t2Asolspr2010

# t2Asolspr2010 - Physics 1112 Spring 2010 University of...

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Unformatted text preview: Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler Solution to PHYS 1112 In-Class Exam #2A Thu. March 18, 2010, 9:30am-10:45am Conceptual Problems Problem 1: A beam of coherent (laser) light of wavelength λ is incident upon two parallel slits with spacing d , and with λ < d , as shown in the figure below. Assume | Δ y | is the distance (in cm) between the two dark fringes (interference intensity minima), observed closest to the central maximum on a screen at a distance L on the other side of the double- slits. This distance | Δ y | will L Double Slit Laser Beam Fig. 2.04 Screen 1st dark fringes Δ y θ (A) decrease if we decrease L (keeping λ and d fixed); (B) decrease if we increase λ (keeping L and d fixed); (C) decrease if we decrease 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: (A) At 1st minimum (closest to central maximum): sin θ = (1 / 2) λ/d . Also, Δ y = 2 L tan θ , by trigonometry (see Fig. 2.04). So sin θ , hence θ , hence tan θ , hence | Δ y | , will decrease if d increases or if λ decreases. Therefore, (B), (C), (D) and (E) are wrong. Because of Δ y = 2 L tan θ (and θ is fixed if λ and d are fixed), | Δ y | decreases if L decreases: hence (A) is correct. Problem 2: If two point charges Q 1 and Q 2 at some distance r repel each other with a force of 10 μ N, what force would they exert on each other if Q 1 is tripled ( × 3) without change of sign; r is doubled ( × 2), and the sign of Q 2 is reversed ? The two charges will (A) attract each other with a force of 13 . 33 μ N 1 Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler (B) attract each other with a force of 15 μ N (C) repel each other with a force of 15 μ N (D) repel each other with a force of 7 . 5 μ N (E) attract each other with a force of 7 . 5 μ N Answer: (E) 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 = 2 r will change F → F = (3 / 2 2 ) × F = (3 / 4) × 10 μ N = 7 . 5 μ N. 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....
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t2Asolspr2010 - Physics 1112 Spring 2010 University of...

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