 2dR cosθ This expression can be used to calculate A1 and A Since Earth does not move very far during one revolution of the moon, we may approximate A − A1 by dA / dt T and T * by T + dA / dt T / c Now b b g gb g Rd sin θ 2vd sin θ dA dθ = = , dt d + R − 2dR cosθ dt d + R − 2dR cosθ b g where v = R (dθ/dt) is the speed of Earth in its orbit For θ = 0, dA / dt = and T * = T Since Earth is then moving perpendicularly to the line from the Sun to Jupiter, its distance from the planet does not change appreciably during one revolution of the moon On the other hand, when θ = 90° , dA / dt = vd / d + R and § · vd T ∗ = T ă1 + 2 â c d +R ¹ The Earth is now moving parallel to the line from the Sun to Jupiter, and its distance from the planet changes during a revolution of the moon (b) Our notation is as follows: t is the actual time for the moon to make N revolutions, and t* is the time for N revolutions to be observed on Earth Then, t∗ = t + A − A1 , c where A1 is the Earth-Jupiter distance at the beginning of the interval and A is the EarthJupiter distance at the end Suppose Earth is at position x at the beginning of the interval, and at y at the end Then, A1 = d – R and A = d + R Thus, t∗ = t + b g d + R2 − d − R c A value can be found for t by measuring the observed period of revolution when Earth is at x and multiplying by N We note that the observed period is the true period when Earth is at x The time interval as Earth moves from x to y is t* The difference is t∗ − t = b g d + R2 − d − R c If the radii of the orbits of Jupiter and Earth are known, the value for t* – t can be used to compute c Since Jupiter is much further from the Sun than Earth, d + R may be approximated by d and t* – t may be approximated by R/c In this approximation, only the radius of Earth’s orbit need be known