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Unformatted text preview: Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler Solution to PHYS 1112 InClass Exam #2B Thu. March 18, 2010, 11:00am12:15pm Conceptual Problems Problem 1: A beam of coherent (laser) light of wavelength λ is incident upon a diffraction grating with line spacing d , with λ < d , as shown in the figure below. Assume  Δ y  is the distance (in cm) between the two 1storder intensity maxima, observed on a screen at a distance L on the other side of the grating. This distance  Δ y  will L Diff. Grating Laser Beam Fig. 2.03 Screen 1st order maxima Δ y θ (A) decrease if we increase L (keeping λ and d fixed); (B) decrease if we increase λ (keeping L and d fixed); (C) decrease if we increase d (keeping λ and L fixed); (D) increase if we decrease λ (keeping L and d fixed); (E) increase if we increase d (keeping λ and L fixed). Answer: (C) At 1storder maximum: sin θ = λ/d . Also, Δ y = 2 L tan θ , by trigonometry (see Fig. 2.03). So sin θ , hence θ , hence tan θ , hence  Δ y  , will decrease if d increases or if λ decreases. Therefore, (C) is correct; and (B), (D) and (E) are wrong. Because of Δ y = 2 L tan θ ,  Δ y  increases if L increases: hence (A) is wrong. Problem 2: If two point charges Q 1 and Q 2 at some distance r repel each other with a force of 160N, what force would they exert on each other if r is quadrupled ( × 4); Q 1 is tripled ( × 3) with unchanged sign; and the sign of Q 2 is reversed ? The two charges will 1 Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler (A) attract each other with a force of 120N (B) attract each other with a force of 2560N (C) attract each other with a force of 30N (D) repel each other with a force of 2560N (E) repel each other with a force of 30N Answer: (C) By Coulomb’s law, the force F = k  Q 1  Q 2  /r 2 ∝  Q 1  /r 2 . Hence, changing  Q 1  →  Q 1  = 3  Q 1  and r → r = 4 r will change F → F = (3 / 4 2 ) × F = (3 / 16) × 160N = 30N. Since Q 1 and Q 2 initially repel each other, they initially have the same sign. Since the sign of Q 1 is unchanged and the sign of Q 2 is reversed, the two charges will have the opposite sign, after Q 1 and Q 2 are changed, and therefore will attract each other. Problem 3: In the figure below, Q 1 and Q 2 are both negative point charges with  Q 1  and  Q 2  being of comparable magnitude. Which arrow drawn at P could correctly represent the electric field vector ~ E generated by Q at P ?...
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This note was uploaded on 04/23/2010 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at UGA.
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