Question

11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal length 8.20 cm. A diverging lens of focal length - 16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 8.442 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? Oreal virtual

Answer 1

Given object hight hos com fie 8.2 cm du – 16cm sequation belumen lenses de 6cm. ob just distance of converging lens O = 4.85cm Image distance of convergin lens is 8.20 4.85CM IL -- 11187 cm Image famed by converging lens acts as object of divenging lens so I - deed do I & 6 + 11.87 = 11.87cm Image distance of divorging lens is l - I - 16cm 17.87000 Position of final image 32 = -8.64 cm Magnification = |- I -22) (11.87 8.44 ) lol oz) = (4.85 17,87) - 10156 Height of final image is his 1.156 ho Image is virtual and wricht? (1•156) (icm) = 10156 em