## anonymous one year ago A 6.0 cm object is located 20 cm in front of a convex mirror with a radius of curvature equal to 120 cm. What is the size of the image produced by the mirror?

1. anonymous

@IrishBoy123

2. anonymous

i get 0.45 and u ?

3. anonymous

@IrishBoy123

4. IrishBoy123

4.5

5. anonymous

how did u do it?

6. IrishBoy123

how did you do it? you first....

7. anonymous

k i found -1.5 for di

8. anonymous

in the end it did (1.5)(6)/20

9. anonymous

to get 0.45

10. IrishBoy123

what is f in this question??!!

11. IrishBoy123

$$R_c = 120$$

12. IrishBoy123

so f = ??

13. anonymous

60

14. anonymous

-60 since it convex

15. IrishBoy123

seems we are in agreement - ish (___), cool. for me: $$\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$ $$f = -60$$ $$d_o = 20 \implies d_i = -15$$ $$\large \frac{h_i}{h_0} = -\frac{d_i}{d_o}$$ $$h_o = 6$$ follows that $$h_i = 4.5$$

16. anonymous

-1/60-1/20=1/di???

17. anonymous

@IrishBoy123

18. IrishBoy123

-1/60 = 1/20 +1/di

19. anonymous

yea so when the 1/20 come to the other side it becomes negative

20. anonymous

ohh wait its 1200 not 120

21. anonymous

omg

22. anonymous

these question are really easy just these silly mistakes ruin them

23. anonymous

btw do u come to site often ? i will be needing lots of help in the coming days since i have excam

24. IrishBoy123

$$\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$ $$f = - 60$$ $$\large -\frac{1}{60} = \frac{1}{20} + \frac{1}{d_i}$$ $$\large \frac{1}{d_i} = -\frac{1}{60} - \frac{1}{20} = -\frac{1}{20} (\frac{1}{3} + 1) = -\frac{1}{15}$$ $$\large d_i = -15$$ $$\large \frac{h_i}{h_o} = -\frac{d_i}{d_o}$$ etc etc

25. IrishBoy123

plus, for "lots of help" on Physics, be sure to tag @Michele_Laino Qualified Helper, and brilliant in every sense of the word

26. Michele_Laino

thanks!! @IrishBoy123