HW09-Solutions - Physics 201 Solution Set 09 CHAPTER 33...

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Physics 201 Solution Set 09 CHAPTER 33: Lenses and Optical Instruments 33[ 4, 11, 12, 13, 17, 20, 21, 22, 25, 87, 41, 45, 48, 49, 53, 55, 99, 61, 72] 1 4. To form a real image from a real object requires a converging lens. We find the focal length of the lens from Eq. 33-2. ( )( ) o i o i o i 1.85m 0.483m 1 1 1 0.383m 1.85m 0.483m d d f d d f d d + = = = = + + Because i 0, d > the image is real. 11. From Eq. 33-3, i o h h = when i o . d d = So find d o from Eq. 33-2. o o i o o 1 1 1 1 1 2 50cm d f d d d d f + = + = = = 12. ( a ) Use Eqs. 33-2 and 33-3. ( )( ) o i o i o 1.30m 0.135m 1 1 1 0.1506m 151mm 1.30m 0.135m d f d d d f d f + = = = = ( ) i i i i o o o o 0.1506m 2.80cm 0.324m 1.30m h d d m h h h d d = = − = − = − = − The image is behind the lens a distance of 151 mm, is real, and is inverted. ( b ) Again use Eqs. 33-2 and 33-3. ( )( ) ( ) ( ) ( ) o i o i o i i i i o o o o 1.30m 0.135m 1 1 1 0.1223m 122mm 1.30m 0.135m 0.1223m 2.80cm 0.263m 1.30m d f d d d f d f h d d m h h h d d + = = = = − ≈ − − − = = − = − = − = The image is in front of the lens a distance of 122 mm, is virtual, and is upright. 13. The sum of the object and image distances must be the distance between object and screen, which we label as T d . We solve this relationship for the image distance, and use that expression in Eq. 33-2 in order to find the object distance. ( ) ( ) ( ) ( )( ) 2 o i T i T o o T o T o i o T o 2 2 T T T o 1 1 1 1 1 ; 0 86.0cm 86.0cm 4 16.0cm 86.0cm 4 21.3cm, 64.7cm 2 2 d d d d d d d d d fd d d d d d f d d fd d + = = + = + = + = ± ± = = = Note that to have real values for o d , we must in general have 2 T T T 4 0 4 . d fd d f > >
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