Finding the Distance for Enlargement with a Thin Lens

[just.karl]

If you want to produce an image of a bulb that is enlarged by a factor of 2, how far from the wall should the lens be placed. Focal length is 40cm.

I know you use 1/d_o + 1/d_i =1/f and m= -d_i/d_o but how do you find d_o? -2/d_o +1/d_o =1/f ? but it comes out to 40cm and not 60cm for d_o. If someone could help me with that bit I can solve it from there. Thanks!

 

[Redbelly98]

Actually it comes out to -40 cm.

Here is a hint: what should m be, if the image is inverted and enlarged by a factor of 2?

 

[just.karl]

it should be -2?

 

[Redbelly98]

Yes. Try that out.

 

[just.karl]

So m d_o = -d_i (-2)(-40)= -d_i so then it comes out to be -80cm but in the back of the book it says it's 1.2m.

 

[nasu]

So m d_o = -d_i (-2)(-40)= -d_i so then it comes out to be -80cm but in the back of the book it says it's 1.2m.

Why (-2)(-40)? do is not 40. The focal distance is 40.
Use both equations and put m=-2 as Redbelly98 told you.

 

[Redbelly98]

So m d_o = -d_i (-2)(-40)= -d_i

No. d_0 is not -40 cm. f is +40 cm.

-2 d_o = -d_i

and as you know, from the thin lens equation:

1/d_o + 1/d_i = 1/(40cm)

Take it from there.

 

[just.karl]

I understand that you use both equations, but the part I'm getting hung up on is how you solve for d_o with 1/do + 1/di = 1/F from my understanding it goes to -2/do + 1/do = 1/40 and the answer from that comes out to be -40 =do I then put that into the mdo=di equation and get 80. I'm assuming I'm solving for d_o wrong?

 

[Redbelly98]

-2/do + 1/do = 1/40

No. Use d_i = 2 d_o, as I said in post #7.

 

[just.karl]

so then it would be 1/do + 1/2do = 1/40?

 

[Redbelly98]

Yes.