# HW3Sol - Solutions to WebAssign HW#3 Giancoli Ch 33 5 10 14...

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Solutions to WebAssign HW #3 - Giancoli Ch. 33- 5, 10, 14, 22, 33, 42, 48, 62; (due 2/11) 5. ( a ) We find the image distance from Eq. 33-2. ( )( ) o i oi o 10.0m 0.105m 111 0.106m 106mm 0.105m df d dd f += = = = = −− ( b ) Use the same general calculation. ( )( ) o i o 3.0m 0.105m 0.109m 109mm 0.105m d == = = ( c ) Use the same general calculation. ( )( ) o i o 1.0m 0.105m 0.117m 117mm 0.105m d = = ( d ) We find the smallest object distance from the maximum image distance. () i max o min i min max max 132mm 105mm 11 1 513mm 0.513m 132mm 105mm d f d f = = = = 10. ( a ) If the image is real, the focal length must be positive, the image distance must be positive, and the magnification is negative. Thus io 2.50 . = Use Eq. 33-2. o o o 1 1 1 1 1 3.50 3.50 50.0mm 70.0mm 2.50 2.50 2.50 ddd d f ⎛⎞ +=+ = → = = = ⎜⎟ ⎝⎠ ( b ) If the image is magnified, the lens must have a positive focal length, because negative lenses always form reduced images. Since the image is virtual the magnification is positive. Thus 2.50 . =− Again use Eq. 33-2. o o o 1 1 1 1 1 1.50 1.50 50.0mm 30.0mm 2.5 2.50 2.50 d f +=− = = 14. For a real image both the object distance and image distances are positive, and so the magnification is negative. Use Eqs. 33-2 and 33-3 to

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HW3Sol - Solutions to WebAssign HW#3 Giancoli Ch 33 5 10 14...

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