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.

 

[Nickie]

2)m=hi/ho=-di/do
-0.30=hi/ho=di/do
-0.30=di/2di
-0.30=1/2
di=-2
1/f=1/do+1/di
1/f=1/2+1/-2
1/f=1/2-1/2
1/f=0
f=infinity which isn't right.

As a scientist, it is important to approach problems systematically and use the appropriate equations to solve them. In this case, we can use the equations provided to find the solution to these questions.

For the first question, we can use the formula 1/f = 1/do + 1/di, where f is the focal length, do is the distance of the object from the mirror, and di is the distance of the image from the mirror. We are given that the focal length is 5.0cm and we want the image to be half the size of the object, so we can set m (the magnification) to be 1/2. This gives us the equation 1/5 = 1/do + 1/(1/2)do. Solving for do, we get do = 15cm, which is the correct answer.

For the second question, we can use the formula m = hi/ho = -di/do, where m is the magnification, hi is the height of the image, ho is the height of the object, di is the distance of the image from the mirror, and do is the distance of the object from the mirror. We are given that the object is placed 2.0cm beyond the center of curvature, so do = 2.0cm. Also, m = 0.30, so we can set this equal to -di/2.0. Solving for di, we get di = -1.7cm, which is the correct answer.

Overall, it is important to carefully read the given information and use the appropriate equations to solve the problem. It is also helpful to double check your calculations to ensure the correct answer is obtained.