Solve this physics problem for me...i'll even give you the answer

[ManBearPig]

How far from a converging lens with a focal length of 23 cm should an object be placed to produce a real image which is the same size as the object?

the formula used to solve it is 1/focal length = 2/distance of object

i have NO idea how to solve it though, this is fucking embarrassing. my answer wont work lol.

 

[GodlessAstronomer]

1.65 jiggafurlongs.

Honestly I'm sorry but I just got done with a several day marathon of studying for my hardest exam so I seriously CBF helping If no one answers the question by tomorrow morning I'll look at it.

 

[ManBearPig]

its due at 11:59

 

[theflyingpig]

I'll post my response at 12 then.

 

[GodlessAstronomer]

Ok I just read the OP, it's very simple.
1/f = 2/d
d = 2f

assuming the formula is correct.

 

[Barfo]

42

 

[BlackTigers]

42

46 actually...

 

[Unmoosical]

"I"ll even give you the answer"

So what is it?

 

[Papagayo]

46

 

[gorcorps]

should be 46cm if your formulas are right... does it ask for it in different units?

 

[Mrvile]

Ok I just read the OP, it's very simple.
1/f = 2/d
d = 2f

assuming the formula is correct.

Is it actually this simple? I remember learning proportional equations back in middle school...

 

[ManBearPig]

i fuckin love you guys. im not sure where i went wrong there, but i feel stupid.

 

[BlackTigers]

Is it actually this simple? I remember learning proportional equations back in middle school...

sin(a)/A = sin(b)/B = sin(c)/C

haha, thats the one i remember the most.

 

[PieIsAwesome]

Ok I just read the OP, it's very simple.
1/f = 2/d
d = 2f

assuming the formula is correct.

This.

For a real image to be the same size as the object, magnification is -1.

m = -1 = -s' / s

So s' = s, where s is the object distance and s' is the image distance.

Substituting into the thin lens equation:

1/f = 1/s + 1/s' = 1/s + 1/s = 2 / s

So s, the object distance from the lens, is 2f = 2*23 = 46cm

 

[BlackTigers]

This.

So s, the object distance from the lens, is 2f = 2*23 = 64cm

oh really now?

 

[PieIsAwesome]

oh really now?

Looks like I am dyslexic.

 

[GodlessAstronomer]

i fuckin love you guys. im not sure where i went wrong there, but i feel stupid.

Hehe don't feel too bad, we've all had our embarrassing brain farts.

 

[ManBearPig]

heres a tough one, this one i cant even offer help on:

A -2.1 D lens is held 12.0 cm from an object 7.3 mm high.

What is the distance of the image?

What is the type of the image?

real, upright
real, inverted
virtual, inverted
virtual, upright

 

[MotF Bane]

Ask the digital camera forum? They might be able to answer lens focusing questions.

 

[GodlessAstronomer]

Ask the digital camera forum? They might be able to answer lens focusing questions.

LOL, most photographers I've interacted with couldn't solve a lens equation to save their lives.

 

[BlackTigers]

lol no

 

[Nox51]

Distance should be 2.18 cm and a virtual upright image if I remember my optics right.

 

[PieIsAwesome]

heres a tough one, this one i cant even offer help on:

A -2.1 D lens is held 12.0 cm from an object 7.3 mm high.

What is the distance of the image?

What is the type of the image?

real, upright
real, inverted
virtual, inverted
virtual, upright

Um, does that -2.1 D means a diverging lens with a focal length of 2.1 cm? I don't recognize it, maybe I forgot already. . .

But if it does mean what I think it means, then using the thin lens equation again:

f = -2.1 cm
s = 12.0 cm
y ( object height) = 7.3 mm = 0.73 cm

1/f = 1/s + 1/s'

1/s' = 1/f - 1/s = 1 / -2.1 - 1/12 = - 0.55952
s' = -1.787 = -1.8 cm

The image is 1.8 cm in front of the lens.

Since s' is negative, the image is virtual, and using
m = -s'/s = -(-1.8)/12

Gives a positive magnification value, so the image is upright.

 

[Nox51]

I thought 2.1 D means 2.1 diopters which would be the power of the lens and so you use power=1/s-1/s'

 

[PieIsAwesome]

I thought 2.1 D means 2.1 diopters which would be the power of the lens and so you use power=1/s-1/s'

You are probably right, I don't remember.

OP should know, he is supposed to be studying this after all. . .