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Physics 2C Summer Session II Quiz #4 Multiple Choice. Choose the answer that best completes the statement or answers the question. 1. A plane wave of linearly polarized light is incident on a stack of 2 linear polarizers. The polarization angle of the incident light is rotated by an angle 3 =8 with respect to the …rst polarizer whose polarization axis is rotated by an angle =8 with respect to the second polarizer. The intensity of the light relative to the incident intensity, I=I0 ; is A) :073 B) 1=16 C) 1=8 D) 1=4 E) :354 From the Law of Malus we have I = cos2 (3 =8) cos2 ( I0 =8) = :125 ! C 2. Red light emitted by hydrogen atoms at rest in the laboratory has a wavelength of 656:3nm. The same spectral line emitted by hydrogen atoms from a distant galaxy has a frequency of 4:16 1014 Hz. The relative velocity of the galaxy relative to Earth is (speed of light is c = 3 108 m= sec) A) 1:41 107 m= sec B) 2:82 107 m= sec C) 2:96 107 m= sec D) 2:70 107 m= sec E) 1:48 107 m= sec The observed wavelength of the light as it reaches Earth is obs =3 108 = 4:16 1014 = 721:15nm From the expression for the Doppler shift of light we …nd s 2 721:15 1 + v=c 1 + v=c = ! = obs em 1 v=c 656:3 1 v=c 1 + v=c = v = 1:2074 (1 2:82 v=c) ! v=c = :2074 = :09396 2:2074 107 m=s ! B 3. In the …gure for problem 3 a light ray is propagating in the plane of the page parallel to one of the mirrors (problem 4 ray in the …gure). Through what angle will it be turned (a normal re‡ection is equivalent to a turn of 180 )? 1 Problem 3 A) 60 B) 120 C) 180 D) 240 E) 300 From geometry the angle between the incident light ray and the lower mirror is 60 : Hence the incident and re‡ected angles are both 30 ; and the angle between the incident and re‡ected rays is 60 : Thus the light ray is rotated through 180 60 = 120 after the …rst re‡ection..After the re‡ection from the lower mirror the light ray and the two mirrors form an equilateral triangle. Hence the above scenario is repeated for the upper mirror and the light is rotated through 2 120 = 240 ! D: 4. Light traveling in water (n = 4=3) is incident upon an water-glass interface. The refracted light makes a 51 angle with respect to the interface. If the glass has a refractive index of n = 1:62; the incident angle of the light is A) 54:8 B) 39:7 C) 35:6 D) 49:9 E) 70:8 The refracted and incident angles are measured from the normal to the interface, hence the refracted angle is 39 : From Snell’s law we …nd n1 sin 1 = n2 sin 1 = 2 ! sin 1 = 3 1:62 sin 39 = :76462 4 49:9 ! D 5. What is the wavelength of the light inside the glass in problem 4, if it has a frequency of 4:82 1014 Hz in water? A) 467nm B) 512nm C) 622nm D) 756nm E) 384nm Since the frequency is the same in all media, the wavelength in vacuum is 108 = 4:82 = c=f = 3 1014 = 622nm The wavelength in the glass is found from =n n ! n = 622=1:62 = 383:95nm ! E 2 6. The prism shown in the …gure is an isosceles right triangle and has an index of refraction of n = 1:76: If the prism is immersed in a liquid, what is the minimum value for the index of refraction of the liquid for the light to cease undergoing total internal re‡ection? Problem 6 A) 1:245 B) 1:250 C) 2:150 D) 1:150 E) 1:075 The incident angle is 45 which in this case is also the critical angle. Hence the index of the liquid is found from p n1 sin c = n2 ! 1:76= 2 = nliq = 1:245 ! A If the index of the liquid is larger than this, then there is some refraction (no internal re‡ection). 7. An object at the bottom of a wading pool appears to be 52cm below the surface. How deep is the pool if the refractive index of water is n = 4=3? A) 39cm B) 45cm C) 52cm D) 60cm E) 69cm The apparent depth is given by `0 = `=n: Hence the actual depth is d=`= 4 52 = 69:33 ! E 3 8. An image is located 36cm behind a spherical concave mirror which has a radius of curvature of 18cm. How much is the image magni…ed (include sign to determine if the image is inverted or not)? A) M = 3:0 B) M = 2:0 C) M = 5:0 D) M = 3:0 E) M = 5:0 The focal length is given by f = R=2: Since the image is behind the mirror its location is `0 = 36cm: From the mirror equation 1 f 1 ` = = 1 1 + 0 ! ` ` 5 !`= 36 3 1 1 1 = 9 ` 36 36 = 7:2 5 Hence the magni…cation is M= `0 = ` 36 = 5:0 ! C 36=5 9. Determine the focal length of a spherical mirror if the image magni…cation is given by M = 2 when the object is 5cm beyond the focal point (measured in cm) of the mirror. A) M = 15cm B) 5cm C) 2cm D) 10=3cm E) 10cm From the mirror equation and magni…cation equations 1 `0 = M = 1 f 1 5 = f +5 f (f + 5) `0 f = = 2 ! f = 10cm ! E ` 5 10. Find the Brewster angle, ; for the interface between air (n = 1) and a material for which the critical angle in air is given by c = 36:87 : A) 37 B) 59 C) 31 D) 53 E) 47 From the critical angle, the index of refraction for this material is found from n sin c = 1 ! n = 5=3: The Brewster angle is found from tan = n=1 = 5=3 ! = 59 ! B u=v) ; f 0 = f (1 1 Some useful formulas f0 n1 sin 1 = f (1 = n2 sin 2; u=v) 1 1 1 = + 0; M = f ` ` 4 ; f0 = f `0 =`; tan s 1 1 u=c u=c = n2 =n1