Question

A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image 2 (the final image), and determine di2 and hi2 (you may want to turn the page sideways to create more space for your diagram). (b) Use the thin lens equation to verify your results from part (a) and calculate the total magnification mt. (c) Classify image 1 as real or virtual, upright or inverted, and reduced or enlarged. Classify image 2, relative to both image 1 and to the original object, as applicable.

Answer 1

Here is the ray diagram

14.41 cm 10.272 cm diverging lens converging tens

Here are the calculations,

using thin lens equation

1/f = 1/i1 + 1/o1

where i is image distance and o is object distance

1/i1 = 1/-6 - 1/5

i1 = - 2.727 cm (means image is on the same side as the object

m = - i1 / o1 = 0.545 cm (smaller than object)

now, this image becomes object for second lens ( converging lens)

so,

object distance o2 = 8 + 2.727 = 10.727 cm

so,

1/i2 = 1/6 - 1/10.272

i2 = 14.4 cm

m = - i2 / o2

m = 1.344 ( larger than object)

total magnification,

M = - 0.732

image 1 is virtual, upright and reduced in size

image 2 is real , inverted and larger in size.