A diverging lens has a focal length of 10.0 cm. An object 1.5 m tall is placed 50.0 cm in front of the lens. Locate the image. Determine both the magnification and the height of the image. Describe the image.
"1\/a+1\/b=-1\/F\\to 1\/0.5+1\/b=-1\/0.1\\to b=-0.083\\ (m)"
"\\beta=-(-0.083\/0.5)=0.167"
"\\beta=h\/H\\to h=H\\cdot\\beta=1.5\\cdot0.167=0.25 \\ (m)"
The image is reduced, virtual and direct.
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