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**Unformatted text preview: **Physics 9 Fall 2008 Final Exam For the final, you may use three sheets of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat ! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the final. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck ! 1. The hydrogen spectrum has a red line at 656 nm and a blue line at 434 nm . What is the angular separation, Δ θ , between the two spectral lines, red and blue, obtained with a diffraction grating that has 4500 lines/cm? (Note - don’t forget about the higher orders! At some angle there will no longer be a red line, and so the spectral lines stop. Determine at what order this occurs, and work out the angular separation described above for all the orders below this maximum order ). ———————————————————————————————————— Solution We know that the diffraction grating obeys the law d sin θ = mλ , where d is the spacing between two slits, m is the order of the bright fringes, and λ is the wavelength of the light. Since the diffraction grating has 4500 lines/cm, this means that the spacing is d = 1 cm 4500lines = 10- 2 m 4500lines = 2 . 22 × 10- 6 m = 2220 nm . So, the red line will be at an angle θ R , where sin θ R = mλ R d , where λ R = 656 nm , while the blue line is at θ B , where sin θ B = mλ B d , and λ B = 434 nm . So, at first order, where m = 1, we have θ R = sin- 1 434 2220 = 11 . 27 ◦ θ B = sin- 1 656 2220 = 17 . 19 ◦ So, the angular separation is Δ θ = 17 . 19 ◦- 11 . 26 ◦ = 5 . 91 ◦ . This is just for the first order! What about higher orders? At the m th order, we have Δ θ = θ R- θ B = sin- 1 ( mλ R d )- sin- 1 ( mλ B d ) . So, at second order we have Δ θ = sin- 1 ( 2 λ R d )- sin- 1 ( 2 λ B d ) = sin- 1 ( 2 × 656 2220 )- sin- 1 ( 2 × 434 2220 ) = 36 . 2 ◦- 23 . ◦ = 13 . 2 ◦ . Moving on to the third order gives Δ θ = sin- 1 ( 3 λ R d )- sin- 1 ( 3 λ B d ) = sin- 1 ( 3 × 656 2220 )- sin- 1 ( 3 × 434 2220 ) = 62 . 4 ◦- 35 . 9 ◦ = 26 . 5 ◦ . What about the fourth order? In that case, the angle for the red line would occur at θ R = sin- 1 ( 4 λ R d ) = sin- 1 ( 4 × 656 2220 ) = sin- 1 (1 . 18), which doesn’t have a real result. So, this means that there is no fourth-order red line ! So, we can stop here. 1 2. Suppose you have two infinite straight line charges, of charge per unit length λ , a distance d apart, moving along at a constant speed v . The moving charge densities constitute a current I = λv , and so the charges behave like current-carrying wires....

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