Convex mirror question how long the image is?

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving convex mirrors, specifically calculating the image location and size of a robin using the mirror equation and magnification. The original poster presents their calculations based on a convex mirror with a diameter of 40 cm and an object distance of 1.5 m.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the mirror equation and magnification formula to determine the image distance and size. There are questions about the calculations and the use of specific values in the equations.

Discussion Status

Some participants are clarifying the calculations and addressing potential typos in the values used. There is an ongoing exploration of the correct application of the formulas, with some guidance provided regarding the interpretation of magnification.

Contextual Notes

There is a noted confusion regarding the values used in the calculations, particularly the object distance and the signs in the equations. Participants are actively questioning assumptions and clarifying the setup of the problem.

batcave1985
Messages
7
Reaction score
0
Shiny lawn spheres placed on pedestals are convex mirrors.
One such sphere has a diameter of 40 cm. A
12 cm long robin sits in a tree 1.5 m from the sphere.
(a) Where is the image of the robin?
(b) How long is the image?
f=R/2 to find radius
1/v=1/f-1/u to find image distance (the mirror equation)
M=v/u magnification and length (maginification equation)
my answer just not to sure if its right.
a)
f=r/2
=20/2
f=10cm
u=150cm

1/v=-1/1-1/50
v= -9.4cm

b) m=v/u
= -9.4/150
= 0.06cm

not to sure if that was the way to go.
 
Physics news on Phys.org
batcave1985 said:
Shiny lawn spheres placed on pedestals are convex mirrors.
One such sphere has a diameter of 40 cm. A
12 cm long robin sits in a tree 1.5 m from the sphere.
(a) Where is the image of the robin?
(b) How long is the image?



f=R/2 to find radius
1/v=1/f-1/u to find image distance (the mirror equation)
M=v/u magnification and length (maginification equation)



my answer just not to sure if its right.
a)
f=r/2
=20/2
f=10cm
u=150cm

1/v=-1/1-1/50?
v= -9.4cm

b) m=v/u
= -9.4/150
= 0.06cm

not to sure if that was the way to go.

I do not understand the line with ?

For question b, you gave the magnification, (with wrong sign) instead of the size.

ehild
 
ehild said:
I do not understand the line with ?

For question b, you gave the magnification, (with wrong sign) instead of the size.

ehild

its suppose to be
1/v=1/f-1/u (trying to find image distance)
1/v=1/-10 -1/50
v= -9.4cm

then with part b i think i use the magnification and multiply it with the height of object to get image size

0.06 x 12cm= 0.75cm
 
batcave1985 said:
its suppose to be
1/v=1/f-1/u (trying to find image distance)
1/v=1/-10 -1/50
v= -9.4cm
Where does 50 come from? The object distance is 150 cm.

batcave1985 said:
then with part b i think i use the magnification and multiply it with the height of object to get image size

0.06 x 12cm= 0.75cm

that is correct.
 
ehild said:
Where does 50 come from? The object distance is 150 cm.



that is correct.

Yes sorry another typo it is 150 i did the calculation with 150 but typed in 50.
 

Similar threads

Replies
19
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
5
Views
3K
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K