How Far Should an Object Be Placed from a Concave Mirror to Halve Its Size?

[skg94]

Homework Statement

1. A student wishes to place an object in front of a concave mirror in order to produce an image one-half the object's size. If the focal length of the mirror is 5.0cm, how far from the mirror should the object be placed? ANS : 15 cm.

2.An object is placed 2.0cm beyond the centre of curvatur of a concave mirror. If the magnification of this object is 0.30 what is the radius of curvature of this mirror. ANS: 1.7cm

Homework Equations

1/f = 1/do + 1/di
m = hi/ho = -di/do

The Attempt at a Solution

1. I believe these are questions where you have to combine two formulas, i don't know how to go about it, I am not asking for full work, if oyu decide to show full steps, I am trying to figure it out and i can't seem to do it, can someone start me off on the right track? Or if I am overthinking it put me on the right track.

1) 1/2ho/ho= f-do/do
1/2do=-f-do
2do=-2f
which would be -5, which isn't right.

 

[ehild]

Homework Statement

1. A student wishes to place an object in front of a concave mirror in order to produce an image one-half the object's size. If the focal length of the mirror is 5.0cm, how far from the mirror should the object be placed? ANS : 15 cm.

2.An object is placed 2.0cm beyond the centre of curvatur of a concave mirror. If the magnification of this object is 0.30 what is the radius of curvature of this mirror. ANS: 1.7cm

Homework Equations

1/f = 1/do + 1/di
m = hi/ho = -di/do

The Attempt at a Solution

1

1) 1/2ho/ho= f-do/do
1/2do=-f-do
2do=-2f
which would be -5, which isn't right.

What do you mean with equation (1)? ho/ho=1 and do/do=1, so eq(1) is equivalent to 1/2=f-1 which is obviously wrong.

ehild

 

[skg94]

well hi is half of ho which makes it 1/2ho, so i replaced hi with 1/2ho
and i replaced di with f-do from

f=do+di

 

[haruspex]

f=do+di

No, 1/f = 1/do+1/di. That's quite different.