3. A 15 cm tall object sits 80 cm to the left of a converging lens with a focal length of f1 = 30 cm. There is a second converging lens with a focal length of f₂ = 20 cm sitting 60 cm to the right of the first converging lens. a. Where is the image from the first lens formed? (find d₁₁) b. What is the object distance for the second lens? (find do2) C. Where is the final image formed? (find diz)

9. A diverging lens (f, = -10.5 cm) is located 54.0 cm to the left of a converging lens (f2 = 20.0 cm). A 2.5-cm-tall object stands to the left of the diverging lens, exactly at its focal point. Determine the distance of the final image relative to the converging lens (including proper algebraic sign). cm

nearsighted man cannot see clearly objects farther than 30.0 cm from their eyes. To prescribe them corrective lenses in the form of glasses that will sit 1.90 cm from their eyes. The corrective lenses will be planoconcave and made of polycarbonate (n = 1.586). a. What is the focal length of the corrective lens?  b. What is the radius of curvature of the concave side of the lens?  A 4.00 cm tall object is placed 66.0 cm in front of the corrective lens. c. Where will the image appear?  d. What is the height of the image?  e. Is the image upright or inverted? Explain how you know.  f. Draw a ray diagram

7. An object is placed 4cm left of a converging lens with a focal length of 10cm. a. Make a sketch of this arrangement and draw 3 principal rays to find where the image should form b. Is the image real or virtual? c. Use the thin lens equation to calculate the position of the image. d. (New situation) An object is placed 30 cm to the left of a diverging lens of focal length 20 cm. Sketch this arrangement by drawing 3 principal rays. Draw where you think the image should form