Question

4. An object is placed 45.0 cm left of a converging OBJECT f = 9.00 cm 12 = 120 cm lens i = +9.00 cm), which is 24.0 cm left of a second converging lens 12 = +12.0 cm). (a) Calculate the position of the image formed by the first lens and sketch an arrow labeled "1st IMAGE" on the diagram to represent its approximate location. 45.0 cm 240 cm (b) Calculate the magnification of the first image. (c) Calculate the overall magnification of the final image, relative to the object. (d) is the final image larger or smaller than the object? Explain, based on the magnification found in (c) - but don't stop at simply quoting its value if you weren't able to get an answer for (c), explain how you'd use the magnification to decide. rimage

Answer 1

For the first converging lens:

Thin lens formula

fi

45.09.00 11.25 cm 1= 45.0 9.00

First image is formed at a distance of to the right side of first lens .

I1.25 cm ст

12 12.0 cm f 9.00 cm OBJECT first image 24.0 cm 45.0 cm

b)

Magnification of first image , negative sign indicates the image is inverted.

11.25 -0.25 m1 45.0

and magnification of image is

0.25

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First image acts as an object to the second lens, Object distance for the second lens

24.0 11.25 12.75 cm

Thin lens formula for the second lens

f2

12.75 12.0 204 cm 2= 12.75 12.0 f2 u2-f2 |

c)

Overall magnification of the final image relative to object is

1 mim2 u2

11.25 204 = 4 mim2 45 12.75

d)

Final image is larger than the object. Final image is 4 times the size of object.

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Magnification of first lens is
0.25
. That is size of first image is one-fourth the size of object and image is inverted. Magnification of the second lens is
-16
. That is size of final image is 16 times the size of first image and it is inverted again. Hence the final image is 4 times the size of object and it is upright relative to object.