Physics 6C Midterm Practice Solutions

Physics 6C Midterm Practice Solutions - Physics 6B Midterm...

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Physics 6B Midterm Practice Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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1. A 5-mm tall object is placed 2 cm away from a lens, and an upright image is formed that is 5 cm tall. What is the focal length of the lens? cm 22 . 2 f cm 20 1 cm 2 1 f 1 S 1 S 1 f 1 cm 20 S cm 2 S mm 5 cm 5 S S y y m + = + = + = = = = = First use the magnification formula to find the image position, then use the other formula to find the focal length. This is a converging lens being used as a magnifier. Answer c) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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2. A light ray traveling in air enters a glass plate with index of refraction 1.62. The ray enters the glass at an angle of 49° to the normal. What angle does the ray make with the normal when it is in the glass? o o 28 ) sin( 62 . 1 ) 49 sin( 1 ) sin( n ) sin( n 2 2 2 2 1 1 = θ θ = θ = θ Use Snell’s Law. Answer a). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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3. A light ray in glass (index 1.62) is incident on an air/glass interface at an angle of 49° to the normal. What angle does the ray make with the normal when it leaves the glass and enters the air? 22 . 1 ) sin( ) sin( 1 ) 49 sin( 62 . 1 ) sin( n ) sin( n 2 2 2 2 1 1 = θ θ = θ = θ o Try Snell’s Law. We get a contradiction – sine and cosine cannot be larger than 1. This means the incoming ray is beyond the critical angle for total internal reflection. There is no transmitted ray. Answer d). This can’t be true. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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4. Light of wavelength 600 nm falls on a single slit of width 0.015 mm. The resulting diffraction pattern is viewed on a screen 10 m away. Find the distance from the center of the pattern to the center of the second bright fringe. ) m 10 600 )( 5 . 2 ( ) m 10 ( y a m R y 3 9 m m = λ = Our single-slit formula gives us the positions of the DARK fringes. The bright fringes are halfway between the dark ones, so we can use m=2.5 in the formula. (Alternately we could find dark fringes #2 and #3, then find the point between them.) m 1 y m 10 015 . 0 m = This is the same as 100cm, so the answer is d). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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5. An oil film floats on water. The refractive index of the oil is 1.4 and the index for water is 1.33. The film appears blue because it reflects light with wavelength 400 nm very well. Calculate the minimum thickness of the oil film. Air n=1
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This note was uploaded on 09/09/2011 for the course PHYSICS 6c taught by Professor Staff during the Spring '11 term at UCSB.

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Physics 6C Midterm Practice Solutions - Physics 6B Midterm...

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