## anonymous one year ago A 2.4 cm object is 15.0 cm to the left of a convex lens with a focal length of +5.0 cm. Part 1: Draw a ray diagram of the setup showing the location of the image. Part 2: Calculate the distance of the image from the lens. Part 3: What size is the image and is it upright or inverted?

1. Michele_Laino

here is the drawing of your problem: |dw:1434906888314:dw|

2. anonymous

whoahh okie!! :)

3. Michele_Laino

here we have to apply this formula: $\Large \frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ where p= 15 cm f= 5 cm please solve for q

4. anonymous

ok! 1/15 + 1/q = 1/5 1/q = 3 q=1/3 ?

5. Michele_Laino

we have: $\Large q = \frac{{fp}}{{p - f}} = \frac{{5 \times 15}}{{15 - 5}} = ...cm$

6. anonymous

ohh oops so we get 75 / 10 = 7.5 cm ?

7. Michele_Laino

that's right!

8. Michele_Laino

now, from my drawing, we get that the image is inverted

9. anonymous

okay! so part 2 is that it is 7.5 cm away from the lens? and part 3 is that it is inverted?

10. Michele_Laino

yes! nevertheless we have to determine the dimension of the image

11. anonymous

yes! we have to calculate the size, too.. how do we do that?

12. Michele_Laino

we can use the similarity property of these triangles: |dw:1434907503742:dw|

13. anonymous

ok!

14. Michele_Laino

so we can write this proportion: $\Large 2.4:15 = x:7.5$

15. Michele_Laino

or, applying the fundamental property of proportion: $\Large x = \frac{{2.4 \times 7.5}}{{15}} = ...cm$

16. anonymous

we get 1.2 cm? that is the size of the image?

17. Michele_Laino

yes! correct!

18. anonymous

yay! so we are finished with this? :O

19. Michele_Laino

yes! we have finished! :)

20. anonymous

yay!! okay, onto the last one!!:) thank you1!

21. Michele_Laino

ok! :)