Problem 9: Suppose a lens has a 48.5 mm focal length and is being used to photograph a flower 62.5 cm away. Randomized Variables do= = 62.5 cm f=48.5 mm

A converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answers

Suppose an object has thickness dp so that it extends from object distance p to p + dp. (a) Prove that the thickness dq of its image is given by (q2/p2)dp. (b) The longitudinal magnification of the object is Mlong = dq/dp. How is the longitudinal magnification related to the lateral magnification M?

In Figure P26.38, a thin converging lens of focal length 14.0 cm forms an image of the square abcd, which is hc = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c, and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P26.38