Calculating Object Distance for Magnified Real and Virtual Images

[iurod]

Homework Statement

(a)How far from a 50.0-mm-focal-length lens must an object beplaced if its image is to be magnified 2.00 x and be real? (b)What if the image is to be virtual and magnified 2.00x?

Homework Equations

m = -d_i/d_o

1/d_o+ 1/d_i = 1/f

The Attempt at a Solution

m = -d_i/d_o solve for d_o
d_o = -di/2

Now i have to use the lens equation:
1/d_o+ 1/d_i = 1/f
-(2/di) + 1/di = 1/f
-1m/di = 1/f
di= -1m/(1/f)

di = (-1 x 2)/(1/50mm)
di = -100

This is what i got for part a, and I know it is totally wrong because to be a real image di>0 which is not what i got.

any help will be greatly appreciated. Thanks!

 

[ehild]

The real image is inverted, so the magnification is negative, -2. ehild

 

[Redbelly98]

Is a real image upright or inverted? What does that tell you about m?

EDIT: oops, ehild beat me to it.

 

[iurod]

so for this part:
Now i have to use the lens equation:
1/d_o+ 1/d_i = 1/f
-(2/di) + 1/di = 1/f
-1m/di = 1/f
di= -1m/(1/f)

di = (-1 x 2)/(1/50mm)
di = -100

it should be:
di= -1m/(1/f)

di = (-1 x -2)/(1/50mm)
di = +100mm

so do = -di/m = -100/-2 = +50mm

in the back of the book it says the answer should be +75mm.

 

[Redbelly98]

so for this part:
Now i have to use the lens equation:
1/d_o+ 1/d_i = 1/f
-(2/di) + 1/di = 1/f

You are still saying that do=-di/2. That will not be true, according to the help we gave you earlier.

 

[iurod]

A real image is inverted and m should be negative

m= -di/do
-2= -di/do
do= -di/-2

If this is correct then
1/do+ 1/di = 1/f
-2/-di + 1/di = 1/f
hmmm I'm stuck after this

 

[iurod]

Any more advice on this?

Thanks!

 

[Redbelly98]

A real image is inverted and m should be negative

m= -di/do
-2= -di/do
do= -di/-2

Correct, but you can simplify "-di/-2"

If this is correct then
1/do+ 1/di = 1/f
-2/-di + 1/di = 1/f
hmmm I'm stuck after this

You are told what f is. Use that value, then you can solve the equation for di.